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28 Oct 12 - 12:34 PM (#3427286) Subject: BS: Monty Hall Problem From: GUEST,Lighter Some of you will have heard of this. On the game show "Let's Make a Deal," host Monty Hall used to present three giant doors. Behind one was a prize like a car. Behind the other two were booby prizes like a goat. The contestant was asked to pick a door. At that point, of course, the contestant's chance of picking the door that hid the car was 1 in 3. (Or 2 to 1 against, if you prefer.) Whichever door was selected, Monty (who knew where the car was) would open one of the remaining two doors to reveal a goat. He'd ask then the contestant whether they wanted to stick with their original choice or switch to the single unchosen and unopened door, which must contain either a car or a goat. Here's the problem: mathematical geniuses say that switching is a good idea because it raises the odds of winning the car from 1 in 3 to 2 in 3. Others, including myself, cannot follow this reasoning. Wikipedia tries to explain (unsuccessfully in my view) at great length: http://en.wikipedia.org/wiki/Monty_Hall_problem Among the skeptics, by the way, are sophisticated engineers and logicians, who evidently are not sophisticated enough. The mathematicians claim, with equations, that switching changes the odds. Obviously (?) eliminating one door increases the odds that your first choice was correct from 1 in 3 to 1 in 2. But how then does switching your choice boost the odds of winning to 2 in 3? You still have no idea where the car is! I'm prepared to believe the math geniuses because they offer incomprehensible equations and I'm not very smart. They also say the answer is obvious and lament the gullibility of the non-mathematician public. But some of the dissenters also are smarter than I am. Can anybody explain CLEARLY and with a minimum of math how seeing the contents of one of the three doors and then switching your choice (from one unknown quantity to another) increases the odds that you'll win the car? Or do I have to sign up for an advanced course in probability theory? |
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28 Oct 12 - 12:47 PM (#3427296) Subject: RE: BS: Monty Hall Problem From: gnu "Can anybody explain CLEARLY and with a minimum of math how seeing the contents of one of the three doors and then switching your choice (from one unknown quantity to another) increases the odds that you'll win the car?" No. Not even unclearly. It's 50/50. |
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28 Oct 12 - 12:51 PM (#3427298) Subject: RE: BS: Monty Hall Problem From: bobad I too see it as 50/50 but then I'm no math genius. |
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28 Oct 12 - 01:15 PM (#3427309) Subject: RE: BS: Monty Hall Problem From: GUEST,Ed Obviously (?) eliminating one door increases the odds that your first choice was correct from 1 in 3 to 1 in 2. I think that this is where you may be getting confused. As you say, in your initial choice of door, the odds are 1 in 3. At this point, you know that at least one of the other doors has a goat behind it. When the host (who knows where the car is) opens one of the other doors to reveal a goat, he is merely confirming what you already know: one of the other doors has a goat behind it. Therefore, and this the counter intuitive part, your odds are still 1 in 3. Nothing else has changed. Hence switching increases your odds of winning to 1 in 2 at this point (or 2 in 3 overall). |
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28 Oct 12 - 01:22 PM (#3427312) Subject: RE: BS: Monty Hall Problem From: DMcG The explanation is not too difficult, but it does depend on understanding how information affects probability. But it is very easy to carry out the experiment: you need, for example one red playing card, two black ones, and a friend to be Monty. Play the game ten times without changing your mind, then another ten where you always change your mind and compare the results. Sometimes doing the experiment is more convincing than any amount of argument! |
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28 Oct 12 - 02:12 PM (#3427343) Subject: RE: BS: Monty Hall Problem From: GUEST,Lighter >Therefore, and this the counter intuitive part, your odds are still 1 in 3. Nothing else has changed. >Hence switching increases your odds of winning to 1 in 2 at this point (or 2 in 3 overall). The "hence" is where you lose me, Ed. Revealing one goat would indeed boost your odds to 1 in 2. But how does *switching* from the original unknown to the remaining unknown raise them to 2 in 3 at the moment you switch? You still have no idea what's behind either remaining door. As you explain it, it almost sounds as though "1 in 3" at the start is somehow mystically equivalent to "2 in 3" later. But "one" never used to equal "two" in my book. I'll do DMcG's experiment later. But if it works, I still won't understand how. Suppose I *wanted* a goat. My chances at the start would be a pretty good 2 in 3. Then Monty opens a door, which I didn't select, with a goat behind it. My chances of winning a goat now have slipped to 1 in 2. Would they increase to 2 in 3 if I switched to the one remaining door? |
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28 Oct 12 - 02:20 PM (#3427349) Subject: RE: BS: Monty Hall Problem From: DMcG I agree doing the experiment won't explain WHY the probabilities are not as you expect, but it should be enough (given enough trials - 10 might be too few) to show the probability is (something like) 2/3 if you change and 1/3 if you do. Once that is clear, you will see that the 'obvious' 50:50 solution can't be right. Then go slowly though your logic to see where you assume it is 50:50 and that's where you are going wrong ... |
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28 Oct 12 - 05:16 PM (#3427427) Subject: RE: BS: Monty Hall Problem From: gnu If you have ONE choice, the odds are 50/50. Previous choices are moot. They are gone, they do not exist. They do not impact subsequent choices. Soooo, you have ONE choice. And the odds are 50/50 that your choice will win or lose. Period. This is not socks in a drawer. This is A OR B and it's defined as such. They are asked to make A choice between TWO doors. One OR... OR... OR... the other door... THAT is 50/50. |
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28 Oct 12 - 05:47 PM (#3427440) Subject: RE: BS: Monty Hall Problem From: DMcG Have you done the experiment? :) Probability is tricky: very subtle changes in definition of the problem can have significant effects. In this case, 50:50 is demonstrably wrong if you spend fifteen minutes or so actually trying it. |
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28 Oct 12 - 05:48 PM (#3427441) Subject: RE: BS: Monty Hall Problem From: GUEST,Ed Previous choices are moot. They are gone, they do not exist. They do not impact subsequent choices But they do here, gnu. If, for example, your initial choice is Door 1, the host can open either Door 2 or 3 to reveal a goat. Let's assume he opens Door 2. You can then choose between Doors 1 and 3. If your initial choice is Door 2 and the host opens Door 1 you can then choose between Doors 2 and 3. In other words, your initial choice has influenced what your subsequent one can then be. It is no longer a simple 50/50 choice. |
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28 Oct 12 - 07:14 PM (#3427478) Subject: RE: BS: Monty Hall Problem From: GUEST,Reverend Bayes To start with, you had a 1/3 chance of picking the car. Opening the door doesn't change that; it's still 1/3 that you have the car. So the probability the other door is the car must be 2/3. If you think this is hard, my friend the Martingale Representation Theorem would like a word. |
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28 Oct 12 - 09:31 PM (#3427520) Subject: RE: BS: Monty Hall Problem From: John on the Sunset Coast http://en.wikipedia.org/wiki/Monty_Hall_problem I know we don't all love and trust Wikipedia, but therein is presented several ways to the solution, from a basic grid to a formal mathematical problem, and they all agree that one should always switch on the second round. This strategy results in winning the car (mathematically) twice as often as standing pat. BTW, you can trust Darrin on the Sunset Coast (if not Wiki) who is a logician/linguist by training. He explained it, sans math, such that even I understood it. |
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28 Oct 12 - 10:54 PM (#3427563) Subject: RE: BS: Monty Hall Problem From: Bee-dubya-ell As I recall it, there were two levels of "real" prizes and one "goat". Let's say the items behind the doors are a toilet brush, a television set, and a new car. There are three possible scenarios: A) If your choice is the door hiding the toilet brush, Monty will reveal the television set you "lost". (He's not going to reveal the car until the last moment, win or lose.) B) If your choice is the door hiding the car, Monty will also reveal the television set. (He's not going to reveal the toilet brush as long as there's the possibility of your changing your mind and getting stuck with it instead of winning the car.) C) If your choice is the door hiding the television set, Monty will reveal the toilet brush. (Again, he's not going to reveal the car until the last moment.) So , there are two scenarios (A & B) in which Monty reveals the television set. Changing your door choice doesn't increase your chances of winning the car in either of them. You're either going to win the car or get conked with the toilet brush. Second prize is out of the running. But, in scenario C, changing doors will DEFINITELY upgrade your prize from the television set to the car. Since you've been shown that you didn't pick the "goat" prize, you know you've picked a winner. But if the winner you'd picked was the car, you'd have been shown the television set, not the toilet brush. So, the probability of your winning the car in scenario A is 50%, whether you change your mind or not. And the probability of your winning the car in scenario B is 50%, whether you change your mind or not. But the probability of your winning the car in scenario C is 100%, if you change your mind. Total those probabilities up and average them out and the overall odds of winning the car if you change your mind are 2/3. |
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29 Oct 12 - 01:03 AM (#3427585) Subject: RE: BS: Monty Hall Problem From: beeliner The best, and clearest, explanation of the problem that I have read is, of all places, in a novel entitled "The Curious Incident of the Dog in the Nighttime" by Mark Haddon. HOWEVER - Haddon's explanation, and all the others that I have read, seem to assume (tho' not state directly) that Monty would offer the choice no matter which door the contestant had initially selected. But what if Monty offered the choice ONLY if the contestant had initially chosen the door with the car? Would that change the percentages? |
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29 Oct 12 - 04:20 AM (#3427619) Subject: RE: BS: Monty Hall Problem From: Rob Naylor Of ther above, Reverend Bayes' explanation is the clearest and most concise. Without any maths, either. As for many of the skeptics mentioned in the original post being "logicians and engineers", if they were *good* logicians and engineers they'd do the experiment suggested by DMcG and see for themselves that they do win more often if they switch....then work on understanding the logic of the statistics. I spend a chunk of my time doing statistical analysis of data to determine its precision and reliability forgeophysical and engineering purposes and it surprises me how often people who are otherwise great engineers have problems understanding statistics. |
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29 Oct 12 - 06:36 AM (#3427648) Subject: RE: BS: Monty Hall Problem From: Megan L Of course the fact is this Monty Hall(whoever that was) did not let them replay the game multiple times so the laws of probability do not affect the outcome it was down to luck. |
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29 Oct 12 - 07:58 AM (#3427675) Subject: RE: BS: Monty Hall Problem From: Nigel Parsons Are we all agreed? On your first pick you have a one in three chance of picking the car. We all agree, good! So on one time in three you would be swapping the car & getting a goat (if you swapped). On your first pick you have a two in three chance of picking a goat. So two out of three times swapping will get rid of the goat, and get you a car. Although swapping may lose the car, this happens only 1/3 of the time. 2/3 of the time swapping will get you the car. I know which way I'd choose! |
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29 Oct 12 - 08:26 AM (#3427690) Subject: RE: BS: Monty Hall Problem From: DMcG Of course the fact is this Monty Hall(whoever that was) did not let them replay the game multiple times so the laws of probability do not affect the outcome it was down to luck. Ummm ... do you want to think that through a little more? Yes, any individual trial is 'luck', but changing your mind is 'luckier' than not. In fact, it is twice as lucky! |
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29 Oct 12 - 08:48 AM (#3427703) Subject: RE: BS: Monty Hall Problem From: GUEST,Lighter > Monty...did not let them replay the game multiple times so the laws of probability do not affect the outcome it was down to luck. Actually even I can understand that no matter how many times you play (i.e., start choosing one door of three) the odds (whatever they are) remain the same each time. Like flipping a coin. It's always 1 in 2 for heads the next time no matter how many times heads has come up in the past. And some of those mathematicians skeptcal of the new 2/3 odds are/were pretty accomplished. According to Wackipedia, which may be reliable in this case, "Decision scientist Andrew Vazsonyi described how Paul Erdős, one of the most prolific mathematicians in history, remained unconvinced until Vazsonyi showed him a computer simulation confirming the predicted result." I'm wondering if those who say the odds increase by switching understand that Monty would allow you to switch your choice of doors *whether or not your first choice found the car.* So revealing one goat did not imply anything about the wheareabouts of the car behind the remaining two doors. He didn't care if you won the car or not, at least for the purposes of this discussion. And where is the flaw in the following reasoning? Regardless of what's just happened, you're left with two doors. The car could be behind either one. Somebody coming fresh to the scene would have a 1 in 2 chance of picking the car, whether he switched or not. (Because there would be no spare goat door to open.) Right? No matter how many times he switched back and forth, his odds of picking the car would be 1 in 2. Right? So how is it that *your* odds can be increased to 2 in 3 by switching, while his can't, in precisely the same situation at the same moment???? Didn't your odds increase from 1 in 3 to 1 in 2 the minute the spare goat was revealed? Why would that change? True, you now know that the car wasn't behind the third door. But so what?! That's in the past, like all those coin flips that came up heads! Two doors, one car, one goat, 50/50! |
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29 Oct 12 - 08:54 AM (#3427709) Subject: RE: BS: Monty Hall Problem From: Howard Jones The first choice is a red herring - it isn't really a choice at all - it's just a bit of TV showmanship. The first choice doesn't matter, because it's not acted on. You may have a 1 in 3 chance of selecting the right door, but your chance of winning is nil. Whichever door you choose, the host will open a different one. The only choice which matters is the second choice, after one of the doors has been taken out of contention so it's then a straight 50/50 chance between 2 doors, one of which must contain the car and the other the goat. It's an entirely new choice, which may or may not be the same as the first choice. The first choice doesn't have a bearing on it, it only appears to. Despite the bewildering (to me) maths, the decision tree and diagrams showing the permutations all show an equal number of wins and losses. |
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29 Oct 12 - 09:04 AM (#3427715) Subject: RE: BS: Monty Hall Problem From: TheSnail Nigel Parsons had it right. Taking the original Car/Goat/Goat scenario, remember that Monty knows where the car is. There are three equally likely sequences of events if you switch. 1) You choose the Car. Monty shows a Goat. You switch and get the other Goat. 2) You choose Goat 1. Monty shows a Goat. You switch and get the Car. 3) You choose Goat 2. Monty shows a Goat. You switch and get the Car. So 2 out of 3, you get the car. If you don't switch, there are another three equally likely sequences of events. 1) You choose the Car. Monty shows a Goat. You don't switch and get the Car. 2) You choose Goat 1. Monty shows a Goat. You don't switch and get the Goat. 3) You choose Goat 2. Monty shows a Goat. You don't switch and get the Goat. So 2 out of 3, you get a goat. |
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29 Oct 12 - 09:13 AM (#3427720) Subject: RE: BS: Monty Hall Problem From: Nigel Parsons The first choice is a red herring - it isn't really a choice at all - it's just a bit of TV showmanship. The first choice doesn't matter, because it's not acted on. You may have a 1 in 3 chance of selecting the right door, but your chance of winning is nil. Whichever door you choose, the host will open a different one. The only choice which matters is the second choice, after one of the doors has been taken out of contention so it's then a straight 50/50 chance between 2 doors, one of which must contain the car and the other the goat. It's an entirely new choice, which may or may not be the same as the first choice. The first choice doesn't have a bearing on it, it only appears to. Yes, given a straight choice between two doors, where your only knowledge is that one hides a goat, the other a car, your chance is 50/50. But... This is not a simple choice. You have already selected a door, and are being asked to choose whether to change. This means that the first choice affects the outcome. The door you have already chosen was a choice from 3, where you knew that it was more likely (2/3) that you would have chosen a door with a goat behind. Someone just seeing the two doors and being offered the choice does not have the additional knowledge, so has a 50% chance of finding the car. With your additional knowledge (by swapping) your chance increses to 66%+. No wonder it confuses the experts! Cheers Nigel (recreational mathematician) |
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29 Oct 12 - 09:15 AM (#3427722) Subject: RE: BS: Monty Hall Problem (other cases) From: Mysha Hi, Two questions above about other cases: - Do the odds change if Mr. Hall only opens another door if at that time you've picked the car? = As it happens, they don't, but your strategy changes slightly. If Mr. Hall shows you a goat, then you know for sure that you have the car, so stick with it. If Mr. Hall does not show you a goat, then you know you yourself picked a goat. You'll now have to pick one of the other two doors for a car, but since he didn't open either of them, you'll only have a fifty/fifty chance of getting the right one. As it happens, this gives you the same two out of three odds the original question had. - If you want a goat, and Mr. Hall opens a door showing one, what door should you pick? = The door Mr. Hall just opened. Bye Mysha |
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29 Oct 12 - 09:25 AM (#3427727) Subject: RE: BS: Monty Hall Problem From: Bobert Seems that after eliminating one of the 3 doors (the goat) and given an opportunity to pick between the two remaining doors presents a new probability as if the game was originally a choice of two... So the "new" probability is 1 in 2 or 50/50... It doesn't, however change the 33.33% chance that was originally offered... The wild card here is that the rules change in that, in essence, if the contestant is allowed to make a 2nd choice then the odds of winning the car are 66.66%... B~ |
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29 Oct 12 - 10:23 AM (#3427751) Subject: RE: BS: Monty Hall Problem From: gnu If you have two doors to choose from, what are you odds of picking the right door? |
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29 Oct 12 - 10:35 AM (#3427757) Subject: RE: BS: Monty Hall Problem From: Jim Dixon You don't need be any equations or mathematical expertise if you have enough patience to use the "brute force" method and list out all the possibilities. To begin with, the prize has already been placed behind one of the doors, so there are 3 possibilities, all equally probable: * The prize is behind door 1. * The prize is behind door 2. * The prize is behind door 3. Now you make a choice, which expands the list to 9 possibilities: * The prize is behind door 1, and you choose door 1. * The prize is behind door 1, and you choose door 2. * The prize is behind door 1, and you choose door 3. * The prize is behind door 2, and you choose door 1. * The prize is behind door 2, and you choose door 2. * The prize is behind door 2, and you choose door 3. * The prize is behind door 3, and you choose door 1. * The prize is behind door 3, and you choose door 2. * The prize is behind door 3, and you choose door 3. As the next step, Monte will open one of the doors, but his choice is not random: he will always open a door that does not reveal the prize. You know in advance that this will happen. So there is no point in waiting to see which door he opens; you might as well decide in advance what your strategy will be. You will either switch or not switch. Suppose you decide that your strategy will be to not switch. We now have enough information to predict the outcome: * The prize is behind door 1, you choose door 1, and you do not switch: YOU WIN. * The prize is behind door 1, you choose door 2, and you do not switch: YOU LOSE. * The prize is behind door 1, you choose door 3, and you do not switch: YOU LOSE. * The prize is behind door 2, you choose door 1, and you do not switch: YOU LOSE. * The prize is behind door 2, you choose door 2, and you do not switch: YOU WIN. * The prize is behind door 2, you choose door 3, and you do not switch: YOU LOSE. * The prize is behind door 3, you choose door 1, and you do not switch: YOU LOSE. * The prize is behind door 3, you choose door 2, and you do not switch: YOU LOSE. * The prize is behind door 3, you choose door 3, and you do not switch: YOU WIN. You can now count the wins and losses and see that your chance of winning is 1/3. On the other hand, suppose you decide that your strategy will be to switch. Here are the outcomes: * The prize is behind door 1, you choose door 1, and you switch: YOU LOSE. * The prize is behind door 1, you choose door 2, and you switch: YOU WIN. * The prize is behind door 1, you choose door 3, and you switch: YOU WIN. * The prize is behind door 2, you choose door 1, and you switch: YOU WIN. * The prize is behind door 2, you choose door 2, and you switch: YOU LOSE. * The prize is behind door 2, you choose door 3, and you switch: YOU WIN. * The prize is behind door 3, you choose door 1, and you switch: YOU WIN. * The prize is behind door 3, you choose door 2, and you switch: YOU WIN. * The prize is behind door 3, you choose door 3, and you switch: YOU LOSE. When you count up the wins and losses, you see that your chance of winning with this strategy is 2/3. |
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29 Oct 12 - 10:53 AM (#3427764) Subject: RE: BS: Monty Hall Problem From: Jim Dixon Here's another thought experiment: Instead of 3 doors, suppose there are 1000 doors. Suppose you choose door number 438. Then Monty opens all the doors except numbers 438 and 722, revealing 998 goats. Now, do you stick with 438, or switch to 722? |
