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BS: Mathematical Probability Query
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Subject: RE: BS: Mathematical Probability Query From: Wolfgang Date: 13 Oct 00 - 05:15 AM More than 100 posts!!
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Subject: RE: BS: Mathematical Probability Query From: Wolfgang Date: 13 Oct 00 - 04:57 AM Murray, I didn't understand the Homer Simpson reference (lack of cultural background) so there's no need at all for an apology. For everybody else: Murray's dead right on Marion's problem and Fionn is wrong. If the house knows where the prize is and alwaysalways offers the claimant to switch then the response to the question what is the probability of winning if I switch is 2/3. If, however, the house always offers a switch and always turns over a cup which (a) was not chosen in the first choice but (b) may contain the prize (e.g., if the house doesn't know where the prize is or knows where the prize is, but nevertheless opens the prize cup in 1/3 of all cases) then the response to the same question is 1/2.
This puzzle is considered technically identical to a number of teasers in conditioned probability, including the prisoners' problem. All of these problems have in common that Other problems which are mathematically nearly identical are
the two dice problem: I throw two dice and tell you truthfully (after I had a look without you looking) "at least one of these two dice shows a six". What is the probability that the other also shows a six? Obvious response: 1/6, very counterintuitive response: 1/11 All these problems can get much more complicated if you allow for unequal prior probabilities (e.g., one prisoner gets executed with prob 1/2, the next with prob 1/3 and the last with prob 1/6) or if you allow for biases (e.g., the father of a mixed pair of kids usually tells about his son, but sometimes about his daughter).
If you are really interested I can only advise you to read : |
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Subject: RE: BS: Mathematical Probability Query From: GUEST,Murray MacLeod Date: 12 Oct 00 - 06:49 PM Fionn, my brain is starting to hurt again. My understanding of Marion's problem depends on the fact that the house DOES have prior knowledge, and is never going to turn over the cup containing the prize. AM I missing something ? Murray |
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Subject: RE: BS: Mathematical Probability Query From: GUEST,Murray MacLeod; Date: 12 Oct 00 - 06:42 PM Wolfgang, I see that I made not one but two extremely stupid arithmetical errors when attempting to calculate your paradox. My lame excuse is that it was 7.00 am, not my best time. I apologize for the Homer Simpson reference. DOH !! Murray |
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Subject: RE: BS: Mathematical Probability Query From: Peter K (Fionn) Date: 12 Oct 00 - 06:23 PM Anyone who's got this far in the thread would appreciate, or will already have appreciated, the first few pages of Resencrantz and Gildernstern are Dead (as well as the rest of the play) by Tom Stoppard. Priceless. Marian, whether the house has prior knowledge about the cups is not a factor, except that if the prize cup is turned over, you lose your chance to reconsider. I know you didn't appreciate Jim's line on this, but it does help at the intuition level, so I'll use that to explain. Suppose the house invites you to pick the Ace of Spades out of a face-down pack of cards. You take a card but don't look at it. The house then turns over all but one of the remaining cards, without turning up the Ace of Spades. The house is not likely to achieve this feat without prior knowledge, but theoretically could. With or without prior knowledge, the house has made a dramatic intervention, which statistically you'd be daft to ignore. When you picked your card, there were fity-one to one chances that the house still had the Ace of Spades. Nothing's changed, except that those 51 to one chances are now vested in the one remained face-down card held by the house. The prisoner analogy has no bearing on the three-cup problem as no-one got to have a guess before the house intervened and changed the odds. |
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Subject: RE: BS: Mathematical Probability Query From: GUEST,Murray MaxLeod Date: 12 Oct 00 - 06:00 PM Bradypus, I love that one too it is a real paradox and MIND magazine had a lot of high-flying philosophers trying to solve it in the 7o's. I think the solution has somethhing to do with the invalidity of the initial premise, but never having completed my logic and metaphysics course, I probably wouldn't understand it anyway. Wolfgang, I am going to recheck the arithmetic on your one. And Marion, the time probably has come for a discussion of what exactly probability means. My take on it is that it is a function of the incompleteness of our information, and not really an attribute of the external world at all. IMHO it is actually incorreect to say "The probability of this coin landing heads is 1/2". The philosophically correct statement is "The probability of my being right if I call heads is 1/2" Murray |
