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BS: Monty Hall Problem
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Subject: RE: BS: Monty Hall Problem From: John on the Sunset Coast Date: 01 Nov 12 - 12:18 AM 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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Subject: RE: BS: Monty Hall Problem From: DMcG Date: 01 Nov 12 - 04:12 AM 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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Subject: RE: BS: Monty Hall Problem From: BK Lick Date: 01 Nov 12 - 04:17 AM 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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Subject: RE: BS: Monty Hall Problem From: TheSnail Date: 01 Nov 12 - 05:45 AM 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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Subject: RE: BS: Monty Hall Problem From: Nigel Parsons Date: 01 Nov 12 - 06:27 AM 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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Subject: RE: BS: Monty Hall Problem From: Nigel Parsons Date: 01 Nov 12 - 06:37 AM Sorry, the extra £10 I'm putting in. |
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Subject: RE: BS: Monty Hall Problem From: DMcG Date: 01 Nov 12 - 07:45 AM 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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Subject: RE: BS: Monty Hall Problem From: GUEST,Lighter Date: 01 Nov 12 - 08:31 AM 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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Subject: RE: BS: Monty Hall Problem From: GUEST,TIA Date: 01 Nov 12 - 08:36 AM A+ in philosophy, but D- in "conditional probability". Look it up. |
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Subject: RE: BS: Monty Hall Problem From: GUEST,Reverend Bayes Date: 01 Nov 12 - 08:38 AM 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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Subject: RE: BS: Monty Hall Problem From: GUEST Date: 01 Nov 12 - 08:55 AM 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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Subject: RE: BS: Monty Hall Problem From: GUEST,DmcG Date: 01 Nov 12 - 08:57 AM That was me, sorry! |
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Subject: RE: BS: Monty Hall Problem From: GUEST,Reverend Bayes Date: 01 Nov 12 - 10:38 AM 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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Subject: RE: BS: Monty Hall Problem From: GUEST,Lighter Date: 01 Nov 12 - 10:48 AM > known to first year undergraduates everywhere. Only if they take the right course! And most won't. |
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Subject: RE: BS: Monty Hall Problem From: John on the Sunset Coast Date: 01 Nov 12 - 10:52 AM 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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Subject: RE: BS: Monty Hall Problem From: DMcG Date: 01 Nov 12 - 11:07 AM 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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Subject: RE: BS: Monty Hall Problem From: GUEST,Lighter Date: 01 Nov 12 - 11:50 AM 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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Subject: RE: BS: Monty Hall Problem From: GUEST Date: 01 Nov 12 - 12:06 PM 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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Subject: RE: BS: Monty Hall Problem From: John on the Sunset Coast Date: 01 Nov 12 - 12:49 PM " 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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Subject: RE: BS: Monty Hall Problem From: GUEST,Lighter Date: 01 Nov 12 - 01:20 PM > 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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Subject: RE: BS: Monty Hall Problem From: TheSnail Date: 01 Nov 12 - 01:44 PM 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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Subject: RE: BS: Monty Hall Problem From: DMcG Date: 01 Nov 12 - 02:08 PM 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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Subject: RE: BS: Monty Hall Problem From: gnu Date: 01 Nov 12 - 02:53 PM "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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Subject: RE: BS: Monty Hall Problem From: John on the Sunset Coast Date: 01 Nov 12 - 03:55 PM 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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Subject: RE: BS: Monty Hall Problem From: Howard Jones Date: 01 Nov 12 - 04:50 PM 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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Subject: RE: BS: Monty Hall Problem From: BK Lick Date: 01 Nov 12 - 05:09 PM "Soooo... I am not as smart as a pigeon?" -- gnu Herbranson and Schroeder, 2010 Embarrassing, ain't it? |
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Subject: RE: BS: Monty Hall Problem From: gnu Date: 01 Nov 12 - 07:06 PM 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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Subject: RE: BS: Monty Hall Problem From: John on the Sunset Coast Date: 01 Nov 12 - 10:46 PM 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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Subject: RE: BS: Monty Hall Problem From: DMcG Date: 02 Nov 12 - 03:14 AM 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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Subject: RE: BS: Monty Hall Problem From: Howard Jones Date: 02 Nov 12 - 05:30 AM 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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Subject: RE: BS: Monty Hall Problem From: GUEST,Lighter Date: 02 Nov 12 - 09:19 AM 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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Subject: RE: BS: Monty Hall Problem From: DMcG Date: 02 Nov 12 - 10:15 AM 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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Subject: RE: BS: Monty Hall Problem From: John on the Sunset Coast Date: 02 Nov 12 - 10:47 AM 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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Subject: RE: BS: Monty Hall Problem From: Jim Dixon Date: 02 Nov 12 - 01:36 PM 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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Subject: RE: BS: Monty Hall Problem From: Howard Jones Date: 02 Nov 12 - 03:13 PM "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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Subject: RE: BS: Monty Hall Problem From: gnu Date: 02 Nov 12 - 06:51 PM 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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Subject: RE: BS: Monty Hall Problem From: Mysha Date: 03 Nov 12 - 09:18 PM 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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Subject: RE: BS: Monty Hall Problem From: DMcG Date: 04 Nov 12 - 03:42 AM 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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Subject: RE: BS: Monty Hall Problem From: DMcG Date: 04 Nov 12 - 03:44 AM Which it isn't, I assert (sigh!) |
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Subject: RE: BS: Monty Hall Problem From: GUEST,Ron Date: 04 Nov 12 - 08:04 AM 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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Subject: RE: BS: Monty Hall Problem From: Mysha Date: 04 Nov 12 - 12:26 PM 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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Subject: RE: BS: Monty Hall Problem From: DMcG Date: 04 Nov 12 - 01:01 PM 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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Subject: RE: BS: Monty Hall Problem From: GUEST,Ron Date: 04 Nov 12 - 04:03 PM 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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Subject: RE: BS: Monty Hall Problem From: Mysha Date: 04 Nov 12 - 10:07 PM 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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Subject: RE: BS: Monty Hall Problem From: DMcG Date: 05 Nov 12 - 05:45 AM 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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Subject: RE: BS: Monty Hall Problem From: GUEST,Lighter Date: 05 Nov 12 - 08:19 AM 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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Subject: RE: BS: Monty Hall Problem From: John on the Sunset Coast Date: 05 Nov 12 - 10:38 AM 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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Subject: RE: BS: Monty Hall Problem From: GUEST,Lighter Date: 05 Nov 12 - 11:16 AM 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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Subject: RE: BS: Monty Hall Problem From: John on the Sunset Coast Date: 05 Nov 12 - 12:44 PM 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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Subject: RE: BS: Monty Hall Problem From: DMcG Date: 05 Nov 12 - 02:31 PM 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. |