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29 Oct 12 - 11:03 AM (#3427772) Subject: RE: BS: Monty Hall Problem From: GUEST,Lighter > The wild card here is that the rules change in that, in essence, if the contestant is allowed to make a 2nd choice then the odds of winning the car are 66.66%... In practical terms it still makes no sense to me. In the end, you still have to choose one door of two. The only probability that seems to count is what exists in fact at the crucial moment of choosing between two doors. If you stick with door 1 it's effectively a new choice. If you switch to door 2 it's another choice. I don't see that the news that door X was a goat is really "new." You knew when you got into this that 2 of the 3 doors had goats. Now you see where one goat was. You still don't know where the car is, or the other goat. You have two doors. What if, given the choice, you switch when given the chance, but then quickly switch back again? Would that increase your odds even further? Obviously not. But how does that hypothetical opportunity for a third choice differ structurally from being granted a second choice? If you grab Monty's mike and switch between the two doors a hundred times, are your chances of winning the car slowly approaching 1 in 1? Please say no. Why is Howard not right when he says that the first choice is a red herring? Whatever door you pick, you'll wind up having to choose between just two doors later. Despite the efforts of well-meaning 'Catters, I don't understand how having made a blind choice from three doors in the past can influence the odds of choosing correctly between two in the present. Are the 2 in 3 odds somehow retroactive? If so, wouldn't this be a vital key to practical time travel? Wait a minute. You know at the start you'll *ultimately* have two choices of three. So in that sense your odds overall are clearly (?) 2 in 3 *at the start* (because you have two shots at 3 doors). But why don't the odds drop to 1 in 2 when the situation resolves into one real car and one real goat? Or is the remaining goat Schroedinger's goat? (To add to the fun, once you made your first or second choice, Monty would offer you a worthwhile amount of cash in lieu of revealing the car or the goat. But let's not get into that. (And, at least in theory, if won a goat, you had to keep it.) |
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29 Oct 12 - 11:16 AM (#3427782) Subject: RE: BS: Monty Hall Problem From: Bobert You are correct, Lighter... If you, in essence, get one bad guess before the count begins then it is 50/50... B~ |
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29 Oct 12 - 11:32 AM (#3427792) Subject: RE: BS: Monty Hall Problem From: Nigel Parsons I don't see that the news that door X was a goat is really "new." You knew when you got into this that 2 of the 3 doors had goats. Now you see where one goat was. You still don't know where the car is, or the other goat. You have two doors. That information is new. You previously new only that the car was behind one of 3 doors. You now know that it is behind one of 2. Before you made your decision you had a one in 3 chance of picking the car. The removal of one door and one goat means that you now only have to choose between two doors. But you know (from the first choice) that your first door has a 1/3 chance of hiding the car, so you accept the swap for a 2/3 chance. Monty Schrödinger's Goat At the start of this act, behind each of three doors is one third of a car, and two thirds of a goat (or one third of two goats). The compere asks you to choose a door ... |
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29 Oct 12 - 11:46 AM (#3427806) Subject: RE: BS: Monty Hall Problem From: GUEST,TIA If you want to use the "nothing new" argument, try this: At the outset, you have a one in three chance. When Monty opens one of the doors, you are correct that you still have a one in three chance that you have picked the car. But there is only one other door at this point, and your door has an unchanged one in three chance of being the car. Now what are the other possibilities *at this point*? There is only one other possibility - the other door! So the single remaining door must have a two in three chance of holding the car! |
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29 Oct 12 - 12:06 PM (#3427815) Subject: RE: BS: Monty Hall Problem From: Nigel Parsons Sometimes I wonder whether some people just don't get it, or whether they're winding us up to see how many different ways we'll try to explain it. |
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29 Oct 12 - 12:29 PM (#3427821) Subject: RE: BS: Monty Hall Problem From: Bee-dubya-ell Perhaps it would help to imagine more doors. What if there were a million doors? You pick one, Monty eliminates 999,998 of the remaining 999,999. Which door is more likely to have the car behind it? The one chosen randomly, or the one left standing after almost a million others have been eliminated? |
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29 Oct 12 - 01:28 PM (#3427836) Subject: RE: BS: Monty Hall Problem From: GUEST,Lighter > Sometimes I wonder whether some people...'re winding us up to see how many different ways we'll try to explain it. If the solution were as obvious as you seem to think, there'd be no controversy. It would be me against the world. And it isn't. Of a million doors, say I pick No. 1. Monty "picks" No. 17. Obviously the car is behind one of those two doors. At the start of the game, with nothing revealed, the chance that the car is behind one of those two doors would be 1 in 500,000, and that it is behind whichever one I choose is 1 in a million. But Monty doesn't know where the car is either. He's thrown doors open at random and left No. 17 closed at random. *If one of 999,998 opened doors had hidden a car, we'd know it and the game would be over. I'd have lost.* And the odds are tremendous that that's what would have happened then and would happen almost all of the time. (It would be a very boring show, with very few winners.) But if we've got this far without finding the car, it must be behind either 1 or 17. No matter what door Monty ultimately "picked," the chance that it hid the car would have been, at the start, precisely 1 in a million, just like the chance that it's behind door No. 1. And now you're left with two doors: odds of 1 in 2. I'm not insisting that I'm right. Far from it. But it would be nice to know precisely what statements or assumptions I'm making that are so obviously incorrect. What reality alters the 1 in 2 odds when Monty doesn't know where the car is either, it hasn't been revealed, and you're down to two doors? Most counterintuitive things become clear when explained, but this hasn't. I presume I'm not the only one still in the dark. And by admitting continued confusion, if that's what it is, I'm probably losing all credibility in every other BS discussion. So be it. |
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29 Oct 12 - 01:32 PM (#3427837) Subject: RE: BS: Monty Hall Problem From: GUEST,Ed But Monty doesn't know where the car is either. But he does! From the Wikipedia article you initially linked to: "the host, who knows what's behind the doors, opens another door" |
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29 Oct 12 - 02:59 PM (#3427873) Subject: RE: BS: Monty Hall Problem From: gnu If you have two doors to choose from, what are you odds of picking the right door? If you have two doors to choose from, what are you odds of picking the right door? Pick one of my questions and answer it. You have a 50/50 chance of picking the right question. |
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29 Oct 12 - 03:36 PM (#3427896) Subject: RE: BS: Monty Hall Problem From: GUEST,Lighter Ed, sorry for the confusion. The W. article is very dense. Stipulating that Monty always knows where the car is changes everything in a way that really is easy to understand. I used to watch the show in the '60s. It appeared that Monty didn't know (as I posted earlier), or at least pretended not to know, and the contestants assumed he didn't know either. Or at least had no way of knowing for sure. I believe he would sometimes open a door and the car would be behind it, without opening a goat door. Probably he knew some of the time - or all of the time, but the contestants didn't know whether he did or not. That was part of the fun. Sure, if he knows, and you know he knows, then obviously if he leaves door No. 17 shut, out of a million, it's pretty suspicious, and that door, selected with insider information, is far more likely to hide the car than your own blindly selected door which would hide the car by pure chance alone. And I hope you're saying that if Monty doesn't know the whereabouts of the car, the 1 in 2 analysis at the crisis point is correct, whether there were a million doors or three. |
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29 Oct 12 - 04:24 PM (#3427925) Subject: RE: BS: Monty Hall Problem From: TIA If Monty doesn't know, *your* odds will not change at all. Sometimes the game will end abruptly when he reveals the car rather than a goat, but *your* odds will remain the same if the opened door reveals a goat. |
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29 Oct 12 - 04:52 PM (#3427942) Subject: RE: BS: Monty Hall Problem From: GUEST,Lighter Exactly. Case closed. Thanks to all. |
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29 Oct 12 - 04:53 PM (#3427943) Subject: RE: BS: Monty Hall Problem From: gnu Oh, BTW... d'ya think it's cheaper for "Monty" to have two goats and one car or to have three goats? Crank out those odds. And, don't tell me Monty showed the car if the wrong choice was made in order to "prove" it was all legit. Ye must be stunned as me arse iffn ya dunno how that ruse was done. Slight of hand is for the slight of mind. |
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29 Oct 12 - 06:34 PM (#3427983) Subject: RE: BS: Monty Hall Problem From: Howard Jones With the first choice, you have no chance at all, because even if you pick the door with the car behind it (1:3) you still don't win it. All the first round does is help the host to choose which of the doors with a goat behind it he opens. But that isn't random - he knows where the prize is. It's just a set-up to raise the tension for the second round, where you have a choice between two doors, one of which must hide the prize. What confuses the issue is that one of the doors must be the door you chose first time. So it feels as if you have to decide whether to switch or stick with the first choice. The host winds up the tension by suggesting that. If you guess right and win you'll feel vindicated, if you lose you'll feel frustrated that your first choice was right all along. But that's all psychology. It's still simply a choice between one door or the other. Why should the choice you made the first time make it more or less likely that you'll guess right the second time? |
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29 Oct 12 - 06:41 PM (#3427990) Subject: RE: BS: Monty Hall Problem From: Bobert What kinda of car is it, anyway??? B;~) |
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29 Oct 12 - 06:43 PM (#3427991) Subject: RE: BS: Monty Hall Problem From: gnu Howard... it doesn't. That's what all the scientists and philosophers and math geeks are hiding behind when they spout their "theories". If this, then that... if that, then this... there ain't none... it's two doors and one choice. Plain and simple. |
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29 Oct 12 - 07:00 PM (#3428003) Subject: RE: BS: Monty Hall Problem From: Howard Jones Ah, I think I've got it. The first round does affect it, because it is twice as likely that your first choice was wrong than right. You're more likely to choose the goat than the car in the first round, and since the other goat is out of contention that makes it more likely that the other door hides the car. |
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29 Oct 12 - 07:02 PM (#3428005) Subject: RE: BS: Monty Hall Problem From: GUEST,Lighter Sometimes it was a luxury car, sometimes a boat, sometimes tickets to Hawaii, etc., etc. It was always valuable. And the "goat" of the puzzle could be any kind of disappointing or worthless item. (Sometimes it was a goat, or a cow, or a dented bowling ball or something like that.) The third door could hide almost anything from junk to something decent to something almost as good as the grand prize. It varied. It's been a long, long time, but as I recall the location of the grand prize was always revealed. Otherwise, of course, the show would just be a variation of three-card monte (get it? "Monty"? "monte"?) with a disappearing "prize." And there was joking and bargaining and costumes and elaborate promotional descriptions of the worthwhile prizes and cash bribes to get people either to switch their choices or not to switch. Silly, but fun at the time. But none of that has anything to do with the odds related to the stylized version of the game we've been discussing. It does seem, though, that many people *don't* think it affects the odds if Monty knows where the grand prize is and doesn't want you to win it. That naivete' *is* a little troubling. |
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29 Oct 12 - 07:14 PM (#3428010) Subject: RE: BS: Monty Hall Problem From: gnu "The first round does affect it, because it is twice as likely that your first choice was wrong than right." There is it! You Got it! Whatwere the chances? |
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29 Oct 12 - 08:40 PM (#3428043) Subject: RE: BS: Monty Hall Problem From: BK Lick I think Jim Dixon's "brute force" method is eminently clear and should be sufficient to convince the most mathematically naive that switching increases the odds of winning from 1/3 to 2/3. Increasing the number of doors in the thought experiment usually makes the result seem more intuitive, but I think it may be more effective (psychologically) to consider increasing the number only to ten; observe that the "not switching" strategy wins the prize one time in ten., while the "switching" strategy wins the prize nine times in ten. Or in the three door problem, consider that switching turns a correct initial guess into a loss but turns an incorrect initial guess into a win. Since the initial guess is twice as likely to be incorrect as correct, switching is twice as likely to win the prize. If you're intrigued by thought experiments that seem to defy intuition, you might lose some sleep over Newcomb's paradox. —BK |
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29 Oct 12 - 09:56 PM (#3428079) Subject: RE: BS: Monty Hall Problem From: John on the Sunset Coast For the game to work properly the two lesser prizes" must be of comparable value", and switching doors wins 67% of the time. If, as has been asserted, there was 1 ea. Megaprize, Goodprize and Clunker, then the game changes. If your first choice is the GP, you will not be be shown the MP...because the game needs to continue; you will be shown the C. Staying Loses If your first choice is the C, you will be shown not your choice or the MP because the game needs to continue. So again you should switch. Staying loses. If you actually pick the MP, you will be shown either GP or C, but you don't know that you actually have chosen correctly. So since you know that in two scenarios out of three switching is correct, you should still switch. Remember you have no way of knowing where the MP is on the first round, but you maximize your chance of getting the MP whenever you are shown a lesser prize. Using the switching strategy you only lose the MP is picked first (1 of 3); you win if you did not pick it first (2 of 3). Where the strategy may break down is if you have picked the MP, but are are shown the C. You might stand because you're sure to get at least |
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29 Oct 12 - 10:08 PM (#3428085) Subject: RE: BS: Monty Hall Problem From: GUEST,Lighter It complicates the game to stipulate that Monty knows where everything is and doesn't want you to win. That's one way of playing. But in the version discussed by me, he doesn't know and he doesn't care if you win or not. And there might be one or two clunkers. I could not understand how, in that situation, switching would increase your odds. And it now looks as if it wouldn't. But if Monty knows and you know he knows, that's a different story. |
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29 Oct 12 - 10:35 PM (#3428100) Subject: RE: BS: Monty Hall Problem From: gnu The contestant picks a door. Monty opens a door with a goat. Then he asks the contestant to pick one of the two doors remaining. Stop changing and twisting the facts to suit your arguements. Two doors, one choice. 50/50. |
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29 Oct 12 - 10:55 PM (#3428109) Subject: RE: BS: Monty Hall Problem From: John on the Sunset Coast For some reason my complete post did not come through. I will repost, and try to remember the last paragraphs: For the game to work properly the two lesser prizes" must be of comparable value", and switching doors wins 67% of the time. If, as has been asserted, there was 1 ea. Megaprize, Goodprize and Clunker, then the game changes. If your first choice is the GP, you will not be be shown the MP...because the game needs to continue; you will be shown the C. Staying Loses If your first choice is the C, you will be shown not your choice or the MP because the game needs to continue. So again you should switch. Staying loses. If you actually pick the MP, you will be shown either GP or C, but you don't know that you actually have chosen correctly. So since you know that in two scenarios out of three switching is correct, you should still switch. But this time you lose. Remember you have no way of knowing where the MP is on the first round, but you maximize your chance of getting the MP whenever you are shown a lesser prize. Using the switching strategy you only lose when the MP is picked first (1 of 3); you win if you did not pick it first (2 of 3). Where the strategy may break down is if you have picked the MP, but are are shown the C. You might stand because you're sure to get at least a GP, thereby settling for whatever your first choice was. That is a 1 in 2 chance, but not as good as 2 in 3 that a switching strategy offers overall. But Hall doesn't want you to settle; that's not a good game. That is why the game must have two Cs so that you won't just settle. Whether it actually did, I don't know. I was not a regular viewer, and really didn't consider odds or probability. No strategy assures winning the car, but an always switching strategy maximizes |
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29 Oct 12 - 11:00 PM (#3428113) Subject: RE: BS: Monty Hall Problem From: John on the Sunset Coast ---maximizes the probability of winning the car. Finally complete, I think. |
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29 Oct 12 - 11:31 PM (#3428121) Subject: RE: BS: Monty Hall Problem From: GUEST,TIA If anyone really believes the pure 50/50 thing, play the game and you will change your mind. Read Jim Dixon's decision tree and you will change your mind. It is not intuitive, but it is entirely and thoroughly provably correct that switching after the reveal gives you much better odds. There are at least three different non-mathematical proofs above for anyone who is truly seeking to understand. But you curmudgeons...please proceed. |
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30 Oct 12 - 12:21 AM (#3428133) Subject: RE: BS: Monty Hall Problem From: BK Lick "Please proceed." Ah yes, as the President succinctly told the challenger. |
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30 Oct 12 - 08:04 AM (#3428252) Subject: RE: BS: Monty Hall Problem From: GUEST,Lighter So, TIA, you're saying that it doesn't matter if *Monty knows or not*? Or if the contestant knows that he knows? If so and you're correct, I have to assume that the plain-English proofs are so badly written that only someone already familiar and at ease with the mathematical proofs can follow and appreciate them. I can't, and I have three degrees. Moreover, many other intelligent people can't either, as the length of the Wikipedia article and of this thread suggests. Or do you mean to say we non-mathematicians are hopelessly stupid and/or putting you on? If we can't understand your explanations, how do we know you're not the hoaxer? And if Monty's knowledge doesn't matter, I'll repeat what I wrote earlier: Which assumptions or statements have I made about possible outcomes are wrong? If switching once increases the odds, why not switching again and again (I realize that's crazy, but it doesn't seem any crazier than increasing the odds by switching once - *if* Monty doesn't know or care.) |
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30 Oct 12 - 08:19 AM (#3428259) Subject: RE: BS: Monty Hall Problem From: Jim Dixon The New York Times has a great simulation of the problem. You can actually play the game: http://www.nytimes.com/2008/04/08/science/08monty.html |
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30 Oct 12 - 08:20 AM (#3428260) Subject: RE: BS: Monty Hall Problem From: DMcG I'll just repeat what I said before. Play the game a few dozen times and it will be clear that the odds of winning double (or within experimental error of that) if you switch. Then go through your own logic and think about every step in turn looking for the (false) assumption where you decide the odds must be equal. If we try to point one out for you, we will simply get into an argument about whether or not that step is valid, or can be rephrased, or otherwise shoe-horned to fit so a similar assumption resurfaces somewhere else: much better if you spot your own error, because you will be alert for similar assumptions you make on other occasions. |
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30 Oct 12 - 08:26 AM (#3428261) Subject: RE: BS: Monty Hall Problem From: DMcG It took me round about 2 minutes to play 10 rounds on the link Jim provided. I always switched and won 80% of the time. By definition, then, I would have won 20% if I hadn't switched. |
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30 Oct 12 - 08:27 AM (#3428262) Subject: RE: BS: Monty Hall Problem From: TheSnail Lighter, from your Original Post - "Whichever door was selected, Monty (who knew where the car was) would open one of the remaining two doors to reveal a goat. Yes, it is fundamental to the problem that Monty knows where the car is and makes an informed choice to open a door to reveal a goat. This gives you information (in the form of a probablity, not a certainty) about what is behind the door he chose not to open. |
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30 Oct 12 - 08:42 AM (#3428266) Subject: RE: BS: Monty Hall Problem From: Jim Dixon Lighter: The Wikipedia article is long mainly because it demonstrates multiple methods of solving the problem, showing that they all come to the same conclusion. You don't have to understand all of them. Pick one. It also gives the history of the problem and considers some psychological reasons why some people come to the wrong conclusion. It's not the only problem that confuses people. Have you ever heard of the Mythbusters airplane takeoff problem? I'll bet they're still arguing about that on the Mythbusters forum. But if people want to discuss that here, I think somebody should start a new thread. |
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30 Oct 12 - 08:47 AM (#3428268) Subject: RE: BS: Monty Hall Problem From: GUEST,Ed Yes, it is fundamental to the problem that Monty knows where the car is and makes an informed choice to open a door to reveal a goat. Not exactly. I think the point that TIA was trying to make was that if Monty is guessing and opens a random door to reveal the car then the game would be voided. If he opens a door with a goat behind it, it doesn't actually matter whether he did it purposely or luckily. The fact that it's a door with a goat behind it is what determines the 2/3 odds of the remaining unopened door having the car. |
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30 Oct 12 - 08:57 AM (#3428271) Subject: RE: BS: Monty Hall Problem From: GUEST,Lighter Yes, I erred in that initial post because in my haste I didn't realize it would make a difference in people's understanding of the game. Obviously it does, crucially. I am embarrassed by my carelessness. No wonder people thought I was winding them up later. A final (I hope) question then: Is it true that, as I infer from TIA's post, that it doesn't matter to the reckoning of the odds, at the moment of the second opportunity to choose, whether Monty knows where the car is and is trying to keep you from winning? (I hope it does, but I'm prepared for the worst.) And a final point: If it doesn't, then someone who understands the real-life principle can make a real contribution by explaining it in plain English in a way that most others can follow. As far as I can tell, this doesn't happen in Wikipedia. Doing the experiment isn't the same as understanding what's happened, which is what I'm trying to get at. Apologies again for carelessly misstating the situation in the IP. |