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Subject: RE: BS: Mathematical Probability Query From: Bradypus Date: 12 Oct 00 - 05:10 PM The prisonner problem reminds me of one of my favourite stories / paradoxes:
A man is in prison on death row. The judge has told him that he will be executed before the end of the week, but won't tell him which day, only that it will come as a complete surprise to him. Bradypus
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Subject: RE: BS: Mathematical Probability Query From: Wolfgang Date: 12 Oct 00 - 11:16 AM Crazy Eddie, here's what you have overlooked in your solution of the three cards problem: The RR card will always slide out red face up, the RW card will slide out red face up only in 50% of all cases. If you take that into account, you'll come to the 2/3 vs. 1/3 solution as well. Your 50/50 solution would only be correct, if the RW card is only shown red face up. Well, this hasn't been explicitely excluded in Murray's post, but my reading of this post tells me that it was fairly obvious that each card could land on each side. Marion, glad to help you here against Crazy Eddie, but I won't help you against his arguments in the prisoners problem. The way you have formulated it, Crazy Eddie is right. Wolfgang |
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Subject: RE: BS: Mathematical Probability Query From: Mary in Kentucky Date: 12 Oct 00 - 10:23 AM Crazy Eddie - good job. This puzzle was designed to illustrate the differences in thinking between right brain and left brain processes. Using right brain thinking, you visualize a man starting at the bottom of the hill and another simultaneously starting at the top, and you quickly see that they must pass. A left brain thinker sometimes gets bogged down in verbal arguments, instantaneous snapshots, rates, etc. He cannot accept the simultaneous picture and has to somehow prove it with a linear prjection of the two trips. You got the correct answer with what appears to be a combination of both. I had never thought about "two" instances of being at the same place at the same time (or time of day). I probably was not clear in the question...I really couldn't remember the exact riddle. |
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Subject: RE: BS: Mathematical Probability Query From: Marion Date: 12 Oct 00 - 10:12 AM Ok, I'll have to think awhile about what it would mean if the house's choice is random. Thanks for the leads. Is anybody going to help me out and argue with Eddie over the red/white cards one? I think I still have to concentrate on my little cups. But about those prisoners... I'll put here my thinking on the problem, and why I think it doesn't contradict my stance on the cups, although at first analysis (that's you, Eddie!) it seems to. You might think that there is a 1/3 probability that you are doomed, and therefore a 2/3 probability that you aren't. Since the doomed person has already been chosen by the jailer, as I specified above, the release of a prisoner shouldn't change your chances, so the other prisoner is stuck with the 2/3 chance of being executed. The problem is, he's thinking the same about you. As Eddie says, there is a contradiction here. However, I would say that the real issue is in Eddie's 5th statement: "We each figure we have a 1/3 chance of being doomed." In fact, if the jailer has already decided who to execute, then each person's probability of being executed is either 1, or 0, not 1/3. Because the prisoners have no way of knowing whether they're a 1 or a 0, they perceive it as randomness, so they think their chances start at 1/3 then go to 1/2. But the execution will not be random, the victim has already been chosen. The probability that they perceive is different from the probability that is actually there. So psychologically, the two prisoners left are likely to be frightened by the third prisoner's release. But logically, if they know that the doomed one has already been chosen, they should consider it indifferent news. Marion
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Subject: RE: BS: Mathematical Probability Query From: Wolfgang Date: 12 Oct 00 - 08:34 AM B'', otherwise it would be no paradox. The odds(!) are 5/6 in A'' and 3/3 in B'', the probabilities are 5/11 and 3/6. This paradox has the awkward consequence that you can have four studies with the results pointing unanimously in one direction and when you pool the results to get a clearer picture the combined results point in the opposite direction. Scientists just hate such a situation.