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30 Oct 12 - 09:24 AM (#3428290) Subject: RE: BS: Monty Hall Problem From: Long Firm Freddie If my first choice happens to be a door with a goat behind it (which will be two times out of every three, on average), then Monty will reveal the other goat, leaving the door with the car behind it which I can get by switching. If my first choice is the door with the car behind it (one time in three), then switching loses whichever goat Monty reveals. So switching wins two times out of every three, on average. I used three coins to represent the goats and the car and pretended to be Monty - it all became clear very quickly! LFF |
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30 Oct 12 - 09:37 AM (#3428296) Subject: RE: BS: Monty Hall Problem From: TIA Ed - Exactly! thanks for clkearing it up. Lighter - I am not calling anyone stupid. There are people who do enjoy playing the curmudgeon, and I simply don't want to argue with such. But if someone really wants to understand, there are many really good explanations in this thread. I will try one more, and if it is still not clear, it is my fault. The contestant's initial choice has a one in three chance of being correct. This never changes at any point for any reason, and does not depend on anything Monty does. The initially-chosen door has a one in three chance of being correct. Period. Now focus on the other doors. Once a door is opened to reveal a goat, that door has a zero in three chance of being correct. There is a three in three chance that there is actually a car behind one of the doors. Your initial choice has an unchanged one in three chance of being correct. So, the other two out of three chances must (I repeat *must*) lie behind the remaining unopened door (not the revealed goat door, not your initial choice, but the remaining unopened door). So please switch - it doubles your odds! |
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30 Oct 12 - 10:21 AM (#3428330) Subject: RE: BS: Monty Hall Problem From: TheSnail You didn't misstate the situation, Lighter. From the Wikipedia article - Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice? My emphasis. Note "opens another door... which has a goat". He can only do that by making a conscious choice. |
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30 Oct 12 - 04:23 PM (#3428507) Subject: RE: BS: Monty Hall Problem From: Howard Jones My own attempt to explain it in plain language has vanished into the ether, so here's another go. I couldn't understand this to begin with. Light dawned when I stopped thinking about chances of winning and thought about your chances of losing. There are 2 goats and 1 car. With your first choice you're twice as likely to pick a goat as the car. Put another way, you're more likely to be wrong than right with your first choice. The host then shows a goat and removes one of the possible wrong choices. Remember, your first choice was more likely to be wrong than right and nothing's happened to change that - if it was probably wrong then it must be probably wrong now. Which in turn means that the only remaining door must be probably right. So switching will mean that you're more likely to win (but doesn't guarantee it) |
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30 Oct 12 - 04:53 PM (#3428527) Subject: RE: BS: Monty Hall Problem From: TheSnail GUEST,Ed Not exactly. I think the point that TIA was trying to make was that if Monty is guessing and opens a random door to reveal the car then the game would be voided. If he opens a door with a goat behind it, it doesn't actually matter whether he did it purposely or luckily. The fact that it's a door with a goat behind it is what determines the 2/3 odds of the remaining unopened door having the car. I've been pondering that and come up with this. The contestant has three choices after which Monty has two giving six possible outcomes. 1) Choose the door that has the Car. Monty chooses Goat 1. Switch and Lose. 2) Choose the door that has the Car. Monty chooses Goat 2. Switch and Lose. 3) Choose the door that has Goat 1. Monty chooses Car. Void. 4) Choose the door that has Goat 1. Monty chooses Goat 2. Switch and Win. 5) Choose the door that has Goat 2. Monty chooses Car. Void. 6) Choose the door that has Goat 2. Monty chooses Goat 1. Switch and Win. Discarding the Void outcomes gives you odds of 1/2. Better than the 1/3 if you don't swap but not as good as the 2/3 if Monty makes a consciuos choice. In that case, the six possible outcomes are - 1) Choose the door that has the Car. Monty chooses Goat 1. Switch and Lose. 2) Choose the door that has the Car. Monty chooses Goat 2. Switch and Lose. 3) Choose the door that has Goat 1. Monty deliberately chooses Goat 2. Switch and Win. 4) Choose the door that has Goat 1. Monty deliberately chooses Goat 2. Switch and Win. 5) Choose the door that has Goat 2. Monty deliberately chooses Goat 1. Switch and Win. 6) Choose the door that has Goat 2. Monty deliberately chooses Goat 1. Switch and Win. |
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30 Oct 12 - 05:30 PM (#3428547) Subject: RE: BS: Monty Hall Problem From: gnu If, if, if, if ad infinitum. Shame! "Yes, it is fundamental to the problem that Monty knows where the car is and makes an informed choice to open a door to reveal a goat. This gives you information (in the form of a probablity, not a certainty) about what is behind the door he chose not to open." "NOT A CERTAINTY". It still boils down to ONE choice... between TWO doors. NO inference can logically be made about "switching" the first choice because there is NO switching... it is now simply, ONLY, a choice between TWO doors. You ask me to play the game to prove your point? I could not be arsed to do so. If you consider that doing so is "proof" you really have too much time on yer hands. Two doors. One choice. You may, in yer mind, THINK you have two choices... hit me or stick... but that is where your argument falls apart... they are the SAME. When you overlook that FACT you are throwing in another IF (which literally has no bearing on the logic on an even an empiracal level) and some of you don't seem to get that. One last time... TWO doors... ONE choice. 50/50. If you don't believe me, flip a coin ten thousand times and report back your results. If you tell me picking heads wins 2/3 of the time, seek help. |
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30 Oct 12 - 06:49 PM (#3428583) Subject: RE: BS: Monty Hall Problem From: DMcG One last time... TWO doors... ONE choice. 50/50. If you don't believe me, flip a coin ten thousand times and report back your results. One last time: that's not equivalent. It doesn't matter that it looks as if it is. Try that link above. Being a bit obsessive I went as far as 100 trials and got 64% wins by swapping. Lewis Carroll claimed 'what I say three times is true', but outside literature just asserting it is 50:50 doesn't beat the experimental evidence |
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30 Oct 12 - 06:54 PM (#3428584) Subject: RE: BS: Monty Hall Problem From: BK Lick Dear Gnu -- try to free up your understanding as Howard Jones did: I couldn't understand this to begin with. Light dawned when I stopped thinking about chances of winning and thought about your chances of losing.which I suggested upthread: ...consider that switching turns a correct initial guess into a loss but turns an incorrect initial guess into a win. Since the initial guess is twice as likely to be incorrect as correct, switching is twice as likely to win the prize.Poor chap, you'll likely be embarassed when light finally dawns, but not so much as Paul Erdős and quite a few other distinguished mathematicians who stumbled over this problem. What fun! —BK |
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30 Oct 12 - 07:15 PM (#3428595) Subject: RE: BS: Monty Hall Problem From: DMcG I am well aware, gnu, that playing the game does not constitute proof. What is does do if provide evidence that swapping does double your chances of winning. That then becomes a fact to be explained. And there are plenty of arguments now in this thread that do explain it. Insisting that an argument predicting no benefit in switching must be right even though it does not fit the evidence is not a good stance. |
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30 Oct 12 - 07:19 PM (#3428597) Subject: RE: BS: Monty Hall Problem From: gnu "Since the initial guess is twice as likely to be incorrect as correct,..." Embarrassed? Me? Not in the least. In YOUR statement as just quoted, you change the parameters of the arguement of logic in an absolutely illogical fashion. YOU put YOUR paramenters on YOUR thought where NONE exist in order to prove YOUR point and tell ME I MUST DO THE SAME and then call me "a poor chap" who should be embarassed? I have tried to explain this in the MOST simple of terms but it appears to me... well, it appears that because I don't deal with anything but the facts at hand that I am correct. No ifs, ands or buts... Me? I don't need any. If you do.... gnightgnu |
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30 Oct 12 - 10:17 PM (#3428657) Subject: RE: BS: Monty Hall Problem From: John on the Sunset Coast Gnu you are incorrect when you say it is 2 doors, 1 choice, therefore 50/50, flip a coin. That is NOT the Monte Hall Game. The game IS 3 doors, 2 choices. The kicker is you will NEVER be shown the car OR your first round door (and sometimes it is the same door) on the first round exposure. Repeat: The kicker is you will NEVER be shown the car OR your first round door (and sometimes it is the same door) on the first round exposure. So if you follow a switching strategy for the second round (many proofs have been offered so I won't repeat them) the chance of getting the car is 67%. Play the game 100 times expect to own about 67 cars, if the expected of probability holds. But if you use a second round coin-toss strategy (as you consider the game to really be), your the probability is only 50/50, and YOU CAN EXPECT 17 FEWER CARS out of 100 games, if the expected probability holds. As I and several others have previously pointed out, no strategy guarantees your winning a car. Chance is a funny thing...you can flip a coin and get heads 20 times in a row, and you could pick the car on the first round 20 times in a row but those are not EXPECTED outcomes. But a 67% chance is still greater than a 50%, and so you should expect to need a larger garage if you switch. --- Please forgive my use of caps; I am not shouting--I simply don't know how to us e HTML for emphases. |
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30 Oct 12 - 10:21 PM (#3428661) Subject: RE: BS: Monty Hall Problem From: GUEST,Lighter Jim Dixon's "brute force" method now convinces me, but only for a few moments at a time. I am not sure why this is. The evidently incorrect 1 in 2 analysis at the end is conceptually overpowering regardless of the math. Let's do something similar for a new contestant who comes on the scene after the three doors have been reduced to two. The car is behind No. 1 and he picks No 1: WIN The car is behind No. 1 and he picks No.2: LOSE The car is behind No. 2 and he picks No. 1: LOSE The car is behind No. 2 and he picks No. 2: WIN The car is behind No. 1 and he picks No. 1 and switches to No. 2: LOSE The car is behind No. 1 and he picks No. 2 and switches to No. 1: WIN The car is behind No. 2 and he picks No. 1 and switches to No. 2: WIN The car is behind No. 2 and he picks No. 2 and switches to No. 1: LOSE For that contestant, any guess has a 1 in 2 chance of winning regardless of switching. But for the original contestant, for whom a third door has been eliminated, the odds of winning are 2 in 3 if he switches, even though both contestants are faced with the same two doors. This seems paradoxical. What am I not getting? If the elimination of the third door creates the paradox, I don't see how, though I now see(I think) that the first contestant gets an extra chance by switching. Can "chance" be different for two observers facing the identical physical situation? Suppose a pitcher throws a fastball. I get one swing. If I somehow had two swings at the same pitch I'd obviously have a greater chance of getting a hit. Like Contestant No. 1? Suppose Contestant No. 2 shows up after I've swung and missed. Miraculously he takes a swing at the same pitch (whose speed, etc., hasn't changed), but only one. His overall chance of hitting the ball is clearly less than mine because I took twice as many swings. But the situation involving the doors seems quite different. Is it? I apologize for my thickness. |
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30 Oct 12 - 10:26 PM (#3428663) Subject: RE: BS: Monty Hall Problem From: GUEST,TIA Gnu- I appreciate your frustration, but it really is not a 50/50. Remember, you chose your door from among three! Not two. The odds that you chose correctly will be one in three no matter what Monty does or says. One in three. Period. Now think about the other doors, and that is where it gets fun. |
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30 Oct 12 - 11:31 PM (#3428685) Subject: RE: BS: Monty Hall Problem From: BK Lick The car is behind No. 1 and he picks No. 1 and switches to No. 2: LOSEThis doesn't make any sense -- "switching" means changing from the initial choice made before Monty opens a door (using the information revealed by Monty's action) to a different choice afterwards. |
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31 Oct 12 - 04:14 AM (#3428723) Subject: RE: BS: Monty Hall Problem From: DMcG Lighter: the reason for the difference between the two people (the original 1 out 3 doors and the newcomer with 1 out of two doors) lies in the fact we are not looking at the probability of which door the car is behind, but the probability of correctly choosing which door the car is behind. That extra indirection is the source of all the confusion, and is, as I said ages ago, why the precise way something is expressed in probability (and statistics) has to be treated with extreme care. |
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31 Oct 12 - 05:06 AM (#3428740) Subject: RE: BS: Monty Hall Problem From: TheSnail Gnu would be right if you had no information about what was behind the doors which is the situation Lighter's second contestant finds themself in. The history so far makes them different. One door is the contestant's original choice; the other is the door that Monty chose not to open. The first has its unchangeable 1/3 chance of hiding the car. The second inherits all of the 2/3 chance of hiding the car once the goat has been revealed. |
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31 Oct 12 - 06:57 AM (#3428772) Subject: RE: BS: Monty Hall Problem From: Howard Jones Your first choice is random, and is probably wrong for the reasons already explained. The host's choice of door isn't random. He knows where the car is, and he knows which door you've picked, so he opens a different door which will always reveal a goat. That leaves a choice between two doors. Your first choice was probably wrong, so the other must be probably right. It's not the same as a random choice between 2 equal doors. Which 2 out of the original 3 are left has been influenced by your first choice. |
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31 Oct 12 - 09:06 AM (#3428818) Subject: RE: BS: Monty Hall Problem From: GUEST,Lighter > Which 2 out of the original 3 are left has been influenced by your first choice. I'm starting to feel that I understand. (It's a strange feeling in this case.) Contestant 2 faces with a simple situation uninfluenced by anything earlier choice or any action that Monty has taken. > we are not looking at the probability of which door the car is behind, but the probability of correctly choosing which door the car is behind. This is an extremely subtle distinction for a nonspecialist to appreciate. You are saying that in the final, two-door situation, the chance of the car being behind either door really is 1 in 2; but the chance of the first contestant being the one to find it (because a door has been eliminated) has become 2 in 3. For the second contestant, the chance is 1 in 2 in each cases. Wow. Another counterexample came to me at 3 a.m. Suppose I buy a ticket in a lottery in which my odds of winning are 1 in 3. I buy a second ticket and double my odds. My 2 in 3 lottery odds rely on my knowing which two numbers I have *at the same time.* In the game show, I give up my original "number" when I switch to the second one and am left with just one. But is it true, in light of more recent posts, that when Monty reveals a goat, *that* "number" (not the one I originally chose) is the second one I "own" or "control" or whatever? Bringing the third door into play gives me a second lottery number, so to speak. I hope that's correct. Regardless, the first-draft principle that one draws from this utterly deceptive situation seems to be, "Always take a second chance in a situation of finite choices that has since altered, even if the result of the first choice remains unknown?" Probably that needs some work. The fog has not completely lifted. |
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31 Oct 12 - 09:44 AM (#3428825) Subject: RE: BS: Monty Hall Problem From: Jim Dixon Lighter: If I understand correctly, by introducing a second contestant, you are stipulating that this second contestant knows nothing of what happened before he joined the game. He sees two closed doors, but he doesn't know which one the first contestant originally chose. This puts him at a disadvantage compared to the first contestant. He lacks important information. Lacking information changes everything. The second contestant would indeed have a 1/2 chance of guessing correctly. |
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31 Oct 12 - 09:46 AM (#3428826) Subject: RE: BS: Monty Hall Problem From: Nigel Parsons Another counterexample came to me at 3 a.m. Suppose I buy a ticket in a lottery in which my odds of winning are 1 in 3. I buy a second ticket and double my odds. You only double your odds if there were only ever 3 tickets for sale. Then you buy one & some other person(s) buy the other two. If you buy two, and some other person buys the remaining one then your chances have doubled from 1/3 to 2/3. Normally in a lottery doubling your entries increases your chance, but does not double it. If you, I & Gnu bought one ticket each, and that was all the tickets, you'd have a 1/3 chance. If Gnu & I bought one ticket each, and you bought 2, and that was all the tickets you'd have a 2/4 (1/2) chance, NOT double the previous chance. |
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31 Oct 12 - 09:49 AM (#3428827) Subject: RE: BS: Monty Hall Problem From: DMcG > we are not looking at the probability of which door the car is behind, but the probability of correctly choosing which door the car is behind. This is an extremely subtle distinction for a nonspecialist to appreciate. You are saying that in the final, two-door situation, the chance of the car being behind either door really is 1 in 2; but the chance of the first contestant being the one to find it (because a door has been eliminated) has become 2 in 3. For the second contestant, the chance is 1 in 2 in each cases. Wow You've got it, although I would generalise 'because a door has been eliminated' into 'because the information he had has changed' Let me give another example. You have a glass bowl in front of you which contains 5 yellow raffle tickets and 5 blue. One of these wins a car. You can select any ticket, and see its colour before you draw it, so when you pick a ticket it is a 1 in 10 chance. Now, before you do, I secretly tell you the colour of the winning ticket, but do not tell another competitor. Your odds are twice theirs, even though nothing in the world has changed at all, including both the colour and number of the winning ticket. It is the extra piece of information that makes all the difference. |
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31 Oct 12 - 10:02 AM (#3428832) Subject: RE: BS: Monty Hall Problem From: TheSnail Let's try again. There are really only three situations. The door that the contestant first chooses has either (1) a car behind it or (2) a goat or (3) another goat giving him a 1/3 chance of getting the car. The remaining two doors then have (1) two goats, (2) a car and a goat or (3) a goat and a car. Monty opens one of these doors (with full knowledge of what is behind the doors) to reveal a goat. The remaining door then has either (1) a goat, (2) a car or (3) a car behind it. If the contestant switches to that door, he stands a 2/3 chance of getting the car. Tell England I died trying. |
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31 Oct 12 - 10:16 AM (#3428842) Subject: RE: BS: Monty Hall Problem From: DMcG That's a neat explanation, Snail - I think it is equivalent to some given above, but more concisely expressed. I went off on my particular tangent to address how it comes about that two people (namely the original player and the newcomer) can both see exactly the same physical situation (two closed doors and one open showing a goat), yet have different probabilities of getting the car. It is subtle, but I think it worth puzzling through because it isn't anything like as obvious(!) as the basic problem, and it does show that information is as important as the physics. |
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31 Oct 12 - 11:37 AM (#3428886) Subject: RE: BS: Monty Hall Problem From: GUEST,Lighter Jim, yes. TheSnail, that is indeed the simplest explanation I've read. Send it to Wikipedia. (If anything like it is there already, it's effectively buried.) DMcG: "Your odds are twice theirs, even though nothing in the world has changed at all." The absence of change in physical reality is probably the source of one's profound disbelief about the final odds in Monty's game. Especially when taken in tandem with the subtle distinction between "where the car is" and "how likely you are to find it." I think I *do* get it at last. But what of my question about the "moral"? Is there a general principle about making the kind of choices embodied in the puzzle? (See my previous post.) |
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31 Oct 12 - 11:42 AM (#3428890) Subject: RE: BS: Monty Hall Problem From: John on the Sunset Coast I think some of the problem for those who don't understand the solution is they don't understand the game. The game is not a variation of the sock drawer; it is not some sort of card game; it is not flipping a coin; it is not a game with infinite choices or chances, nor one where your choices are affected by any other player. And you probably will only have one opportunity to play the game for real. But they analyze the Hall game as if it were something it isn't. The game is simple: 2 separate chances to pick a prize from 3 choices; the second chance is given only after one choice is eliminated. To make the game more exciting, the contestant is given the option of changing choices before a remaining door is opened. One opportunity to play. * While it is unknown at first which door contains the prize, the written explanations cause us to think, perhaps sub-consciously, that they do. So the analyze the problem as if they do. * They don't understand what the game really is and what it consists of. It is one person (not multiple people in sequence), choosing for a prize from 3 doors. * But that is only the set-up round. The host knows where the prize is, and the door you chose. For suspense and the entertainment value of the show, the Host will never show the prize or your choice immediately; he will only show the non-prize from the remaining door (either of two doors if your choice happened to contain the prize.). * The Host will not immediately open one of the remaining doors to determine the game's outcome. If he did, you would always only have 1 chance in 3 of winning. Pure luck, and not a very exciting game. * So there is a second round to the game. The host now offers you the opportunity to stay with your original choice, or switch to the remaining door. Stay/switch, switch/stay--it seems like you now have a fifty-fifty chance of getting the prize. This would be true if you had no other information, but you do have other information--the elimination of one door. That information results in your actually having 2 out of 3 chances of winning the prize if you follow the math. The math is the math is the math, and it dictates switching in order to give you the better chance of getting the prize. No strategy offers a sure thing, as has been often pointed out. But which would you rather have, 1 chance out of 2 (50%), or 2 chances out if 3 (67%)? You don't have to understand the math, but you have act as if you do. |
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31 Oct 12 - 12:12 PM (#3428909) Subject: RE: BS: Monty Hall Problem From: TheSnail I don't see these subtle distinctions. I think it is entirely a matter of the probability of the car being behind any particular door. You have to work this out from the information available. Monty's shenanigans have provided you with a little more information than "There is a car behind one of these three doors." The "general principle" is to analyse each situation as you find it. |
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31 Oct 12 - 01:20 PM (#3428963) Subject: RE: BS: Monty Hall Problem From: John on the Sunset Coast 'The "general principle" is to analyse each situation as you find it.' There is nothing to analyze; that has already been done [see above posts or Wikipedia], resulting in the optimal strategy being to switch EVERY time. If YOU DO NOT SWITCH EVERY time, you minimize your probability of winning by 17%age points--50% v 67%. You cheat yourself! That does not seem rationale to me, assuming you hope to win the prize. |
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31 Oct 12 - 01:29 PM (#3428970) Subject: RE: BS: Monty Hall Problem From: TheSnail Yes, John, I know. See my previous posts. I think Lighter was looking for a general principle that could be applied in similar circumstances. People are not just interested in how to play this game, which they will probably never do, but ARE trying to understand the maths. |
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31 Oct 12 - 05:42 PM (#3429047) Subject: RE: BS: Monty Hall Problem From: GUEST,Reverend Bayes Still at it, I see? You still have a 2/3 chance of being wrong on the first step, and the open door doesn't change that. But gnu, you're in very good company in your error. Certain problems in mathematics are fascinating for being intuitively obvious and yet dead wrong. In other news, folks, did you know that 0.99999999999999999999.... is equal to 1? |
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31 Oct 12 - 05:59 PM (#3429052) Subject: RE: BS: Monty Hall Problem From: BK Lick But which would you rather have, 1 chance out of 2 (50%), or 2 chances out if 3 (67%)?Oy! This is not helpful -- the choice is between 1 chance out of 3 (33%), or 2 chances out of 3 (67%). |
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31 Oct 12 - 06:55 PM (#3429067) Subject: RE: BS: Monty Hall Problem From: gnu "Gnu- I appreciate your frustration, but it really is not a 50/50. Remember, you chose your door from among three! Not two. The odds that you chose correctly will be one in three no matter what Monty does or says. One in three. Period. Now think about the other doors, and that is where it gets fun. " ********************************************************************* But, in the "second round", I am choosing between TWO doors. NOT THREE! Three doors do not exist after one door is gone. ONLY TWO exist... and ONE must be picked. Good lord! There are only two doors on the second choice. ONE goat. ONE car. ONE CHOICE. Stick or switch is ONE choice. Two options... ONE CHOICE! ONLY! If you must make a choice between TWO doors, by empirical definition (common sense), ya got a 50/50 chance. The eliminated door has NO impact upon your RANDOM decision in the second round. Your justifications that it does are based on assumptions and conjecture and not upon fact. You can keep trying to tell me I just don't understand, but... you don't understand that 3 - 1 = 2 and 2/1 = 50 % |
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31 Oct 12 - 07:26 PM (#3429077) Subject: RE: BS: Monty Hall Problem From: DMcG What would it take to convince you, gnu? You have an opinion, sure, but there is a lot of argument saying that is mistaken. You have been given the chance to do experiments to see what happens, but you declare you can't be bothered. I reported my experience with those trials (on 100 trials I got 64% wins if I switched), but you ignore that. Have you such faith in your own opinion it is inconceivable to you that you might be mis-understanding the situation? There is a growing similarity to some of the fundamentalist threads: a solid faith in a point of view which admits of no challenge by argument or evidence. |
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31 Oct 12 - 08:28 PM (#3429105) Subject: RE: BS: Monty Hall Problem From: John on the Sunset Coast 1."Still at it, I see? You still have a 2/3 chance of being wrong on the first step, and the open door doesn't change that." 2" The eliminated door has NO impact upon your RANDOM decision in the second round." 1-Yes! But the game does not give a prize on the first step, even if you choose the prize door. There is NO open door at the time you make the first choice; it is only after you choose that a door is opened. 2-But your choice SHOULD NOT BE RANDOM (which is 50/50); the choice must be DELIBERATE, following a switching strategy (giving 67/33 winning expectation). This outcome is proven at Wikipedia and by the lower half of TheSnail's (and others') previous post at 4:53pm yesterday. |
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31 Oct 12 - 08:35 PM (#3429109) Subject: RE: BS: Monty Hall Problem From: gnu "There is a growing similarity to some of the fundamentalist threads: a solid faith in a point of view which admits of no challenge by argument or evidence." You give me NO logic, no arguement and no evidence. You say you conducted a (random) experiment and have results after 100 trials and that proves your postualte? That is your evidence? You disregard logic based on chance? I can flip coins and come up heads that often in one hundered tosses... OR tails... THAT don't mean shit. What do you people not understand? Two doors. Pick one. I pick reality. |
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31 Oct 12 - 08:58 PM (#3429120) Subject: RE: BS: Monty Hall Problem From: gnu Oh... BTW... for all who have posted logical arguements, truth tables (yeah, I got an A+ in Philosophy at uni), and whatever, you all base your arguements on caveats... on assumptions. I do not. I base my arguement on the FACT that it boils down to ONE CHIOCE. Maybe that is where MY argument ends? Indeed, if there is ONE choice, how can that be anything but 50/50... NO MATTER any of YOUR odds however you put them forth? You taint the basic logic of human thought and intuition. I have made my arguements. You have made yours. gnightgnu |
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31 Oct 12 - 09:12 PM (#3429134) Subject: RE: BS: Monty Hall Problem From: GUEST,Lighter Yeah, it's starting more and more to make sense. Question: As soon as you pick a door, your chance of finding the car is 1 in 3. And the chance of the car being behind your door is also 1 in 3. Suppose Monty then reveals nothing but just gives you the option of switching. You switch to, say, door No. 2. The chance of the car being behind door No. 2 is still 1 in 3, exactly the same as being behind either remaining door. Because the car doesn't move. But the chance that *you'll have found it* is 2 in 3 because *your choice* has "moved." You've been given two swings at the same pitch, if I understand correctly. And it doesn't matter if you "hit" the first time and never find out: your chances once you switch remain 2 in 3, but you can still lose. But I stipulated that Monty doesn't have to do anything but offer you a second chance. Does it still work without seeing a goat? (Right now I think it does.) A second contestant, facing two doors, has a different situation. There's a 1 in 2 chance that the car is behind either door. His odds of finding it are also 1 in 2. Switching won't increase them because with only two doors and only four possible outcomes there are no available odds between 1 in 2, and 2 in 2 (certainty). Certainty is impossible, so (as anyone would expect)the odds stay at 1 in 2 no matter how many times he switches between closed doors. Contestant 1 has the odds advantage because he started with 3 doors and nine possible outcomes; he could build from odds of 1 in 3 to odds of 2 in 3. Certainty was not the sole alternative to odds of 1 in 3. Switching works with 3 doors but not with 2. This made sense as I was writing it, but now I'm not so sure. The accepted solution so defies (my) common sense that it's hard to transfer the principle to situations that are similar but not identical. |
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31 Oct 12 - 11:56 PM (#3429170) Subject: RE: BS: Monty Hall Problem From: John on the Sunset Coast Gnu-- You speak of Random Chance in playing the Hall game as being the same as a coin toss. A coin toss is a single event. The Hall game is a dual event. The first event, your choosing a door, is also random, but the exposition of that event is not random. As I pointed out before, Hall will neither show you your choice nor the car at that time. Now you have a second related event, another choice. If you treat that second choice as a coin toss, heads change-tails stay, you have, indeed, reduced the game to a 50/50 proposition. But the math and the bottom part of the grid at TheSnail - Date: 30 Oct 12 - 04:53 PM [or at Wikipedia] show a switch door choice increases the probability (not a guarantee) of winning to 67%. That is a better chance than the pure chance of a coin toss, is it not? Of course you can deny the accuracy of the grid, but that would just be bull-headedness, gnu. (Hmmm, is that sort of a pun?) |
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31 Oct 12 - 11:57 PM (#3429173) Subject: RE: BS: Monty Hall Problem From: GUEST,Guest from Sanity 100...and I'll take the prize behind skirt number 2!! GfS |
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01 Nov 12 - 12:18 AM (#3429177) Subject: RE: BS: Monty Hall Problem From: John on the Sunset Coast GUEST,Lighter Lighter-- You said, "A second contestant, facing two doors, has a different situation." There is no second contestant in the Hall Game. There is only the contestant and Monty. There is a new contestant playing the exact same game (except perhaps for the location of the car) on the next show; his outcome is independent of what you did. But he and you should chose a switching strategy to maximize the chance of getting a car. |
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01 Nov 12 - 04:12 AM (#3429211) Subject: RE: BS: Monty Hall Problem From: DMcG You give me NO logic, no arguement and no evidence. Yes we have. It is logic that you do not accept, but that does not mean it is not logic. You insist that a simple analysis is adequate. We insist it is not. You say you conducted a (random) experiment and have results after 100 trials and that proves your postualte? That is your evidence? You disregard logic based on chance? I can flip coins and come up heads that often in one hundered tosses... OR tails... THAT don't mean shit. What do you people not understand? No, I make no such claim, nor have I ever. The purpose of the experiment is not to prove or disprove anything. It is to test your theory. That's what one does in the scientific world: build theories and test them. The purpose of the experiment is not to prove anything, but to reduce the chance one ends up with egg on your face because you forgot or misunderstood something when you built your theory. I pick reality. No you don't. You pick your belief but refuse to verfiy in any way at all. That's what I mean by the similarity to the fundamentalist threads. So I repeat my question from the original post. Is there anything that would lead you to reconsider your stance? |
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01 Nov 12 - 04:17 AM (#3429215) Subject: RE: BS: Monty Hall Problem From: BK Lick Gnu is very persistent -- I like that in a bloke. Against all odds, I believe I may have devised a thought experiment that might actually convince him to reexamine his so firmly held conviction. In this scenario:
Second, Monty examines the deck and turns over 50 of the remaining 51 cards not including the Ace of Spades, leaving one card face down. At this point, clearly the Ace of Spades is either the card Gnu chose or the one Monty has left face down. Monty now invites Gnu to change his choice from the card he originally chose to the other unknown card. TWO cards... ONE choice. Will Gnu still insist that either one of these cards has a 50-50 chance of being the Ace of Spades? Or will he see that the chance of his original card being the Ace of spades is only 1 in 52, while the chance of the other card being the Ace of Spades is 51 in 52? —BK |
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01 Nov 12 - 05:45 AM (#3429232) Subject: RE: BS: Monty Hall Problem From: TheSnail Lighter Does it still work without seeing a goat? (Right now I think it does.) No it doesn't. The contestant can change his mind as many times as he likes on his first choice. The odds will still say 1/3. He may have changed his mind several times in his head before declaring his choice. Why should changing his mind publicly make any difference? It is only Monty opening a door to reveal a goat that gives you extra information to recalculate the odds. I will leave gnu to more patient souls. |
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01 Nov 12 - 06:27 AM (#3429239) Subject: RE: BS: Monty Hall Problem From: Nigel Parsons As I'm a generous sort, I'll make the following offer to anyone who still believes the odds are 50/50. If we ever meet, we each put $100 (5*$20) on the table. I'll add an extra $20 (20*$1) for luck. I'll supply 3 identical envelopes, put the $200 in one,and $10 in each of the others. I'll shuffle the envelopes below the table (keeping track of which is which), and then offer you the choice of 1 in 3. Once you've chosen I'll show you tha one of the other envelopes contains £10, and pocket that envelope. You are then left with two envelopes (one of which you have chosen). I will pocket one envelope.As you're now down to 2 envelopes, I'll take the one you haven't chosen (as clearly it's a 50/50 chance, and you've made your choice already). I should win often enough that the extra $50 I'm putting in won't bother me. Sound fair? |
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01 Nov 12 - 06:37 AM (#3429244) Subject: RE: BS: Monty Hall Problem From: Nigel Parsons Sorry, the extra £10 I'm putting in. |
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01 Nov 12 - 07:45 AM (#3429265) Subject: RE: BS: Monty Hall Problem From: DMcG I can flip coins and come up heads that often in one hundered tosses... OR tails... THAT don't mean shit. What do you people not understand? Apologies for not commenting on this one earlier. Actually, it means a great deal and it is a branch of testing scientific papers that is being used more and more. Without going into too much detail you set up two hypotheses 'this is a fair coin' and 'some skullduggery is going on'. When you have tossed your hundred heads it is indeed possible it is sheer chance, but it is far more likely you are using a double-headed coin, or using various tricks that distort the result. There has been a major kerfuffle in psychology recently which used exactly this sort of technique to detect certain papers contained fraudulent data. |
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01 Nov 12 - 08:31 AM (#3429284) Subject: RE: BS: Monty Hall Problem From: GUEST,Lighter John, I was speaking only of a hypothetical second "contestant" who appears when there are only two doors left. If the word "contestant" is misleading, "person" will do. His odds are different from the first person's, though faced with exactly the same reality: two doors, one car. That may be the most difficult conceptual hurdle for common sense to get over. The addition of a second person highlights the difference between the odds on where the car is and the odds on someone's finding it, which happen to be identical for the second person but not for the first.(Doesn't it?) At one moment it makes perfect sense to me. The next, none at all, in spite of Jim's irrefutable chart and The Snail's summary. |
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01 Nov 12 - 08:36 AM (#3429285) Subject: RE: BS: Monty Hall Problem From: GUEST,TIA A+ in philosophy, but D- in "conditional probability". Look it up. |
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01 Nov 12 - 08:38 AM (#3429287) Subject: RE: BS: Monty Hall Problem From: GUEST,Reverend Bayes Folks, there might be a point to arguing with someone who isn't right. However, gnu isn't even wrong. One would think a philosophy PhD would have learnt a little humility by now, but nevertheless the Internet is not an appropriate medium for teaching it. Move on. |
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01 Nov 12 - 08:55 AM (#3429293) Subject: RE: BS: Monty Hall Problem From: GUEST OK, Rev., I agree. I was interested to read a few years years ago that your work was a long forgotten branch of mathematics which some entrepreneur was reviving for the first time in nearly a century. Funny, that ... or maybe I'm older than I think. Do you think we should consider 'the unexpected egg paradox'? Maybe not. |
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01 Nov 12 - 08:57 AM (#3429294) Subject: RE: BS: Monty Hall Problem From: GUEST,DmcG That was me, sorry! |
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01 Nov 12 - 10:38 AM (#3429336) Subject: RE: BS: Monty Hall Problem From: GUEST,Reverend Bayes No, my theorem is a fairly fundamental result in probability, known to first year undergraduates everywhere. It has an interesting application in brute-force numerical calculations which has only really become possible since the advent of sufficiently powerful computers. People occasionally make a fuss about this. |
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01 Nov 12 - 10:48 AM (#3429342) Subject: RE: BS: Monty Hall Problem From: GUEST,Lighter > known to first year undergraduates everywhere. Only if they take the right course! And most won't. |
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01 Nov 12 - 10:52 AM (#3429344) Subject: RE: BS: Monty Hall Problem From: John on the Sunset Coast I'm sorry, GuestLighter, but I do not understand the gist of your post (at 8:31 today). But what you describe is not the Hall game, so the analysis is moot. |
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01 Nov 12 - 11:07 AM (#3429350) Subject: RE: BS: Monty Hall Problem From: DMcG known to first year undergraduates everywhere Ah, yes, I remember it well from my first year undergraduate days. It just amused me how the reporter took this assertion about its obscurity from a snake-oil salesman, without any obvious use of his journalistic critical facilities. |
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01 Nov 12 - 11:50 AM (#3429382) Subject: RE: BS: Monty Hall Problem From: GUEST,Lighter John, DMcG asserted that there's a difference between the odds on where the car is and the odds on someone's finding it. Someone else seems to have disputed that. I'm trying to understand, in a way that I can fully comprehend, whether there is a such difference. I'm also trying to fully comprehend the apparent fact that (regardless of how the game was played on TV), a second person coming to the two remaining doors, after the original contestant switched, has only a 1 in 2 chance of guessing the car's whereabouts, while that original contestant has a 2 in 3 chance. I realize the original contestant has more information, but it's hard to grasp how intangible "information" can affect the physical presence of 2 doors and 1 car. Both people are faced with the same material reality. Yet their odds of finding the car are quite different because the first chooser has seen a third door and a goat, but the second hasn't. This seems like a paradox. Information obviously affects *behavior*, but here it affects *odds* whether or not we know what the meaning iof what it is we've learned. (The odds increase for the first contestant even if he's switched doors for no reason but a whim. The whimsical contestant knows only he's seen a goat; he doesn't understand its mathematical significance; his odds still increased when he switched.) The math may be perfectly straightforward to those who frequently calculate probability; but for the rest of us, including some PhDs, the workings of chance, in this case, can seem almost uncanny. It was easier for me to grasp, as a child, that the earth goes around the sun despite the evidence of my own eyes, than it is for me to grasp as an adult just how the odds of finding the car change. I don't doubt it; I'm just having a hard time comprehending it and its implications (not that I'm entirely sure what they are, beyond "Don't believe everything you think"). |
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01 Nov 12 - 12:06 PM (#3429396) Subject: RE: BS: Monty Hall Problem From: GUEST I didn't actually point this out before, because I thought it might be another unnecessary source of confusion, but the phrase 'the odds where the car is' should really be 'the odds of choosing the car given that you have no information beyond that observable in the immediate physical situation'. It is not really meaningful to talk about the odds of where the car is without reference to the information you have available. But as Rev B suggested, it is probably (*grin*) a good idea if I bow out of this thread. |
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01 Nov 12 - 12:49 PM (#3429417) Subject: RE: BS: Monty Hall Problem From: John on the Sunset Coast " I'm also trying to fully comprehend the apparent fact that (regardless of how the game was played on TV), a second person coming to the two remaining doors, after the original contestant switched, has only a 1 in 2 chance of guessing the car's whereabouts, while that original contestant has a 2 in 3 chance." The second person has NO choices. One goat has been revealed, the contestant has switched doors, the person then gets the original door by default. The contestant has a 2/3 chance of getting the car, therefore the person has only a 1/3 chance of getting the car. Conversely, if the contestant does not switch he has reduced his chance to to 1/3 by staying [see the grids]. The person, still having NO choice, gets the unpicked door. But he has the better chance of getting the car, 2/3. It is important to realize that the second person has "no choices" no matter what the contestant does, but his chances of winning the car are totally dependent on the contestant's action. |
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01 Nov 12 - 01:20 PM (#3429427) Subject: RE: BS: Monty Hall Problem From: GUEST,Lighter > The second person has NO choices. In the setup I described, he does. That's just the point. There are two closed doors in front of him. He can choose either one. The two people are not competing. Both see two closed doors. The first chooser, who has seen a goat and switched, now has a 2 in 3 chance of finding the car by guessing. The newcomer, who's just shown up and hasn't seen or done anything, has only a 1 in 2 chance. With the same two doors. It sure feels paradoxical to me. Recall DMcG's bowl of tickets. In that situation, knowing the color of the winning ticket ahead of time obviously increases your odds of picking it from a bowl with your eyes open. Your odds are dependent on your actual knowledge. But in the case of the now-revealed goat, the chooser doesn't have to know or understand anything. Just switching increases his odds - accidentally. "Information" in the ordinary sense appears to be irrelevant. And the uninformed second chooser is stuck with 1 in 2 no matter what he does or thinks. I envy anyone who thinks it's conceptually straightforward. Maybe it's all elementary, if you've had the right course. |
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01 Nov 12 - 01:44 PM (#3429442) Subject: RE: BS: Monty Hall Problem From: TheSnail Perhaps what hasn't been spelled out, is that, by revealing the goat behind one of the doors, Monty has actually changed the odds of the remaining door. Up until then, it was 1/3 just like the other two. After the goat has been revealed, the odds of the car being behind the last door become 2/3 (See my post of 31 Oct 12 - 10:02 AM). The two doors are different. The advantage that the original contestant has over Lighter's late arrival is that he knows which door is which. |