To use a real life example: When you are a parent of now adults kids or just watch the young generation growing you are not surprised to learn that the average size (height) of male German students has increased since the 1950s. Similarly, the average size of female students in Germany has increased during the same time. However, if you pool the results to get the average size of all students in Germany, both male and female together, you'll find that the average size has decreased. The problem is most times not as easy to spot as in this example. There was even a court case in California in which a woman whose application to a Californian university was rejected sued for violation of equal opportunity (or whatever the term is) when she found out that the university did reject more females than males on a percentage base. She lost, because she fell prey to Simpson's paradox, when the university could show that (though the data pooled for the university showed a higher percentage of males being accepted) for each faculty separately there was even a small bias in favour women. Jurists must hate that: If they want to do justice to equal opportunity on the university level, they necessarily do injustice to males at the faculty level. If they want to do justice on a faculty level, they must do injustice to woman at the whole university level. And all that for purely mathematical reasons that do not bow to justice. Wolfgang |
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Subject: RE: BS: Mathematical Probability Query From: GUEST Date: 12 Oct 00 - 08:00 AM ..and sometimes 2 + 2 = 5, for extremely large values of 2. |
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Subject: RE: BS: Mathematical Probability Query From: GUEST,Muray MacLeod Date: 12 Oct 00 - 07:09 AM OK Wolfgang, after the cups I am prepared to be wrong again, but it looks to me like urn A now has a 5/6 probability of black, and urn B has 3/4 probability of black. So urn A is still the preferred option. So where is the paradox? Sounds more like Homer Simpson's paradox to me. Murray |
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Subject: RE: BS: Mathematical Probability Query From: AndyG Date: 12 Oct 00 - 06:53 AM Marion, If the house move is random then;
1) Player selects a cup. (sticks with choice, 1/3 chance to win)
If case three arises, and the player always switches, his chance to win is 2/3, but only in 2/3 of all the games played. (The house wins 1/3 of the games when it selects at (2).)
I hope that;s right ;)
Wolfgang,
I hope that's right too ;)
AndyG |
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Subject: RE: BS: Mathematical Probability Query From: Crazy Eddie Date: 12 Oct 00 - 06:52 AM Mary In Kentucky You said <.
Good one. To begin with, assume no rest breaks & constant speed for each journey (simplified version) Draw a line segment on a page, label one end A the other B [A is the botton of the hill B the top] Above the line,mark the monks position at zero point, Day one, 12 hours Day one 24 day one, etc. using 24 hour clock. BELOW The line do the same for the return journey. As you can see, the times increase Left to Right above the line, and Right to left. Times actually cross at two points: With rests, sleep etc, we cannot calculate exactly WHEN he will cross, (Lets take a ridiculously extreme example, where the hill is really short. Up hill he walks five minutes, rests almost 3 days, walks five minutes end! Return journey he rests for one day less!). However, if you draw the diagram for the simplified version above, it is clear that he MUST cross. Normally, he must cross TWICE, however, you can bring it down to crossing only once, if he stays in one place for a full day. |