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01 Nov 12 - 02:08 PM (#3429450) Subject: RE: BS: Monty Hall Problem From: DMcG But in the case of the now-revealed goat, the chooser doesn't have to know or understand anything. Just switching increases his odds - accidentally But he does. He needs to know which door he picked. Suppose we have a slightly different situation: You pick a door, Monty opens the door to show the goat ... and the fire alarm goes off. Everyone is evacuated for several hours and by the time you come back you have forgotten which door you picked. Now you can now longer 'switch' in a meaningful sense, because you don't know which to switch from: you simply have the random choice of the two doors, exactly like the newcomer. |
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01 Nov 12 - 02:53 PM (#3429469) Subject: RE: BS: Monty Hall Problem From: gnu "Interestingly, pigeons make mistakes and learn from mistakes, and experiments show that they rapidly learn to always switch, unlike humans (Herbranson and Schroeder, 2010). Soooo... I am not as smart as a pigeon? |
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01 Nov 12 - 03:55 PM (#3429509) Subject: RE: BS: Monty Hall Problem From: John on the Sunset Coast Lighter-- The contest takes advantage of probability, and switches doors. I think you agree that his probability of winning the car is 2/3. If we do not agree, stop here. The door which the contestant keeps is now taken away; the exposed door is closed, leaving two closed doors for a new, unaware person to choose from. But the new person may start his choice with no opportunity to win the car because it has, in fact, already been won. This leaves the following sets. Pick A has Goat and B has Goat Lose Pick A has Goat and B has Car Lose Pick A has Car and B has Goat Win Pick B has Goat and A has Goat Lose Pick B has Goat and A has Car Lose Pick B has Car and A has Goat Win Out of these possible combinations 2 of 6 win; therefore the new person has 1/3 probability of winning the car. This is the expected probability since the contestant has 2/3. QED |
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01 Nov 12 - 04:50 PM (#3429538) Subject: RE: BS: Monty Hall Problem From: Howard Jones It's the difference between the probability that a particular door conceals a car and the probability that you will make the correct choice. The first remains fixed (depending on the number of doors to choose from), the second depends on how much information you have. Someone coming along at the second stage of the game where there are only 2 doors to choose between has no information on which to base his choice. He faces an entirely random choice between two doors. His chance of winning is therefore 50/50. The original contestant is not making a random choice at the second stage, because he already has some information. From the situation which applied to the first round, he knows that there is a 2:1 chance that his first choice was wrong. By using this knowledge (which is not perfect, but better than none at all) he can therefore improve his chances of winning to 2:1 by switching. Suppose someone had tipped him off which door the car was actually behind. With that knowledge his chance of winning increases to 100%. There's still only a 1 in 2 chance that a particular door hides a car. |
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01 Nov 12 - 05:09 PM (#3429548) Subject: RE: BS: Monty Hall Problem From: BK Lick "Soooo... I am not as smart as a pigeon?" -- gnu Herbranson and Schroeder, 2010 Embarrassing, ain't it? |
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01 Nov 12 - 07:06 PM (#3429598) Subject: RE: BS: Monty Hall Problem From: gnu Not at all. I am rather proud that I made a (good) joke of it. You ceratinaly faild to do so unless that wasn't meant to be funny and if that's the case, Lick me. |
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01 Nov 12 - 10:46 PM (#3429705) Subject: RE: BS: Monty Hall Problem From: John on the Sunset Coast Howard-- You said, "Someone coming along at the second stage of the game where there are only 2 doors to choose between has no information on which to base his choice. He faces an entirely random choice between two doors. His chance of winning is therefore 50/50." You are wrong in that assumption. Lighter change the game such that the second person sees two doors, and has no other information. The game is not set up to give away two cars, so if the original contestant did win the car, there will only be goats left for the second person to choose from. In that case, he would have zero chance of winning the car, and that possibility must be accounted for in determining the probability of winning the car. I showed, earlier today, that the new person has only a 1/3 chance of winning the car. Now you can change the game further by only having a second player if the first contestant fails to win the car. The new person facing two doors with the assurance there is a car available, and no other information, does have a 50/50 chance. We have come a long way away from the original Monty Hall Problem. |
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02 Nov 12 - 03:14 AM (#3429756) Subject: RE: BS: Monty Hall Problem From: DMcG RE those pigeons I would say yes, it is embarrassing. Not for gnu specifically, but for all humans. We set great store by our intelligence and it is good for us to occasionally be reminded that it is not always as useful as it is made out to be. There is little doubt that the students in the experiment were using their intelligence: they were just using it wrongly. And, time and again, we all do that. We are not talking about simple mistakes here, but the absolute insistance that we always know best. |
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02 Nov 12 - 05:30 AM (#3429791) Subject: RE: BS: Monty Hall Problem From: Howard Jones I was making the comparison between someone faced with a choice of only two doors without having had a first choice out of three. I was assuming that one of the doors still conceals the car. What confuses people, including me to begin with, is that there is a difference between (1) the probability that a particular door conceals the car and (2) the probability that you will choose that door. (1) simply depends on the number of cars and doors - in the first round it is 1 in 3, in the second round 1 in 2. (2) depends on your level of knowledge. For a contestant in the game, the results of the first round increase his chances, because he knows his first choice is probably wrong. For someone who actually knows the answer it increases his chances to 100%. None of this alters probability (1) This assumes the rules of the original game. If you change the rules you alter the probabilities because you alter the level of knowledge which can be obtained from the first round, but the principle remains. |
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02 Nov 12 - 09:19 AM (#3429851) Subject: RE: BS: Monty Hall Problem From: GUEST,Lighter In the disagreement between Howard and John, Howard seems to me to be correct, especially since he understood what I was getting at with the second player. So two perceptive people who understand the calculations still disagree about various additional issues shows. That shows just how confusing the whole set-up really is. I'm tempted to press some of my unanswered questions, but since they're expressed in words rather than numbers they may be too troublesome to fool with. Yes, we have come a long way from the original problem, because to the naive observer (me) the solution raises additional questions about what kinds of situations resemble it. As for the pigeons, it's silly to suggest that they're "smarter" than the students. The only thing that "smart" could mean in a sentence like that is "far more efficient in learning to respond rightly to stimuli, unconsciously and without preconceptions." If the students were like most people, they were hampered by the years-long conscious conviction (rarely if ever challenged) that two doors must always give a 1 in 2 chance, regardless of additional factors. The pigeons, so far as we know, weren't. And unlike some people, they can't explain why they're right. But it's conscious awareness and the ability even to have preconceptions (or any conceptions) that makes people generally smarter than pigeons. |
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02 Nov 12 - 10:15 AM (#3429880) Subject: RE: BS: Monty Hall Problem From: DMcG So two perceptive people who understand the calculations still disagree about various additional issues shows. That shows just how confusing the whole set-up really is. Not necessarily quite that, I'd say. The original game had, as we all remember, just one player. When you involve a second player, there are a whole slew of ways you could do so, and many of these have different consequences: i) you can assume the player arrives entirely after the whole game has played. He sees one closed door and two open ones. The open ones may show goats and/or a car. ii) you can assume the player arrives entirely after the whole game has played. He sees one closed door and two open ones. The open ones are now both empty because the goat has wandered off and the car has been driven away. ii) you can assume that the game delivers at most one car, so that gives different results whether the car has or has not already been won. ii) you can assume that the game can deliver more than one car: whichever door the original person picked is closed again and a fresh goat or car is placed behind it, matching the original contents. iii) you can assume the original player and the new player pick their doors at the same time with no knowledge of what the other picks, and if the pick the same door they have the job of deciding how to share the car or goat. iv) you can assume the original player and the new player can only open one door each but they negotiate to decide who gets which door .... and so on. So the disagreements may simply be due to playing a different 'extended game'. |
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02 Nov 12 - 10:47 AM (#3429890) Subject: RE: BS: Monty Hall Problem From: John on the Sunset Coast GuestLighter -- Howard and I do not fundamentally disagree. He said, "I was assuming that one of the doors still conceals the car," in asserting the second person's chances were 1 in 2 Where we differed: I assumed that the new person was getting to choose from what was left by the original contestant, and that his choice was influenced by that action. You and Howard, it turns out, assumed that the contestant had lost the game, leaving the car in play--but you did not clearly make that a part of the proposition. The paragraph I wrote following my analysis reflects the problem you meant to describe. "Now you can change the game further by only having a second player if the first contestant fails to win the car. The new person facing two doors with the assurance there is a car available, and no other information, does have a 50/50 chance." So, while I analyzed the problem as it seemed you were posting it, I also took into account the problem you say you were posing. On that, the three of us agree. I'm not sure whether we agree on my analysis when there is no car to be had for the second person. |
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02 Nov 12 - 01:36 PM (#3429974) Subject: RE: BS: Monty Hall Problem From: Jim Dixon I have started a new thread on a new problem. I hope you enjoy it: BS: The Mythbusters airplane takeoff problem |
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02 Nov 12 - 03:13 PM (#3430027) Subject: RE: BS: Monty Hall Problem From: Howard Jones "Where we differed: I assumed that the new person was getting to choose from what was left by the original contestant, and that his choice was influenced by that action. You and Howard, it turns out, assumed that the contestant had lost the game, leaving the car in play--but you did not clearly make that a part of the proposition." I obviously expressed myself badly - what I had intended to convey is someone facing the same choice of two doors, one concealing the car and the other a goat, but without the first contestant's advantage of having participated in the first round. Without prior knowledge of how those two doors came to be selected, this person can only make a random choice and has a 1 in 2 chance of winning. The first contestant, however, knowing what the first choice was and that it is probably wrong, faces a different set of odds. I had originally fallen into the trap of believing that the second round of the game is an entirely new choice between two equally likely doors, and hadn't understood what information could be derived from the first round. Light eventually dawned, but it took me a while! |
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02 Nov 12 - 06:51 PM (#3430111) Subject: RE: BS: Monty Hall Problem From: gnu It's definitely not embarrassing for me because when I am wrong and proven to be wrong, I admit it and apologize. Now, I am not gonna make an excuse, although I could, as to why I missed... Subject: RE: BS: Monty Hall Problem From:Jim Dixon - PM Date: 29 Oct 12 - 10:35 AM ... but I missed it and I feel terribly about the consternation I caused people with each of my posts. In my defense, I really have no defense that is actually valid. I have an "excuse" that is pathetic so, like I said, I won't make such an excuse. I'll just say I am sorry and in future, hopefully, I'll be able to curb "my excuse". No guarantees, mind you. Sorry in advance. I guess you'll just have to take all that with a rim of salt. >;-( |
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03 Nov 12 - 09:18 PM (#3430628) Subject: RE: BS: Monty Hall Problem From: Mysha Pfffff, look away a moment and ... Well, here I go again: John on the Sunset Coast: The fact that switching is good in two cases doesn't require you to switch in the third case, as with three distinct prizes, the cases are recognisable. If you are indeed always shown the good prize unless you picked it, then it doesn't matter what you do if it's shown to you. The other two doors have equal chances of being clunker and being mega prize. But if you're not shown the good prize it would mean you picked the good prize, so you should switch to get the mega prize. If you're always shown the clunker, unless you picked it, likewise, it doesn't matter what you do if it's shown to you. But, again, if you're not shown the clunker, it would mean you picked the clunker, so you should switch to get the mega prize. If you are shown a random not mega prize. then you should indeed always switch. That's not because in two of the three scenarios switching is correct, but because in either case, the scenario where switching is correct is twice as likely as the one where it isn't. E.g.: You're shown a clunker, so you either picked the good prize or the mega prize. However, for the good prize, you're always shown the clunker, and for the mega prize, you're only shown the clunker half the time (the other half you're shown the good prize). So, it's twice as likely that you're shown the clunker because you picked the good prize than that you're shown the clunker because you picked the mega prize. Thus, two out of three, switching will give you the mega prize. Lighter: If Mr. Hall doesn't want you to win, his best strategy is to give you as little information as possible. If he let's you pick a door and he doesn't touch the others, you're stuck with a 1 in 3 chance of winning. If he gives you more information and allows you to choose anew, he'll always improve your chances. Lighter: It does matter whether Mr. Hall knows where the car is, as otherwise he may accidentally reveal its position. In that case: if Mr. Hal reveals the car, switch to the open door; if he reveals a goat, it doesn't matter whether you switch or keep. Curmudgeon Gnu: It does indeed boil down to one choice between two (closed) doors. (Actually, one of three such choices, but anyone who cares about that distinction is a curmudgeon.) And picking a door at random will give you a 50% chance of getting a goat. What's being discussed in this thread is that the events before that choice will allow you to determine with greater certainty than random chance which door will hold the car. That such strategies are possible can be seen from the fact that anyone trying to win a goat has a 100% chance of winning, by not selecting one of the two closed doors, but rather the open door showing a goat. Lighter: Your swinging example seems obvious. There's only one way to get to swings two bats at the same pitch: You swing two bats at the same time. Because of that, your hit chance is indeed better than those of contestant 2, who swings with only one bat [must be a lefty, to be able to swing at the same time]. With Mr. Hall, the initial case is reversed: Here you have initially two chances to miss, pick door with a goat, and one to hit, pick a door with a car. Then Mr. Hall takes away a choice so the final choice is between a car and a goat. This way, switching reverses the results, getting you back to two chances to hit and one to miss, like in your two-bat swing. England: The Snake died trying. Lighter: Buying double the tickets only doubles your odds in a lottery if there's a fixed number of tickets. Also, since we're aiming for one car, there has to be only a single prize, so the chance of you winning two prizes doesn't come into play. Basically, your example would be a lottery with three tickets, of which you buy two. So, just before the lottery, you talk with the owner of the third ticket, and he says: "I'm a curmudgeon and I believe you either win or lose, that must be fifty-fifty. We might as well switch our hands of tickets." And you say: "No way, Jose! I'm not switching, I have two chances in three to win." Now the next lottery, Jose is quicker in buying the second ticket, and you get only the one. And he says: "I'm a curmudgeon and I believe you either win or lose, that must be fifty-fifty. We might as well switch our hands of tickets." What do you say? Jose gives you a choice to keep or switch between two hands of tickets. But he's wrong in thinking that those hands are fifty-fifty. Of course you would switch now, as your hand has the chance of one ticket, and his hand has the chances of two. Now Mister Hall sells tickets to three doors. You get only the one ticket, but before he opens the door belonging to your ticket, he says: "I heard from Jose, who has the other two doors, that he is a curmudgeon and that he believes that you either win or lose, which must be fifty-fifty, and that he would as well switch doors with you. Would you?" What do you say? "No wait!", he says before you reply; "Let me tell you that one of the doors that Jose has does not have the prize." And you say: "Two tickets, and only one prize; obviously one does not have a prize." "Yes, quite right", he says, and he opens one of the doors. "That was one of the doors of Jose, and see: No prize." Now, do you want to switch to Jose's remaining door, or do you want to keep your own?" Now it seems like he has reduced your chances to one in two, but it's really just like before: You have the chance to switch from your one chance to Jose's two chances. The fact that Mr. Hall has demonstrated what you already know, namely that it's 100% certain that Jose holds at least one ticket that won't win the prize, does not change that Jose holds two tickets against your single ticket. The next step would be the basic problem that started this thread. John on the Sunset Coast: "Not a card game", he says. Draw a card, any card, of the three I hold in my hand. Right, don't look at it yet, but let me tell you one of the three is a King, the other two are jacks. If you show me the king, you win the game. Look, of the two cards in my hand, one is in fact a jack. Now I have one unknown card left in my hand, and you have one unknown card left in your hand. Would you rather show me the card you're holding, or the one I'm holding? And, yes, there is a strategy that offers a sure thing: Picking the open door is sure to get you a goat. Reverend Bayes: No, I didn't. Would you settle for very similar to 1? Lighter: Yes, it does matter that Mr. Hall opens a door. Let's approach it from that side: Why does he open that particular door? Because the prize is behind one of the other doors. It's two in six that the prize is behind the door that was picked, but only one in six that Mr. Hall then would have opened that door he now opened, and one in six that he would have opened the remaining door. It's two in six that the prize is behind the remaining door and then he would always open that door he now opened. (It was also two in six that the prize was behind the door he now opened, and he would have opened the remaining door had it been so.) So, it was three in six that he opened this particular door, and of those three, only one is for the door that was picked, while two is for the remaining door. One in three for staying with the door that was picked, or two in three for the remaining door; make your choice. The asymmetry is in the fact that Mr. Hall never opens the door that was picked. You of the three doors know that. Your hypothetical second contestant who joins you when you're down to two doors, would have a fifty-fifty chance of picking the right door because he does not know about the asymmetry. I know that information is not strictly influencing chance and that to describe it as such may be confusing. This is one of the reasons why in cases like these sometimes the word "certainty" is used. E.g.: Choosing between the two doors at random gives you a 50% chance of getting the car. But if Mr. Hall was in a good mood, and actually told you, you would have 100% certainty. Likewise, the situation we discuss here isn't actually wholly a matter of chance any more; knowing the history of the situation gives you a 67% certainty where the car is. Your second contestant, not having such information, has certainty no larger than the chance; he's only 50% certain, commonly known as "he's in doubt". DMcG: It's not our intelligence that is worse than that of pigeons; it's our memory for small details that is not as good as the pigeons' memory for food. (It's not the problems that I have trouble with, but the paradoxes.) Bye, Mysha |
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04 Nov 12 - 03:42 AM (#3430714) Subject: RE: BS: Monty Hall Problem From: DMcG Reverend Bayes: No, I didn't. Would you settle for very similar to 1? Not sure amout the Rev, but I wouldn't. 0.9999999 recurring (i.e. going on for ever) is indeed exactly the same as 1. If it were not it would introduce serious problems in mathematics. But it is not as much of a problem as it seems, because there is a distinction between the mathematical quantity 'one' and how you choose to represent it. All the statement is actually saying is that the mathematical quantity has multiple possible representations, which no-one should really baulk at, since they are quite comfortable with 1 and one. Of course, a mathematician by heart will start talking about what we mean by equality and epsilons of difference, but I think we can spare the world that one. As for the relationship between pigeons, intelligence[whatever that is], memory and goals (which might be food or course credits depending on the species): it is an interesting subject but probably better discussed elsewhere. But as a taster the idea that intelligence can be better or worse assumes you can in principle set up an ordered sequence of intelligences, which in its turn assumes intelligence is single dimensional. What it isn't, I assert. |
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04 Nov 12 - 03:44 AM (#3430715) Subject: RE: BS: Monty Hall Problem From: DMcG Which it isn't, I assert (sigh!) |
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04 Nov 12 - 08:04 AM (#3430769) Subject: RE: BS: Monty Hall Problem From: GUEST,Ron It is impossible to assess the odds unless the exact rules are known. If the host knows where the car is and the other two doors hide both hide goats, you only ever get the choice of a car or a goat. Your first choice has no meaningful effect unless you are particular about which goat you get. The only 'unknown' involved is which goat may be eliminated, he will always eliminate one. Your choice is always 50/50 between a car or a goat. If the two 'other' prizes are of different values and he always eliminates the lesser remaining prize depending on your first choice, your second round strategy should be dependent on the outcome of the first round, so could be better than 50/50 in the second round. If he doesn't know where the car is in either of the above cases, it becomes a totally different game. Your odds of winning a car will be zero if he's randomly chosen that door. So your odds are less than 50/50. |