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Subject: RE: BS: Mathematical Probability Query From: Escamillo Date: 12 Oct 00 - 06:18 AM The extrapolation to 1000 cups helped me too. The event is divided in two stages. In the first, you have a probability of 1/1000 to choose the prized cup. The house is OBLIGATED to take 998 cups out and leave to you the alternative: "Did you win in the first stage, with 99.9 % probabilities against you, AND will win AGAIN with 50% against you, in two consecutive choices ? Or did you loose first, with 99.9% probability of loosing, and will now win with this 50%, in two consecutive choices ?" Undoubtedly, and very intuitively, I would change ! Moreover, I imagined a situation in which the house is allowed to change places of the prize before you make your second choice. This would become TWO games, one with 33.3% probabilities of winning, and another one with 50%. Since the house cannot change the place of the prize, and always disclose all remaining cups minus one, it is giving you a very valuable information that changes your probability to win in the second stage. You can be sure that if I open a casino, I'll hide out the three cups for a moment before you attempt your second choice. Un abrazo - Andrés (and thanks to all for taking the trouble to explain this so well )
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Subject: RE: BS: Mathematical Probability Query From: Wolfgang Date: 12 Oct 00 - 05:12 AM A new problem known in statistics as Simpson's paradox: Imagine you get a prize for drawing a black bead from an urn (without looking) and you always get the choice of two urns to draw from. In the first choice you have urn A with a single black bead (and no other bead) and urn B with 2 black beads and 1 white bead. You'd not hesitate long to see that urn A is the better choice for you. In the second choice you have urn A' with 4 black beads and 6 white beads and urn B' with 1 black and 2 white beads. You take a little longer to find that 4/10 is larger than 1/3 but again you'd opt for urn A'. Now the total contents of the A urns (without missing a bead) are put together in one urn A'' and the total contents of urns B are put together in urn B''. Which urn do you prefer now? Wolfgang |
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Subject: RE: BS: Mathematical Probability Query From: Wolfgang Date: 12 Oct 00 - 05:01 AM Marion, I don't mind you asking. I'm professor of psychology and the work I mentioned was a recent diploma thesis by one of my students. She explicitely told the subjects that she knew where the prize was and would use this knowledge to make sure that she only opened cups with no prize under them. If she hadn't told this before the experiment the subjects might have been right assuming that they could as well stick to their first choice. Why?
I try to make it understandable without mathematics (but you could calculate it by using the tables of events as has been done above). Often it works if you make the case more salient by using more extrem numbers. Let's take Jim Dixon's example from above: The very same problem arises in your prisoners example (the most often and best studied problem in conditioned probabilities). You have not told us enough to make the response unambiguous. First, the decision which prisoner has to die has not just to be random but also random with equal probabilities (how goes random with unequal probabilities? E.g., you'd have an urn with 100 red beads (you die), 700 black beads (coprisoner B dies) and 200 green beads (coprisoner C dies)). The two responses below are only correct under that assumption, otherwise it is much more tricky. But that's only a minor quibble. The major is this. You (as prisoner) have to know beforehand that the next day definitely not you but another prisoner will be released to come to the response you (Marion) think of, namely, you should not be worried. If, however, you know that the next day one of the three prisoners (not necessarily you, but possibly you) will be released and the one released is not you, you should be worried for now the probability of you dying goes up from 1/3 to 1/2. Wolfgang |
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Subject: RE: BS: Mathematical Probability Query From: Crazy Eddie Date: 12 Oct 00 - 04:37 AM Marion, I am glad you put in the prisoner thing. I figured there was no advantage to switching, but some of the switch theories were a bit confusing. The prisoners thing makes it clear.
BTW the following is not a joke, or an ethnic slur. Just an easy way to for me to keep track.
[1] There are three of us in jail. Englishman, Scotsman, Irishman (me).
No problems with the logic so far.
[7] If I'M better off to switch, then clearly HE'S also better off (because from HIS point of view HE is the one who is switching.)
[8]How can BOTH of us improve our odds? If my odds improve, his must disimprove, since 1 of the 2 of us must hang?