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04 Nov 12 - 12:26 PM (#3430860) Subject: RE: BS: Monty Hall Problem From: Mysha Hi Ron, "It is impossible to assess the odds unless the exact rules are known." Actually, though we explored other avenues, the rules are quite explicit in Lighters initial post. They are basically your first case. One could argue that they allow your second case, but that would not change the outcome. "If the host knows where the car is and the other two doors hide both hide goats, you only ever get the choice of a car or a goat. Your first choice has no meaningful effect unless you are particular about which goat you get. The only 'unknown' involved is which goat may be eliminated, he will always eliminate one. Your choice is always 50/50 between a car or a goat." Well, since these are the rules Lighter gives, some ten solutions treat this case. They already show you the outcome, but I find the observation "unless you're particular about which goat you get" rather appropriate, so I'll go with that. Let's call the goats "Daisy" and "Bonny". Whether you care for which goat you get or not, your first choice is now between Bonny, car, and Daisy. - If you pick Bonny, Mr. Hall will reveal Daisy, and you should switch to the remaining door, the car. - If you pick car, Mr. Hall will reveal a goat, and you should keep your door, the car. - If you pick Daisy, Mr. Hall will reveal Bonny, and you should switch to the remaining door, the car. You don't know what's behind the doors that you're choosing between, and choosing at random gives you a 50% chance for the car. But as you can see from the three cases above, there are two ways to pick a door with a goat as your first choice, namely either Bonny or Daisy, and in those two cases you should switch. That's opposed to the one case where you picked a car as your first choice, and only in that one case you should keep. Thus, as you knew from the beginning, the odds favour the gamble that there's a goat behind your door, and in that situation you should switch. "If the two 'other' prizes are of different values and he always eliminates the lesser remaining prize depending on your first choice, your second round strategy should be dependent on the outcome of the first round, so could be better than 50/50 in the second round." We also explored that avenue above, and the fact that the goats are not of equal value does not influence your chance of winning the car. (It does allow alternative strategies, though, and in some cases gives a higher certainty.) If you'll check for Bonny and Daisy in the treatment of your first assumption, you'll see that Mr. Hall will only have to choose between Bonny and Daisy, described as "Mr. Hall will reveal a goat", in the case where your first choice is the car. And in that case, regardless of the way Mr. Hall selects which of the other doors to open, you should keep what you have. Thus, like with two equally-valued goats, the strategy to always switch yields the best results, winning you the car two times out of three. (And fortunately, since Mr. Hall eliminates the lesser prize, even if you don't get the car, you'll at least get the most valuable goat.) "If he doesn't know where the car is in either of the above cases, it becomes a totally different game. Your odds of winning a car will be zero if he's randomly chosen that door. So your odds are less than 50/50." That one has been given some thought above as well, on the assumption that you can choose any door. But indeed, when sticking strictly to the description given in the first message, but ignoring that Mr. Hall knows where the car is, the situation becomes completely symmetrical: Regardless of which door you pick, one of the other doors is eliminated at random. The fact that you can see what was behind it doesn't matter, as it 's purely random, and you're limited to the remaining, closed doors. Thus always keeping or always switching is jut as random as your initial choice, which is one in three, indeed lower than 50:50. DMcG: You didn't ask for a definition of "very similar". I'll define by example: In the first explanation above, Bonny and Daisy are very similar. Bye Mysha |
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04 Nov 12 - 01:01 PM (#3430869) Subject: RE: BS: Monty Hall Problem From: DMcG You didn't ask for a definition of "very similar". I'll define by example: In the first explanation above, Bonny and Daisy are very similar I interpret that to be interchangable but nevertheless distinct in themselves, not just in the label we have used. The values denoted by 0.99999... and 1 are not distinct except in the labels, in the same way that at the moment 'The man occupying the post of the president of the US' and 'Obama' are two distinct labels for the same thing. The goats, on the other hand, are two distinct goats, whatever we choose to call them, even if we choose to confuse everyone by giving them the same name. |
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04 Nov 12 - 04:03 PM (#3430969) Subject: RE: BS: Monty Hall Problem From: GUEST,Ron Mysha, I explored all three scenarios because they had all been previously mentioned. While each one had been examined, they had not been compared. Oh dear! I didn't expect or intend anybody to take my rather tongue in cheek comment "unless you're particular about which goat you get" seriously. As it is unlikely that the contestant has met either goat (and possibly doesn't even want a goat) they would be unlikely to be concerned with the which they got. So I'll move away from goats and use loo brushes instead (same make, same colour, same value, no names). The reality of the first scenario is that your first choice is irrelevant, and is pure showmanship as somebody has already said. Whether you make a choice or no choice at all, you know that Mr Hall will eliminate one loo brush. Therefore your only meaningful choice is in the second stage, and will always be 50/50. Your part in the first stage was meaningless, and did not improve or lessen your ultimate chance of winning. It is always a straight choice, a loo brush or a car, the rest is window dressing. The second case is very different. If the prizes are a loo brush, a TV, and a car, your first choice influences what Mr Hall eliminates. If you choose the loo brush, then the TV will get eliminated, you are then guaranteed to get the car by changing your choice. If your first choice is either the car or TV, then the loo brush is eliminated. You are then left with a 50/50 chance of getting the car. |
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04 Nov 12 - 10:07 PM (#3431121) Subject: RE: BS: Monty Hall Problem From: Mysha Hi Ron, No, I didn't take seriously the possibility that anyone might want a particular goat. But you introduced the possibility of distinguishing between the two goats, and I used that to again demonstrate that in the case of two goats, you can pick the door with the car with approximately 67% certainty. And since you still appear unconvinced, counter to Gnu who seems to have found the answer while I was writing that long message, for you I'll do it again. Now that we no longer have goats, as it's rather ridiculous to give loo brushes names (One is "Lou" ...? Naah.), let's just say that one brush is red and the other is blue. So, for your first choice, we now see three possibilities: - If you choose the red brush, Mr. Hall will reveal the blue brush, and you should switch to the remaining door, the car. - If you pick car, Mr. Hall will reveal a brush (selecting one by a method that may or may not be known), and you should keep your door, the car. - If you pick the blue brush, Mr. Hall will reveal the red brush, and you should switch to the remaining door, the car. Now, let us assume that Mr. Hall has opened a door showing a red brush. There are two possible reasons for him to do so. The first reason would be that the door you chose had the blue brush (You're right that your initial choice is not a real choice; it functions more as protection for that door from being eliminated.) meaning he can't open the blue brush door and has to open the red brush door. This reasons is valid in one in three cases, the chance that you pick the door with the blue brush. The second reason would be that you chose the door with the car, which allows Mr. Hall to pick between the two brushes in whatever way he likes. He may pick a brush at random, he may like the red brush and not want you to win it, or he may just be eliminating the worst prize and may know that the blue brush is a high quality brush, making it a good prize, as brushes go, while the red is a clunker. Unless we know his method, the chance can be anything, as long as it's above 0% and no more than 100%. So, in one case out of three, the case where you selected the blue brush, Mr. Hall will always show the red brush, and the remaining door will always be the car. In another, the case where you selected the car, there's only a chance that Mr. Hall shows the red brush, and the remaining door would be the blue brush. But whatever his method of selecting between the red brush and the blue brush, even if it's 100% chance that he shows the red brush, it's still 100% of one in three cases. The chance that he shows the red brush because you picked the car, and you should keep your door, thus can never be larger than the chance that he shows the red brush because you picked the blue brush, and in the latter case you should switch as the remaining door has the car. It can be considerably lower, however, e.g. 50% if he chooses at random when he has access to both the doors holding brushes. So, the strategy to switch if Mr. Hall reveals a red brush will never be less effective than the strategy to keep, and at best be considerably better. (It can be almost a certainty if e.g. it is know that when the contestant picks the door with the car, Mr. Hall will only display the red brush if he encountered the Chinese president on his way to the studio.) Now, you can read that part again with the colours reversed, and it would again hold true. Furthermore, in the worst case where switching is only equally successful as keeping, that brush is always shown if Mr. Hall has a choice between the brushes, so if the other brush is shown, which is never selected in that situation, switching will then guarantee a car. So, regardless of whether Mr. Hall reveals a red brush or a blue brush, the strategy of switching will never be worse and usually be better than the strategy to keep. And that's independent of the method Mr. Hall uses for deciding which prize to reveal in the case that you picked the car, hence independent of whether the non-cars are equally valuable or whether one is a loo brush and the other a goat. All this because whether Mr. Hall has a choice depends on what's behind the door you pick first. So, your first choice is relevant after all. I still like the goats version better, though. DMcG: Then I'd have to content that since 1 is a natural number and 0.99999... is a fraction, they are clearly distinct in themselves, even if they are interchangeable in the context of value. But I expect that this would quickly lead us to philosophy, and it's not the topic of this thread anyway. So let's leave it at me not being convinced but feeling no particular desire to change that. Bye, Mysha |
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05 Nov 12 - 05:45 AM (#3431187) Subject: RE: BS: Monty Hall Problem From: DMcG I fear it might get into philosophy and as about two years ago I went to a series of Philosophy lectures where the keynote discussion was precisely on the philosophy of naming, I am convinced I could get out my depth very rapidly. So we will leave it beyond telling you a tale. In the early seventies I was told of an experimental computer that had been built where two floating point numbers were reported as equal if they were sufficently close. It turns out to be a nightmare to program beause A=B and B=C did not imply A=C. The moral is that blurring the distinction between equal and very similar gets you onto pretty unstable ground |
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05 Nov 12 - 08:19 AM (#3431244) Subject: RE: BS: Monty Hall Problem From: GUEST,Lighter Oddly, I'm getting re-confused. Everything might be falling into place, if I had my twenty-year-old brain back. |
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05 Nov 12 - 10:38 AM (#3431326) Subject: RE: BS: Monty Hall Problem From: John on the Sunset Coast Lighter says, "Oddly, I'm getting re-confused." Why? Mysha is just rehashing a closed post which pretty much had reached consensus, and threw in a few red herrings for extra measure. The only thing to remember: If you want to maximize you chance to win the car in the classic Hall game, you always must switch doors. If you decide on a whim whether or not to switch, you lower your odds to equal chance. Sometimes you will lose by playing correctly; less often you will win if you play incorrectly. End of story. Period, Finish. Live long and prosper. |
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05 Nov 12 - 11:16 AM (#3431351) Subject: RE: BS: Monty Hall Problem From: GUEST,Lighter What I said about whims only applies to a decision to switch. A contestant can switch because they understand the odds or just on a "hunch." In other words they can improve their odds by sheer accident and without knowing they've improved them. This has no practical consequences; people often succeed by accident. It just shows trivially that, like the pigeons, you can improve your chances without "knowing" the significance of the revealed goat or fully understanding the situation. As for pigeon intelligence, I'd be interested in hearing any sensible argument that pigeons are really "smarter than people." I'd be more inclined to entertain the *possibility* if a pigeon told me in person and then explained why switching raises the odds to 2 in 3. |
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05 Nov 12 - 12:44 PM (#3431401) Subject: RE: BS: Monty Hall Problem From: John on the Sunset Coast Ligher, "In other words they can improve their odds by sheer accident and without knowing they've improved them." JotSC, "Sometimes you will lose by playing correctly; less often you will win if you play incorrectly." You do not improve your odds by accident, you just get lucky after decreasing the odds. |
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05 Nov 12 - 02:31 PM (#3431432) Subject: RE: BS: Monty Hall Problem From: DMcG As for pigeon intelligence, I'd be interested in hearing any sensible argument that pigeons are really "smarter than people." 'Smarter' is not a simple term: if you mean better at doing a specific task it is one thing ("that was a really smart answer to that question"), but if you mean over a collection of tasks, it means another ("she's a really good architect"), and if you mean the chances of being able to solve some problem never encountered before, it is yet another. There's no doubt that there are some tasks pigeons are better than us at, such as finding your way home from a random place. You could say the same for virtually any creature you like. Even if you restrict this to 'mentally-based tasks', such as learning, a lot of creatures can learn (specific things) better than us. So can pigeons be smarter than us? Yes they can, providing you are clear what precise meaning of 'smarter' you are using. |
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05 Nov 12 - 03:18 PM (#3431451) Subject: RE: BS: Monty Hall Problem From: GUEST,Lighter I think you're playing with words without realizing it. I mean "smart" in a the ordinarily understood, general way. To claim without qualification that pigeons are smarter than people (as some have claimed) is to use the word in a most inclusive sense. Ask a stratified random sample of a million native speakers of English whether pigeons, as a matter of fact, are smarter than people (yes or no), and you'll get nearly a million noes. Plus, quite as significantly, much laughter at the silliness of the question. Those respondents would not be ill-informed. Nor would they be unfamiliar with the ordinary meaning of "smart." Say flatly, "Pigeons are smarter than people" and some would undoubtedly say, "Yes, in some ways," because there are certain human attitudes and practices they don't think very highly of. But I'm not speaking of "some ways." I'm speaking of general intelligence. Pigeons are more efficient at learning to choose doors by trial and error in a setup like "Let's Make a Deal!" They're also better at finding their way home and at flying by flapping. Those abilities don't make them smarter than people, unless, like Humpty Dumpty, you wish to stipulate a special definition of "smart" (for example, "able to outperform people in certain non-intellectual tasks"). |
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05 Nov 12 - 05:04 PM (#3431519) Subject: RE: BS: Monty Hall Problem From: gnu "There's no doubt that there are some tasks pigeons are better than us at, such as finding your way home from a random place." Buddy, I am here ta tell ya, that ain't true. I can pick my way outta anywhere. The ONLY reason I can't do it faster that a pigeon is on accounta I can't fly. If I could, I could knock the feathers off a pigeon in that scrap. Beat it home and have a pot on the boil for it's arrival. Better? I say nay. |
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05 Nov 12 - 05:09 PM (#3431523) Subject: RE: BS: Monty Hall Problem From: DMcG Well, like several other topics raised recently, this is one is distinctly off-thread, so we don't need to pursue it too far. I think the charge of playing with words "without realising it" is a little harsh, though. I have certainly not claimed pigeons are smarter than humans in the ordinary sense of the phrase ... But I do say there are several distinct 'ordinary senses' of the phrase, not just one. You may be aware of the Turing Test for artifical intelligence. Roughly speaking it says if a machine and a human are given the same tasks and another human without direct observation can't say which is which, then we have no grounds for asserting the machine is not intelligent (within the tested realm). I would say that applies to animals versus human as much as machine versus human. Now think of some test for 'smartness in a specific context'. If the machine or animals outperforms the human it seems odd to me to be prepared to use the word 'outperform' but object to 'smarter in that context'. |
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05 Nov 12 - 07:06 PM (#3431570) Subject: RE: BS: Monty Hall Problem From: John on the Sunset Coast "Pigeons are more efficient at learning to choose doors by trial and error in a setup like 'Let's Make a Deal!'" Q. What is the difference between a pigeon and a Hall contestant? A. The pigeon gets to do the task over and over and over. Usually the set-up for pigeons or mice is to do something having only two choices, which are the same all the time, I believe. If there were three chioces could the pigeon learn to always switch? I don't know pigeon behavior; can make complex choices or only simple ones?. But a contestant only gets to play the game one time (tomorrow there is a different contestant), so he/she must play the best odds for getting the car. and there is a rational behavior for that. Switch!!!! |
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05 Nov 12 - 07:24 PM (#3431578) Subject: RE: BS: Monty Hall Problem From: gnu And, let's get one thing straight. Pigeons cannot answer my arguement that each of the remaining two doors in the "second round" had the same initial probability of being a winner in the first round. Truth and outcome tables? Opening a goat door changes the rules and, ergo, changes the entire situation. I contend, yet again, that if one changes the rules and the situation, one changes the application of logical analysis. Food for thought or pigeon feed? |
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06 Nov 12 - 02:01 AM (#3431656) Subject: RE: BS: Monty Hall Problem From: DMcG John: in the normal game a contestant played once, but in the experiment referred to the humans could play as often as they liked. Now, pigeons being greedy and students being ... students I guess it is quite likely the pigeons played more often, but we don't have the data. Gnu: there's a difference between learning something and being able to explain or justify it. Ask any sportsman with a pair of lucky socks ... |
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06 Nov 12 - 02:02 AM (#3431657) Subject: RE: BS: Monty Hall Problem From: GUEST,Jack Sprocket g = goat c = car x = open door W = win without change w = win with change L = lose without change l = lose with change. C = contestant P = presenter. Three possibilities (123): ggc gcg cgg ggc: C chooses 1 P chooses 2 - gxc (forced choice) (i) C sticks - L (ii) C twists - w C chooses 2 P chooses 2 - xgc (forced choice) (i) C sticks - L (ii) C twists - w C chooses 3 (a) P chooses 1 (i) C sticks - W (ii) C twists - l (b) P chooses 2 (i) C sticks - W (ii) C twists - l Total outcomes for ggc: 8 Total wins if Stick (W): 2 Total wins if Twist (w): 2 C must either Stick or Twist so the probability of win is 25%. Repeat for the other two combinations, merely shifting the situation where P has a choice to one of the other positions. Which is in line with what you might intuit: if the door C chooses is opened at stage 1, the probability would be 1/3. Needing to get two guesses in a row right must reduce the probability. Now repeat the exercise where all 8 combinations of g and c are permitted. |
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06 Nov 12 - 10:45 AM (#3431815) Subject: RE: BS: Monty Hall Problem From: John on the Sunset Coast GUEST,Jack Sprocket - Sorry, but your analysis and, therefore, your conclusion are totally wrong. See either or both: Jim Dixon, 29 Oct @ 10:35am TheSnail, 30 Oct @ 4:53pm (a little easier to follow) |
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06 Nov 12 - 02:20 PM (#3431934) Subject: RE: BS: Monty Hall Problem From: gnu DMcG... "Ask any sportsman with a pair of lucky socks ... " Can't agrue with that. I NEVER went huntin without wearin green socks. It MUSTA worked on accounta I got a whack of ruffed grouse in my huntin days. |
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06 Nov 12 - 04:04 PM (#3431995) Subject: RE: BS: Monty Hall Problem From: John on the Sunset Coast Guest,Lighter-- It is good that you don't design games for a living. They'd be boring. Game shows do want to give prizes, else no one would watch, but they don't want to give the best prize every time. If Monty does not know where the car is, he may expose the car on the first round...game over. Big whoop! But he wants to generate excitement so he opens a door with a goat. It is important to understand that the door he exposes will not reveal the car, and it will not be the door you choose on that round. The door he shows is eliminated, and you go to the second part of the game. That is the game, and those are the rules. Monty has to give you information or there is no game, only pure chance. Gnu-- We agree, I think, on the first round, your chance to of picking a goat is 2/3, picking the car is 1/3. On the second round, if you stay with your door, you still given yourself only a 1/3 overall because you took the same available door twice. But by making a change, you have now given yourself 2/3 chances to win the car; that is you chose two different of the three doors available in the game, thereby you have doubled your chance of finding the car. The only times this strategy fails (for the umpteenth time) is when you pick the car's door on the first round, but that chance was only 1/3. |
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06 Nov 12 - 04:16 PM (#3431998) Subject: RE: BS: Monty Hall Problem From: gnu "The only times this strategy fails..." Well, there ya go, eh? Change the situation, change the rules, change the choice. |
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06 Nov 12 - 04:58 PM (#3432016) Subject: RE: BS: Monty Hall Problem From: GUEST,Lighter > Monty has to give you information or there is no game, only pure chance. Correct. However, if you don't appreciate the information (and the controversy here and elsewhere shows that few contestants could have), and switch on a whim, and boost your odds accidentally, that part *is* pure chance, at least as I understand chance. Or, equally, you could stay on a whim: a "whim" implies no consequential thought or deliberation. And, win or lose, you would never know whether your decision had boosted your odds or not. All you would have is a goat or an automobile or Monty's proffered cash (which complicates your strategy to come away with something valuable, based on the principle of the bird in the hand). |
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07 Nov 12 - 02:27 AM (#3432285) Subject: RE: BS: Monty Hall Problem From: DMcG And, win or lose, you would never know whether your decision had boosted your odds or not. I'm not sure follow you there, Lighter. By definition the odds are 'if you played the game oodles of times what proportion would you win'. You know that switching improves your odds every time. An individual game on the other hand wins or loses but does not affect the odds: that's true of any game without an 'end-stop' (For example, actual gambling games rarely allow you to continue once you have run out of cash/shirts) |
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07 Nov 12 - 07:57 AM (#3432362) Subject: RE: BS: Monty Hall Problem From: Wolfgang Problems with conditioned probabilities are among the most difficult to grasp. A 90 minutes lecture is usually not enough to teach them sufficiently. Even one of my heroes, the great Martin Gardner, once got another variant ("one of my two children is a son") wrong on the pages of Sci. Am. I have since long given up using intuition for such problems. I simply use Bayes' formula to arrive at the sometimes very surprising correct solution. The verbal formulation of the problem is often underspecified (more formally, the "event space" is not stated unambiguously) and therefore, there is sometimes more than one correct solution. However, even if the problem is described in a way that has only one correct solution, often people don't see it. Two easy extremes: (1) If Monty knows where the prize is and is always mean, and only gives you, after opening a door without the big prize, a new choice in the case your first choice happened to be correct, the probability of winning by switching is nil. (2) If Monty knows where the prize is and is always benevolent, and only gives you, after opening a door without the big prize, a new choice in the case your first choice happened to be wrong, the probability of winning by switching is 1. The two more common solutions, one of which is counterintuitive. (1) Monty knows where the prize is, and always (a) opens a door without the big prize and (b) gives you the choice to switch, then you should switch (counterintuitively), for the probability of winning then is 2/3. (2) Monty himself doesn't know where the prize is and (a) always opens one of the two doors that were not your first choice and (b) in those cases in which the big prize was not behind the now open door always offers you the choice to switch, the probability of winning (whatever you do) is 50%, the intuitive solution. Since you don't know what Monty knows you should switch unless you think he is mean, for even in the worst case, your probability of winning by switching does not increase. Of course, in Monty's case, one can retrospectively find the correct solution the empirical way. Someone has done that and has looked at all available videos of the show. The clear result: the majority of the people has not switched, but in the majority of the shows switching would have been better. However, across all shows the winning probability for switchers was less than 2/3 (though of course higher than 50%) which clearly hints that sometimes Monty actually was mean spirited. Learning by doing it repeatedly is not so successful, by the way, as someone has posted. After 50 repetitions (in the switching wins in 2/3 of the cases variant) still half of the participants in an experiment did not switch, though the difference between 2/3 and 1/2 should be obvious after so many repetitions. I once found a way to make the majority of the participants switch in an experiment even at the first opportunity by increasing the number of "doors" to ten. Actually, the student in her diploma thesis used ten upside-down cups under only one of which was a prize. After the first choice, she opened eight of the ten cups but never the cup with the prize (the participants were fully informed about the procedure). Then she offered the participants to switch and most of them did (but only a tiny minority in the three cups control). The probability of winning for switchers in the ten cups condition was 9/10, of course. Though most of the participants did switch, when asked about the probability of winning by switching, even most of the switchers said stubbornly 50%, same as in the three cups control condition. (So much for intuition about why we do what we do.) As I said above, most of the conditioned probability problems suffer from an insufficiently specified event space. A last example: If someone throws two dice in a dice box (you cannot look at the result) and then takes out one of the two dice from under the box showing a 6, what is the probability that the other dice also shows a 6? The intuitive response, 1/6, is only correct, if the other guy has not looked at the dices before taking one out from under the box, that is if the probability for taking a dice showing a 6 was not higher then chance. If however, your opponent had a look under the dice box before taking the dice out that shows a 6 and will always show you that one of the two dices has a 6 whenever he can (like in a well known German game of dice, in which the double 6 is the second highest throw), then the probability that the other dice also shows a 6 is 1/11 (hint for those that do not use Bayes' theorem to find the solution: the probability of a double 6 is 1/36, the probability for any of the other five combination with one 6 is 1/18) . It pays to know that when playing this game. Wolfgang |
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07 Nov 12 - 11:22 AM (#3432474) Subject: RE: BS: Monty Hall Problem From: John on the Sunset Coast Wolfgang-- You wrote, "The two more common solutions, one of which is counterintuitive. (1) Monty knows where the prize is, and always (a) opens a door without the big prize and (b) gives you the choice to switch, then you should switch (counterintuitively), for the probability of winning then is 2/3." This is basically correct, but not complete...Monte will will never reveal your Goat Door under any circumstance, or the game is instantly over. Not an exciting game for a television audience. I don't know if that means Monte is mean or benevolent as he always acts same manner. You also wrote that the actual game has been retrospectively analyzed, and the results showed that results did not match exactly the expected probabilities, no matter the strategy the contestants followed. This, of course, is to be expected because each game is a new event wherein even the least probable outcome can occur. Since there were hundreds of games played, it would be interesting to know how close outcomes came to expected probability throughout the run of the show. |
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07 Nov 12 - 12:16 PM (#3432501) Subject: RE: BS: Monty Hall Problem From: Murray MacLeod I have to say that I am totally amazed by gnu's inability to grasp the essence of this problem. I know from his postings on Mudcat over the years that he is intelligent, and has a well developed grasp of maths and engineering, so it is bizarre that an elementary probability problem such as the Monty Hall problem continues to elude him. Then again, I got to grips with probability theory at a very young age, and since doing so have been aware that the vast majority of people have difficulty in reconciling the mathematical facts with their initial intuitive assessment. You can test this by asking the average person, "How many people would you have to assemble in one room to make it more likely than not that at least two of them share the same birthday ?" Once you understand factorials, the answer is devastatingly simple, but approached intuitively, the solution seems incredible. ( The answer is 23 btw) |
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07 Nov 12 - 09:53 PM (#3432806) Subject: RE: BS: Monty Hall Problem From: Mysha Hi Jack, Yes, I was wondering when someone would come up with that one. You're right that there are four possible outcomes for a given distribution of a car and two goats, but they're not equally likely. The first two choices for the presenter were forced, so they will always occur for that particular choice of the contestant. The other two, however, are true choices, and both appear for the same choice of the contestant. This means that those latter two outcomes together are as likely as one of the other outcomes alone. For the sake of creating equality, let's assume Mr. Hall flips a coin once the contestant chooses a door. If he has a real choice, for heads he'll take the leftmost choice, and for tails he'll take the rightmost choice, which is still a random choice as it's by coin toss. The fact that he tosses a coin splits the outcome of each forced choice up into two outcomes, but they remain the same forced choices as the flipping has no influence on them. (I'll use your term "Twist", rather than the "switch" this thread otherwise uses.) ggc: C chooses 1 (a) P tosses: heads P chooses 2 - gxc (forced choice) (i) C sticks - L (ii) C twists - w (b) P tosses: tails P chooses 2 - gxc (forced choice) (i) C sticks - L (ii) C twists - w C chooses 2 (a) P tosses: heads P chooses 2 - xgc (forced choice) (i) C sticks - L (ii) C twists - w (b) P tosses: tails P chooses 2 - xgc (forced choice) (i) C sticks - L (ii) C twists - w C chooses 3 (a) P tosses: heads P chooses 1 (i) C sticks - W (ii) C twists - l (b) P tosses: tails P chooses 2 (i) C sticks - W (ii) C twists - l Total outcomes for ggc: 12 Total wins if Stick (W): 2, out of 6 Total wins if Twist (w): 4, out of 6 C must either Stick or Twist so the overall probability of win is 6 out of 12: 50%. C's optimal choice is to Twist, whose probability of win is 4 out of 6: 67%. Analogue for the other two starting positions. Wolfgang: I suspect that the switchers in the ten cups experiment are a result of the last-man-standing effect, rather of people seeing how the cups not chosen would effect their odds. Bye, Mysha |
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08 Nov 12 - 08:30 AM (#3432962) Subject: RE: BS: Monty Hall Problem From: GUEST,Lighter DMcG, I'm speaking of one's own odds of finding the car. You would only know that you had boosted your odds if you understood the calculation of the odds and the value of switching. Most contestants could not have known this. Indeed, part of the fun of the game was the excitement generated by the belief that sticking or switching was based entirely on "intuition" or a straight 50/50 chance. As I may have said before, part of what makes the 2/3 odds counterintuitive is that the two remaining doors are visible and present. The odds are an intangible mathematical abstraction, which in this case seem more "real" (to those unable to grasp them correctly) than the evidence of one's own untutored eyes. ("Two doors, 50/50.") It's as though the odds reach down from some rarified plane of existence to affect those of us below - though, as you say, the "effect" is notional rather than material (they don't turn possible failure into certain success, they only nudge it in that direction). ...if I'm making myself clear. |
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08 Nov 12 - 11:57 AM (#3433077) Subject: RE: BS: Monty Hall Problem From: John on the Sunset Coast Lighter, you cannot boost your odds (or as I prefer, probability), you can only use them to advantage. Of course, to do that you must understand what the probabilities are. Or you can ignore them to your thereby disadvantaging yourself. I'm not sure I understand what you are saying in paragraph two. I think you mean the math seems more UNREAL than the evidence of your own eyes. If so, I agree that the odds seem counter intuitive in an actual game situation (mostly because of the excitement and psychology of the moment) causing a contestant to make disadvantageous decisions. Also, they do not understand that the rules of the game clearly take it out of the realm of pure chance. I can understand folks on stage having no, preparation not understanding the best strategy. What amazes me that gnu and some others don't grasp the strategy even after the posting of several grids, and narrative descriptions of the best play which they can analyze at leisure. The only reason I can think of for this: Since the best strategy dos not guarantee a win, it seems no better than any other strategy. But the numbers are the numbers, and that makes those folks wrong. Jack's analysis is wrong for at least two reasons. P will never choose the same door as C, which he indicates could happen ["C chooses 2/P chooses 2-gxc (forced choice)]. Before that, he states there are "Three possibilities (123): ggc gcg cgg" But each 'g' is a separate entity which needs to be accounted for. If he differentiates them by some means (for example G1, G2) he will easily see there are six possibilities for ordering the prizes behind the doors. If the premises are wrong, the analysis is wrong, the conclusion is wrong--GIGO. And Mysha, who seemingly agrees with Jack (but not quite), is also wrong. He has made up his his own game where Hall/Host/Presenter must flip a coin. Since it is not THE game, it is moot. |
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08 Nov 12 - 12:19 PM (#3433099) Subject: RE: BS: Monty Hall Problem From: DMcG Well, Lighter you are getting into really difficult areas there, because you are moving out of the realm of mathematics and into psychology, judgement and ad-hoc generalisations. It is indeed interesting but includes such questions as why anyone ever buys a lottery ticket, when the odds of them winning are so minute, and how this compares to buying a raffle ticket for your fok club or favourite charity, when winning is pretty much irrelevant to your motivation. Complex and intriging but nothing to do with Monty, except insofar as that is an example |
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08 Nov 12 - 01:19 PM (#3433141) Subject: RE: BS: Monty Hall Problem From: GUEST,Lighter Agreed. The point is that the mathematical probability, when not truly obvious (e.g., head or tails on one toss, 1 in 2), can seem not only counter-intuitive but also almost uncanny. |
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08 Nov 12 - 03:58 PM (#3433239) Subject: RE: BS: Monty Hall Problem From: gnu Murray... "I have to say that I am totally amazed by gnu's inability to grasp the essence of this problem." Likewise, I am sure. You are talking about probabilities based on a simple ruse... offering two choices, the first of which DOES NOT COUNT and therefore CANNOT influence determination of subsequent probabilities. You can think it does, you can "prove" it does, but it simply cannot. Allow me, again... just one last time... THERE ARE ONLY TWO DOORS! Now, IF Monty doesn't know where the car is, there ARE three doors. But, Monty KNOWS where the car is, so truth and probability tables and fancy theorems based on there being three doors (FALSE!) are inapplicable. You cannot apply conditions and analyses from the first choice to the second choice because they are inherently different... there are only two doors. Anyone who thinks there are three doors should be locked in a dark room with a small black and white TV with poor reception and forced to watch EVERY Monty Hall car-goat segment ever aired over and over and over and... Ya know, if I ever get a chance to win that car, I am gonna switch... and if I get the goat, I am gonna tie it up in front of yer house. Don't worry... there is only a 50/50 chance of that happening. |
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08 Nov 12 - 05:44 PM (#3433309) Subject: RE: BS: Monty Hall Problem From: GUEST,Lighter Gnu, I feel your pain. The key is that there's a distinction between the odds of where the car *is* (1 in 2 for either of the two final doors, regardless) AND the odds that you, the contestant, will *find* the car. It's too easy for a non-mathematician (like me) to confuse the two. You're thinking of where the car *is* in the final setup. It's got to be behind one of the two doors (odds for either door, 1 in 2). But the real issue is the odds of your *finding* the car when there were three doors, Monty knowingly eliminated one, and you get to stick or switch. A more complicated situation (odds if you switch, 2 in 3). Remember, the 2 in 3 probability is still no guarantee that you'll win the car. It isn't anything tangible like a door or a goat. You just have a better chance of finding the car if you switch. It still feels weird, but I'm now convinced. It's almost as though something spooky must be migrating from your first choice to your second, but that's not what's happening at all. All that changes is the intangible mathematical relationship among the choices. You can't detect it without having worked out the possible results one by one. The mathematical relationship is real even if not detectable by straight observation. Check out Jim Dixon's chart last week of all possible results. It will take a while to absorb all the permutations, but the 2 in 3 odds are right there. |
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08 Nov 12 - 06:18 PM (#3433326) Subject: RE: BS: Monty Hall Problem From: TheSnail GUEST,Lighter The key is that there's a distinction between the odds of where the car *is* (1 in 2 for either of the two final doors, regardless) AND the odds that you, the contestant, will *find* the car. I think I'm going to cry. |
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08 Nov 12 - 06:27 PM (#3433331) Subject: RE: BS: Monty Hall Problem From: DMcG Since something I said earlier is likely to be dragged back out, remember I also pointed out that the phrase 'the probability where the car is' is an informal gloss, and is better expressed as 'the probability of finding the car given no information beyond that immediately observable'. It is not really possible to talk about the probability of where a car is independently of the information you have access to. |
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08 Nov 12 - 06:53 PM (#3433338) Subject: RE: BS: Monty Hall Problem From: DMcG Let me flesh that out with a different example. In dealing with probability outside mathematical notation, it is very common to omit the information bit. People will say that the odds of heads or tails is 50:50. In fact, that's wrong. What you need to say is (statement:) the odds are 50:50 (information:) if you are using a fair coin. Without the fair coin information, the odds are not clear. Suppose I show you a coin and, having told you it is fair, I then tell you the last 100 tosses were all heads. Then, if you accept my claim of fiarness the odds remain at 50:50. Now suppose I had told you I had doubts about the fairness of the coin and the last 100 were all heads. Then is is reasonable for you to also have doubts that the odds are truly 50:50 and you would be well advised to bet the 101st head turns up. |
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08 Nov 12 - 07:01 PM (#3433342) Subject: RE: BS: Monty Hall Problem From: gnu "It is not really possible to talk about the probability of where a car is independently of the information you have access to." Correct. Someone tell me that is incorrect and I'll tell you... In the second round, the ONLY information you have is that it is behind ONE of TWO doors. There are only TWO doors. Monty has negated your "probabilities" by the elimination of one door... it ALWAYS was happens because it's a TV show! Your analyses, predictions, probabilities, and predictions are based on false assumptions because they do not apply to the ACTUAL situation in the second round. Yeah, I am just a good ol Kent County boy. Yeah, I took philosophy at uni. Yeah, I hold a Master of Science degree in engineering. Yeah, I am ALWAYS a guy who will accept when he is wrong and apologize... even done it at times just to quell the waters, so to speak, but THERE ARE ONLY TWO DOORS. Please... could ANY of you explain to me why you think there are three doors? |
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08 Nov 12 - 07:29 PM (#3433362) Subject: RE: BS: Monty Hall Problem From: John on the Sunset Coast gnu-- I find hard to believe that a person trained in philosophy who has a second level degree in engineering is not pulling our collective chains by stubbornly asserting that this is a two door problem. |
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08 Nov 12 - 08:21 PM (#3433392) Subject: RE: BS: Monty Hall Problem From: GUEST,Lighter The two concepts "where it physically is " and "how likely I am to find it" seem to be sufficiently distinct. If Snail can stop crying, maybe he can explain why no difference exists. It's a car, not an electron. |
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08 Nov 12 - 10:51 PM (#3433440) Subject: RE: BS: Monty Hall Problem From: John on the Sunset Coast Gnu asks, "Please... could ANY of you explain to me why you think there are three doors?" I'll try. It is asserted on the first round that there are three doors...because there are. Whether the car is won by the contestant (you) is determined by two separate, but connected actions. The first action is picking a door. Your chance of picking the door with the car is 1/3. Since there are two goats available, the chance of picking a goat is 2/3. You don't win the car on the first round if you chose that door, but neither do you lose if you didn't choose that door. By rule, you will never be shown the car's door; you will be shown a goat's door; neither will you be shown the goat's door you chose. The door exposed is then taken away. You need a second action to complete the game. Nobody is asserting there are three doors on the second round. There are two doors left. These are the same two doors that were part of the available three doors in the first round. They are in the same relative position, and the prizes are still behind the doors they were behind in the first round. The prizes are always one goat and the car. [If the car were eliminated, you would have no chance to win the car, but Monty has showed you that it is still available, because he exposed one of the goats.] The action you need to take in the second round is to choose a door. Your choice can be your original choice, in which case you have kept your chance of winning at 1/3, because you chose only one door out of the three that were available to you for the complete game. You have simply made the same choice twice! But if you avail yourself of the chance to switch, you have then chosen 2 out of the 3 doors that were available at the beginning of the game, thereby increasing your chances to snag a car to 2/3!! BTW, you may have a degree in Philosophy, and a Masters in Engineering, but the scion of JotSC has a PhD in Philosophy, emphasis in Logic and Linguistics. He has read this explanation, and the grid of TheSnail; he agrees with both. Actually, he wonders why the grid is not sufficient, in and of itself, to convince you of the proper probabilities and outcomes. |
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08 Nov 12 - 11:13 PM (#3433450) Subject: RE: BS: Monty Hall Problem From: GUEST,Lighter > But if you avail yourself of the chance to switch, you have then chosen 2 out of the 3 doors that were available at the beginning of the game, thereby increasing your chances to snag a car to 2/3!! This is another counter-intuitive aspect for the following reason. If I hold two lottery tickets out of a total of three, my chances of winning are likewise 2 in 3. In that case, however, I have physical control over both tickets at once. I can see with my own eyes that I have two of the three chances. In Monty's game, by way of contrast, I am *giving up* my claim on the first door I select when I switch to another. It feels the same as tearing up a ticket. Giving up a claim on the first door would thus seem by everyday reasoning to make a difference in the final calculation; but in fact it does not. |
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09 Nov 12 - 02:17 AM (#3433470) Subject: RE: BS: Monty Hall Problem From: John on the Sunset Coast Lighter -- Forget about things being counter-intuitive; that only means that you haven't considered all of the components of the problem. When I said that you were choosing 2 of the 3 doors, that was over two rounds, and with the knowledge that the car was still available on the second round. That is not what your lottery scenario implies, and it therefore has no analogy to the Hall Problem. You have been given the components you need fortyteen different ways by narrative and by grid with the correct analysis. You seem to be trying to reinvent the wheel instead of trying to understand the Game under the rules followed in the Hall Problem. There are six possible outcomes in the Hall problem. Four of those outcomes result in getting the car if you switch at round two. Two of those outcomes result in not getting the car. There is no way to change that. If you don't ever switch, you can only win in two outcomes. And if you stay or switch on a whim you only win half the time. So the order of descending likelihood of winning the are are: Always switch -- 2/3 wins Alternately switch or stay -- 1/2 wins Never switch -- 1/3 wins I believe someone(s) has posted all of this during the last week in one form or another, and it is really easy to set-up you own grid to prove it to yourself. |