Since [7] & [8] are mutually exclusive, switching cannot improve the odds for either of us. Therefore to switch or not to switch makes no difference! Q. E. D. |
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Subject: RE: BS: Mathematical Probability Query From: Crazy Eddie Date: 12 Oct 00 - 04:14 AM Marion, (are you also "Guest Marion?") (1)Re the Cards, Guest Marion said: << There are three red sides. Of those three, two have red on their opposites, and one has white on its opposite. >> There are TWO cards which can have Red as the top surface. Of these, ONE has white on the bottom, ONE has Red on the botton.If you can see red side, on top, you can eliminate the W/W card. i.e. you KNOW this is NOT the W/W card. Therefore it must be EITHER the R/R card, or the R/W card. 50/50 chance. If it is the R/R card, then the side you cannot see is RED, If it is the R/W card, then the side you cannot see is WHITE. So again, 50/50
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Subject: RE: BS: Mathematical Probability Query From: Marion Date: 12 Oct 00 - 01:12 AM Just to throw in a little twist, here's an open question for anybody who agreed with me from the beginning on the cups problem, or was convinced in this thread: Imagine that you and two other people are thrown in jail and told that one of you (and that one has already been decided by the jailer) will be executed and the other two released. You have no basis for guessing which one of you is doomed, so as far as any of you can tell, it's a random decision. The next day, one of the other prisoners is released. Assuming that you only care about yourself, are you relieved, worried, or indifferent about the other prisoner's release?
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Subject: RE: BS: Mathematical Probability Query From: Marion Date: 12 Oct 00 - 01:00 AM Wolfgang, you've done real life experiments on this? Do you mind if I ask what your profession is? I'm not sure of this statement of yours: "If just any cup is opened by chance either you have lost at once (for this cup contains the prize) or you have the choice between the two remaining cups (in the other two third of the cases) then it does not matter at all whether you change your first choice or not." If the dealer didn't know where the prize was and showed you one cup randomly, then sometimes he would show you the one with the prize so the game would be over for you. But if he shows you an empty cup, I still think it would be better to switch. Why would the dealer's knowledge or ignorance affect the game from your end of things? It's still true that the prize is more likely to be among the two cups that you didn't pick than in your first choice, and the dealer is still (although unintentionally) showing you which of those two has the prize if either of them does. Marion |
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Subject: RE: BS: Mathematical Probability Query From: Marion Date: 12 Oct 00 - 12:31 AM Murray, re: "And thank you to all who still dont understand the correct solution for inducing a warm glow of smug self-satisfaction in those who do." I am much amused, thank you. It seems like a number of converts have come over to the light side here... I'm curious to know if anyone, ever, sees the right answer right away. I certainly didn't. Jon and Andres, thank you for making tables. You both made the same mistake; you just COUNTED the possibilities when you have to WEIGH them. Remember that having a certain number of possibilities doesn't necessarily mean that each of those possibilities is equally probable; in this case, they are not. I will demonstrate this by editing Andres' table so that the probability of each outcome is weighed (Jon, I don't dare mess with your table). You remember, of course, that the probability of a complex outcome is the product of the probabilities of the events that compose it. I will assume that the player is some silly person who believes that it doesn't matter if he switches or not, so the player's second choice is random. And I will code each outcome: A means a switch resulting in a win, B means a switch resulting in a loss, C means a stay resulting in a win, and D means a stay resulting in a loss.
PRIZE IS IN CUP 1 (1/3)
PRIZE IS IN CUP 3(1/3)
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PRIZE IS IN CUP 1(1/3)
PRIZE IS IN CUP 2(1/3)
PRIZE IS IN CUP 3(1/3) (This is the same in essence as Andy's fixing of the table, but with everything spelled out.) OK, so if you add up the probabilities of the all the A scenarios, and all the B scenarios, and so on, you get: Probability that player will switch and win by doing so: 6/18 or 33% Probability that player will switch and lose by doing so: 6/36 or 17% Probability that player will stay and win by doing so: 6/18 or 33% Probability that player will stay and lose by doing so: 6/36 or 17%
I think I can now say, with deep sincerity: QED. I have proved my case. Marion, secretly hoping there'll be more holdouts so I'll have to learn another way to explain it |
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Subject: RE: BS: Mathematical Probability Query From: Mary in Kentucky Date: 11 Oct 00 - 11:47 PM OK, SA, no trick here, let's say he begins each journey at precisely 7 AM. Have I ever posted the recipe for a mint julep? I'll look for it. Those from Kentucky have probably heard it. |
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Subject: RE: BS: Mathematical Probability Query From: GUEST,Murray MacLeod Date: 11 Oct 00 - 11:01 PM Mary, it is late at night and you haven't phrased the question too well. Too many Mint Juleps, I'll be bound (isnt that what Kentucky ladies drink ?) You have to be specific about the monk's departure time on each occasion. "Sunrise" isn't enough, otherwise some smatass mudcatter is going to pull you to pieces. Murray |
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Subject: RE: BS: Mathematical Probability Query From: Mary in Kentucky Date: 11 Oct 00 - 10:55 PM OK Jon, it's late but I'll try to reconstruct this one. A monk started walking uphill on a winding mountain path at sunrise. It took him three days (with lunch breaks and rest stops) to reach the summit. He then began the journey down the mountain, same path, left at sunrise, no trick that I can think of, and it took him two days to complete the downhill journey. Question: Is there any point on the path where he is at the same time of day? |
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Subject: RE: BS: Mathematical Probability Query From: Jon Freeman Date: 11 Oct 00 - 09:20 PM I'm loving this Murray, even though I get things wrong and confuse myself.... Anyone got any more? Jon |
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Subject: RE: BS: Mathematical Probability Query From: GUEST,Murray macleod Date: 11 Oct 00 - 09:13 PM yes Jon, correct answer, faultless reasoning, but on the street corner you gotta think on your feet ..... Murray |
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Subject: RE: BS: Mathematical Probability Query From: Jon Freeman Date: 11 Oct 00 - 07:07 PM This is what I get with the cards.
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| Card | Red Shows | Red/ Red | ||
| 1 | W/W | FALSE | 0 | 0 |
| 2 | R/W | TRUE | 1 | 0 |
| 3 | R/R | TRUE | 1 | 1 |
| 4 | W/W | FALSE | 0 | 0 |
| 5 | W/R | FALSE | 0 | 0 |
| 6 | R/R | TRUE | 1 | 1 |
| 3 | 2 | |||
| %Win | 66.66667 |
Maybe one day I'll learn the maths...
Jon
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Subject: RE: BS: Mathematical Probability Query From: GUEST,Murray MacLeod Date: 11 Oct 00 - 06:12 PM Narion, you are of course perfectly correct regarding the red/white problem. Martin Gardner , who writes (wrote?) for !Scientific American" published this problem in the 60's and was immediately heckled by a so-called professional gambler who insisted that the probability of red was evens,("only two possibilities , either red or white"). He even offered to meet Gardner and play for real money. Gardner refused, reasoning that his available capital was so much less than the gamblers capital that he could not afford the possibility of a freak run which as all statisticians know is certain to occur sooner or later. Going back to the cups problem, my indebtedness to Jim comes by my extrapolating the problem from 1000 cups ( as he postulated) all the way down to four. In each case it is obvious that , when the house knows the correct cup, it makes sense to switch. I have to say that I still do not see it intuitively at the three-cup level, but I am familiar with mathematical patterns and I know your solution is correct. I would just like to add that this is one of the most enjoyable threads i have ever encountered on the net.