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09 Nov 12 - 06:33 AM (#3433540) Subject: RE: BS: Monty Hall Problem From: TheSnail Lighter The two concepts "where it physically is " and "how likely I am to find it" seem to be sufficiently distinct. Quite right but that isn't what you said that reduced me to tears. What you said (with my emphasis) was The key is that there's a distinction between the odds of where the car *is* (1 in 2 for either of the two final doors, regardless) AND the odds that you, the contestant, will *find* the car. Physically, the car is behind one, and only one, of the doors. It is not fifty percent behind one and fifty percent behind the other. The trouble is, we don't know which. That's where the odds come in. From the information available, we can calculate the probabilities. For your late comer, there are two doors with a car behind one and a goat behind the other so the 50:50 option is the best he can do. But, we have more information than that so, if we analyse it properly, we can calculate more accurate probabilities. In this case, that the odds are 1/3 for the door the contestant chose and 2/3 for the one that Monty didn't open. The 50:50 odds have now gone. In the first part of your distinction above, you seem to think that the odds have some sort of physical reality. They do not. They only exist in our heads as a result of the analysis we have made of the information. As a bit of an aside. It might help you to think, not of the door that Monty chose to open, but of the one he chose not to open. In two situations out of three, depending on the contestants first choice, he chose not to open it because it had the car behind it. |
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09 Nov 12 - 08:45 AM (#3433593) Subject: RE: BS: Monty Hall Problem From: GUEST,Lighter > Forget about things being counter-intuitive; that only means that you haven't considered all of the components of the problem. It also means that the correct solution is inconsistent with everyday experience. Its counter-intuitiveness is the chief reason that the problem generates so much controversy. That makes it of interest. For decision-makers, the psychology is just as important as the math. So the analogy to the lottery situation is entirely relevant. It partly explains why the Hall problem *is* so counter-intuitive that a vast number of people (according to Wikipedia) angrily refuse to believe the correct solution even after it's been explained to them. That's a very unusual response, even when something as bizarre as advanced physics is being explained, Quantum mechanics makes most people shake their head and wonder how it can be: but they don't shout that the numbers are wrong and that the physicists are obviously idiots. I find that difference fascinating and possibly very important. One reason for it is that Joe Blow has never peered into an atom but he certainly has had to make choices between two or three options. > You seem to be trying to reinvent the wheel instead of trying to understand the Game under the rules followed in the Hall Problem. I'm trying to grasp fully the implications of the wheel and of the rules themselves. Jim Dixon's chart/grid/summary shows that the 2 in 3 outcome is correct. I've pointed this out several times. But that doesn't make it any less counter-intuitive, for reasons I've already explained. Those reasons are of interest. |
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09 Nov 12 - 03:48 PM (#3433808) Subject: RE: BS: Monty Hall Problem From: gnu Lighter.. "The action you need to take in the second round is to choose a door." Yup. And there are two doors available. ONE choice between TWO doors (options). The fact that there were three doors at the start simply cannot influence ONE choice between TWO options in the second round. If you stick, that is choosing one option. If you switch, that is choosing the other option. Basic, nitty-gritty of philosophy... the base of all human analyses of logical thought... A OR B. Positive or negative. Night or day. Where the arguements proffered fail MY logic is that they fail to equate switching and sticking as ONE choice between TWO options. Oh... re me having a go? Having "fun" by sticking to my arguement... Nope, not a hope. I am quite serious. And, until someone shows me "different", I will stick. When anyone can prove me wrong, I shall buy them a pint and apologize profusely. Then again, I bought a Looto Max (yes, typo intended) ticket today for the $40M prize tonight and that is quite illogical. >;-) |
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09 Nov 12 - 04:42 PM (#3433830) Subject: RE: BS: Monty Hall Problem From: Nigel Parsons Gnu: Yup. And there are two doors available. ONE choice between TWO doors (options). The fact that there were three doors at the start simply cannot influence ONE choice between TWO options in the second round. Yes, you have a choice between two doors, but not two identical doors. What differentiates them? A) On has a car behind it the other has a goat. Unfortunately, we cannot know for sure which is which (yet) B) One of the doors was selected earlier, at which time we knew there was only a one in three chance of it being the one with the car. If Monty had left you with a choice of three doors (the one he has shown to contain a goat is not removed) how would you decide? Your choice is still 1/3 now, so does the car still have an equal chance of being behind each door? Your previously chosen door still has a 1/3 chance of hiding the car. You know that the door Monty opened has a 0% chance. Is the chance if you switch 1/3 (the same as your chosen door)? If so you have three options which only add up to 2/3. Should the odds on both the remaining doors have increased, or has Monty given you enough information to increase the likelihood that the unchosen door hides the car? |
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09 Nov 12 - 06:04 PM (#3433854) Subject: RE: BS: Monty Hall Problem From: GUEST,Rev Bayes >>Yup. And there are two doors available. ONE choice between TWO doors (options). The fact that there were three doors at the start simply cannot influence ONE choice between TWO options in the second round. When you first pick a door, you have a 1/3 chance of picking the car. Yes? If I now open a goaty door, the probability is still 1/3 I have a car. Yes? So if I have two options, and the option I've got is 1/3, the other must be 2/3. Yes? |
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09 Nov 12 - 06:05 PM (#3433855) Subject: RE: BS: Monty Hall Problem From: gnu Nigel... "If Monty had left you with a choice of three doors (the one he has shown to contain a goat is not removed) how would you decide?" Ummm... I guess I wouldn't pick the goat? I kinda thought that would be obvious, no? Did I miss something? Maybe I did, on accounta that's as far as I read your post. Ya kinda lost me there so maybe you can explain that before I read any further. Fair enough? BTW... I may not entertain any more discussions regarding logic this eve as I have old friends dropping by. Alfred, Tommy, and the Three Amigos... me, moi and Jimmy Suis. We shall play tunes and sing as best we can and solve the world's problems. Too bad Monty Hall isn't joining as I would put his goat on the BBQ and make hime eat the whole fuckin thing. So, I was at the gym today and a gorgeous young thing started working out next to me. I asked my trainer what machine I should use to try to impress her and he replied, "The ATM in the lobby." You ain't wrong. I ain't wrong. I understand that. Few do. No matter where this goes or ends up, I just want to emphasize, NO, I ain't shittin. I truly believe I am correct in my assumptions and my analysis. The ONLY thing that bothers me (and it bothers me a lot... A LOT!!!) is that some have posted that I am having a lark... teasing... whatever... being a... it's hard for me to even type it... a troll. Again, sorry for the consternation... if I am wrong. |
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09 Nov 12 - 07:24 PM (#3433895) Subject: RE: BS: Monty Hall Problem From: John on the Sunset Coast "The fact that there were three doors at the start simply cannot influence ONE choice between TWO options in the second round." Gnu, you can tell yourself that until the day you die, and you will be just as wrong on that day as you are now. You either don't know the rules of the Hall game, or you are pretending not to know them. I have my suspicion which it is. Shame on us for humoring you. At any rate, I think Monte Hall would get a kick if he knew we have devoted so much bandwidth discussing his long ago game |
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09 Nov 12 - 07:41 PM (#3433900) Subject: RE: BS: Monty Hall Problem From: gnu "You either don't know the rules of the Hall game, or you are pretending not to know them." Let me explain the rules as I see them... AGAIN! Three doors. Contestant chooses one. Monty opens another door behind which is a goat. Monty asks contestant to pick either of the two doors that Monty has not opened. Are those the rules? Yes or no? |
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09 Nov 12 - 08:03 PM (#3433906) Subject: RE: BS: Monty Hall Problem From: John on the Sunset Coast No. For the reasons why, read some of the correct analyses posted over the past week. |
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10 Nov 12 - 02:51 AM (#3434025) Subject: RE: BS: Monty Hall Problem From: DMcG At any rate, I think Monte Hall would get a kick if he knew we have devoted so much bandwidth discussing his long ago game I'm sure you are right. But as I see it, the issue isn't really the game, and has, I suppose, been what Lighter has been concerned about for the last third of the thread or so. It is how it comes about that we as a species are so poor when it comes to dealing with probabilities and since the whole of life is highly dependant on making such judgements to what extent we can improve of this. Now, it is of course possible in some circumstances to go away and think about it for a week and carry out formal analyses. But in most cases we have a second or two to make up our mind. Do we think anything can be done to improve our decision making in such areas? My guess, is that the only way is to become aware how fallible we are in such circumstances and try to hold our judgement very lightly, being prepared to drop them when we have time to think things through. I suspect gnu is either teasing us or, despite his protestations otherwise, finds it difficult to change from his original opinion. For example, gnu has not responded to the post from BK Lick @ 01 Nov 12 - 04:17 AM as far as I can tell. That in its turn was a variation of Jim Dixon's post of 29 Oct 12 - 10:53 AM which also seems to have been overlooked. |
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10 Nov 12 - 04:05 AM (#3434031) Subject: RE: BS: Monty Hall Problem From: BK Lick "Three doors. Contestant chooses one. Monty opens another door behind which is a goat. Monty asks contestant to pick either of the two doors that Monty has not opened. "Are those the rules? Yes or no?" Yes, those are the rules. I cannot fathom why JotSC said they are not. |
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10 Nov 12 - 08:38 AM (#3434110) Subject: RE: BS: Monty Hall Problem From: GUEST,Lighter Thanks, DMcG. That's part of it. But beyond the deceptiveness of such situations is a second issue: why the correct solution is not just counter-intuitive to so many people, but also infuriating (and possibly a little scary). |
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10 Nov 12 - 09:12 AM (#3434123) Subject: RE: BS: Monty Hall Problem From: John on the Sunset Coast "Yes, those are the rules. I cannot fathom why JotSC said they are not." Perhaps my original answer was too hasty or too flippant. Perhaps I should have written, 'Yes, but...'. The "but" or "buts" have been noted several times in the past week. |
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10 Nov 12 - 11:00 AM (#3434171) Subject: RE: BS: Monty Hall Problem From: BK Lick What gnu needs is clarification, not mystification. |
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10 Nov 12 - 11:29 AM (#3434194) Subject: RE: BS: Monty Hall Problem From: DMcG Why is it infuriating? All I can do is make wild guesses with not a hint of evidence. But with that caveat I would suggest it is because we set up all sorts of patterns in our brain/mind/suppositions and in most cases these work well. As they need for us to survive. Others are relatively little used so we don't mind to much as the are deeply trotton paths. For most of us whatever we know of quantum physics so we do not have any dependance on it. As a result if we are wrong it is not too painful. On the other hand 'which judgement is best?' Is the kind of decision we might make dozens of times of a day. So if our basis is shown to be faulty is is almost literally painful because we must doubt many decisions of our life. That's my guess, I don't doubt others may have better ideas |
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10 Nov 12 - 11:37 AM (#3434200) Subject: RE: BS: Monty Hall Problem From: DMcG Apologies for all the typos. That's what happens when I use my phone 6or this :( |
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10 Nov 12 - 05:31 PM (#3434399) Subject: RE: BS: Monty Hall Problem From: gnu DMcG... "I suspect gnu is either teasing us or, despite his protestations otherwise, finds it difficult to change from his original opinion. For example, gnu has not responded to the post from BK Lick @ 01 Nov 12 - 04:17 AM as far as I can tell.rom his original opinion." Addressed, quite adequately and VERY truthfully I might add, except for BK Lick's post. I dunno why you think I am supposed to respond to that post. It seems obvious that he addressed it to you. Did you respond? If not, why not? After all, this IS a discussion about logic and reason and truth. What you write is what you get. BK... "What gnu needs is clarification, not mystification." You gotta be shittin me. JotSC, re his response re "rules".... "Perhaps I should have written, 'Yes, but...'. The "but" or "buts" have been noted several times in the past week." No shit eh? "But" just don't cut the grass. Facts cuts the grass. Those ARE the rules. PERIOD. FULL STOP. NObody but me seems to fully grasp this simple ruse. Some even think I am mystified and some cast aspersions upon my intelligence and, far worse, upon my character. Yup, I AM mystified. But it's not because *I* don't understand a simple parlour game invented far before that great Canuck, Monty Hall, was even born, in order to entertain children. There are only two doors. Ask Monty. |
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10 Nov 12 - 06:09 PM (#3434422) Subject: RE: BS: Monty Hall Problem From: DMcG I said: gnu has not responded to the post from BK Lick @ 01 Nov 12 - 04:17 AM as far as I can tell. gnu said: I dunno why you think I am supposed to respond to that post. It seems obvious that he addressed it to you. --- Well, since it starts with "Gnu is very persistent -- I like that in a bloke. Against all odds, I believe I may have devised a thought experiment that might actually convince him to reexamine his so firmly held conviction" and finishes with "Will Gnu still insist that either one of these cards has a 50-50 chance of being the Ace of Spades? Or will he see that the chance of his original card being the Ace of spades is only 1 in 52, while the chance of the other card being the Ace of Spades is 51 in 52?", it certainly seems to invite a response from 'gnu'. In fact if you look at the original it mentions 'gnu' four times, and sadly fails to mention DMcG at all. So that's why I thought you might wish to respond. If you think it will remove ambiguity, I will invite you to respond. I have to say, though, that I do not expect you to answer the question about what you believe the odds to be in the card game BK Lick describes. |
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10 Nov 12 - 06:49 PM (#3434446) Subject: RE: BS: Monty Hall Problem From: gnu "it certainly seems to invite a response from 'gnu'." It certainly does not. Read the question and prove otherwise. "Will gnu..." was the question. If the question is CLEARLY not addressed to me, how can you claim that my lack of response in any way credits your false arguements? I am growing weary of this. Anyone care to offer any real arguements? There are only two doors. It's a parlour trick... a ruse. |
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11 Nov 12 - 12:36 AM (#3434592) Subject: RE: BS: Monty Hall Problem From: BK Lick Dear Gnu -- I'm sad that you misunderstood my remark that you need clarification, not mystification. You had correctly restated the rules and asked for confirmation. JotSC then stated that you had got the rules wrong for some unstated, mysterious reason. I was saying that JotSC was mystifying, instead of clarifying, the situation in his response to you. |
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11 Nov 12 - 03:20 AM (#3434614) Subject: RE: BS: Monty Hall Problem From: DMcG I am growing weary of this. Aren't we all? I explicitly invited you answer the question about the odds on the card game in my last post and predicted you would not. Sure enough, that's what happened. Now its up to other people, not you or I, to decide why you avoid answering. |
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11 Nov 12 - 12:59 PM (#3434803) Subject: RE: BS: Monty Hall Problem From: John on the Sunset Coast BKLick wrote, "I was saying that JotSC was mystifying, instead of clarifying..." That is an idiotic statement. You may not agree with what I answered, but I referred gnu back to ten days worth of explanations, the re-reading of which might help him in clarifying his understanding of the problem. What's mystifying is that you used the word mystifying. BTW, it is nice of you to take up for gnu...I know him to be the shy, incoherent, retiring type who never speaks up for himself. |
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11 Nov 12 - 02:20 PM (#3434844) Subject: RE: BS: Monty Hall Problem From: GUEST,Rev Bayes >>Those ARE the rules. PERIOD. FULL STOP. Actually, no, you got them (subtly) wrong. After you have made your initial choice, you do not pick between two doors. You choose to stick or switch. Yes, the point is slightly semantic, but this seems to be where you're going wrong. It is *because* you can stick to the door you originally chose that the probabilities are not equal - because the probability you get a car to start with is 1/3. |
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11 Nov 12 - 06:00 PM (#3434927) Subject: RE: BS: Monty Hall Problem From: gnu DMcG... card game? What card game? I thought we were talking about Monty and goats, no? Why would you talk about card games and not about goats? I say there are two doors. Others still tell me I do not undertsand that there are three doors. I asked you all to ask Monty. I think you should. Oh, BTW... DMcG... I don't care if you wanna change the subject. If you do wish to do so, start another thread. gnightgnu |
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11 Nov 12 - 06:41 PM (#3434941) Subject: RE: BS: Monty Hall Problem From: DMcG Still not answering then gnu? I'm not changing the subject. There are a whole series of games, played with 3 doors and showing 1; 4 doors and showing 2; 5 doors and showing 3 and all the up to and beyond using 52 doors and showing 50, which is isomorphic to the deck of cards example. Whatever logic applies to the 52 applies to 51, 50, 49 ... all the way down to 3 doors and showing 1. All that adding extra doors does is make it more and more obvious that your argument is flawed. Which I am convinced you already know. Hence the somewhat wild attempts to say a question in a format commonly used in real-life and online forums is not one you were invited to answer, or that the logic differs whether we use a goat or a playing card. So, no, thanks for the invitation, but while we are discussing Monty and its implications I will stay here. |
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11 Nov 12 - 09:46 PM (#3434996) Subject: RE: BS: Monty Hall Problem From: BK Lick JotSC: "What's mystifying is that you used the word mystifying." Well,let me try to demystify that. In response to my saying Yes, those are the rules. I cannot fathom why JotSC said they are not.JotSC replied Perhaps my original answer was too hasty or too flippant. Perhaps I should have written, 'Yes, but...'. The "but" or "buts" have been noted several times in the past week.which I characterized as mystification. |
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11 Nov 12 - 11:53 PM (#3435034) Subject: RE: BS: Monty Hall Problem From: John on the Sunset Coast BKLick, you mystify easily, or we have a different definition of the word. If gnu and you and any others can't get it by now, that's on you. The rules, and the precise way they work, and how they affect the chances of gaining the car have been explicated in many different ways, within over 200 posts. There have been grids showing the precise permutation of results, and the probability of ultimately winning the car. There have been numerous ways of showing how the contestants first choice affects the the forced action by the host, and why that takes the second choice out of the realm of pure luck. Finally, it has been shown how making consistent choices affect getting the car (and what that choice is), as opposed to random choices. There is nothing mysterious nor mystifying about either the reasoning of the problem, nor in my referring gnu (but most probably YOU) back to the various posts on the subject. So, BKLick, read, study, learn and be well. |
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12 Nov 12 - 02:30 AM (#3435057) Subject: RE: BS: Monty Hall Problem From: BK Lick I have been trying from the beginning of this long thread to help poor gnu overcome the difficulty he's been having in understanding this problem. It is not helpful when, having stated the rules of the game absolutely correctly and then asking for confirmation that he has indeed understood the problem correctly, gnu is then told by you that no, he had not stated the rules correctly. JotSC, let me tell you that I find your last post insulting -- you think I don't get it, do you? I'm a Ph.D. mathematician whose research was largely in mathematical logic. I suggest that you refer back to the various posts on the subject, paying particular attention to yours and mine. |
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12 Nov 12 - 10:51 AM (#3435229) Subject: RE: BS: Monty Hall Problem From: John on the Sunset Coast That, sir, is how I found your previous two posts, and responded in kind...although I usually try not to do that. I don't expect to be responding more to this thread, nor re-reading it any longer. It's been cutting into my other activites. I probably will, I admit, I may visit it just to see how many ways folks deny the logic and probability of the winning car, and how long it takes for the thing to close. BTW, congratulations on your degree. I know that required a lot of time and more effort. But I know that the Hall Problem does not require a PhD to correctly understand it. For the record, I worked for several years at a job, test development, which required statistical analysis. I got an A in each of two UCLA courses, and the understanding of combinations and permutations, regression, correlation, confidence limits etc. That's not needed, either, except for permutations, to solve the Hall problem. Also, I read some on Game Theory, which, I admit, I did not fully understand beyond the basics, and I still don't. But I know enough for the Hall Problem. |
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12 Nov 12 - 11:05 AM (#3435240) Subject: RE: BS: Monty Hall Problem From: Nigel Parsons One problem is that the 50%ers insist on viewing it as a choice between two identical doors. They are not identical, the contents behind them differ. They also differ in that the competitor has already chosen one, at a time when he may, logically, have believed it had a 1/3 chance of being the one with the car. |