Thank you FMaj7 for instigating it and thank you Marion for enlivening it. And thank you to all who still dont understand the correct solution for inducing a warm glow of smug self-satisfaction in those who do
Murray |
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Subject: RE: BS: Mathematical Probability Query From: Penny S. Date: 11 Oct 00 - 03:50 PM It seems to me that if you have a basket which can hold 10 apples, there is no evidence that you have any apples to take. Penny |
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Subject: RE: BS: Mathematical Probability Query From: Mary in Kentucky Date: 11 Oct 00 - 02:54 PM Marion - I do absolutely as little as possible because I WORK FOR MY HUSBAND! I try to get fired everyday, but so far can only manage it at about one minute before closing time. I help in the vet clinic, and yesterday was such a hard day I need to take a nap now on my afternoon off. |
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Subject: RE: BS: Mathematical Probability Query From: Jim Dixon Date: 11 Oct 00 - 02:02 PM The following is not so much a mathematical problem as a psychological problem, but it has to do with the subjective evaluation of outcomes. PROBLEM 1: You start out with $60 in your wallet. You spend $20 on an advance general-admission ticket to a concert. You arrive at the concert hall, open your wallet, and find that the ticket is missing and presumably lost forever, but you still have $40. What would you do? Would you spend another $20 to buy another ticket? Or would you skip the concert? (There is no right or wrong answer here, but please consider your answer before you read the next paragraph.) PROBLEM 2: You start out with $60 in your wallet. You head for the concert hall, planning to buy a ticket at the door. When you arrive, you open your wallet and find that $20 is missing and presumably lost forever, but you have $40 left. Now what would you do? Someone did an actual survey. I don't remember the numerical results, but a large number of people - let's say half - when problem 1 was posed, said that they would skip the concert. When problem 2 was posed instead, nearly everyone said they would buy a ticket. I will leave it to you to figure out what this says about human nature. |
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Subject: RE: BS: Mathematical Probability Query From: GUEST,Marion Date: 11 Oct 00 - 01:59 PM Wow, I'll have to come back here when I have more time and study the new responses to the cups problem, but let me just say this for now: Mary, you're not suppsed to think while you're at work? What do you do exactly? :) Andres, you were right in the beginning that the house does know where the prize is, and will always uncover an empty cup. And I've thought about the red/white problem, which I'll copy here since it's way up there: "This problem seems to be related to one which was first aired in Scientific American in the 60's, and which stirred a fair amount of debate. There are three cards, one is red on both sides, one is white on both sides, and one is red on one side and white on the other. The dealer places them in a bag, shakes them then slides out one card onto the table so that only one face is visible. The face is red. What is the probability that the other face is red? " Here's my answer: the probability is 2/3 that the side you can't see is red. A little counterintuitive, but simpler (I hope!) to explain than the cups problem. I think the key is to look not at the number of cards but at the number of sides. There are three red sides. Of those three, two have red on their opposites, and one has white on its opposite. Are you still around, Murray? Is this what the Scientific American said? Marion |
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Subject: RE: BS: Mathematical Probability Query From: Wolfgang Date: 11 Oct 00 - 12:38 PM First, for those who don't believe (yet), test it yourself. It won't take too many trials to convince you that switching is the correct strategy. Jeri, however, has pointed to the correct detail. It matters if the one who opens the first cup, knows (and cares in opening) whether this cup contains the prize or not. This detail was not completely clear in Marion's first post. If just any cup is opened by chance and either you have lost at once (for this cup contains the prize) or you have the choice between the two remaining cups (in the other two third of the cases) then it does not matter at all whether you change your first choice or not. Like in a variant of that problem, when a father of two comes up to you and says: 'I've two children, one of them is a boy' and you are asked what is the probability that the other child is a boy too. Depending upon conditions I have deliberately not mentioned the correct response is either 1/2 or 1/3. Yes, there are experiments done on the Monty Hall dilemma. I've done one, for instance. In the classical situation Marion has described, only about 8 % of our subjects switched (made the experiment cheaper, for there was a real prize under there to take home). If we had 30 cups, however (28 with no prize opened after the first choice), about 2/3 of our subjects switched which shows that they were somehow sensible to the odds (in this case about 97 % probability for winning, if you switch). The best recent article on these problems is by R. S. Nickerson in the Psychological Bulletin (circa 1997/98). Sorry, I don't find the exact reference right now. If you've understood the problem so far, you're ready for the Russian Roulette variant (only one miss, all other options are prizes): 3 cups with two prizes and one miss. After your first choice an other cup that is known to contain a prize is opened, leaving you with two cups, one containing a prize. Should you switch??? Or does it not matter? Of course it matters, you should not switch under these conditions, for you have a 2/3 probability of winning if you stay. It is an open question for science (yes, we do have open questions, lots of), why in this situation when switching is bad many more subjects switch than in the standard version in which switching is good. Well, people do have many difficulties with conditioned probabilities (even math profs and PhD's are on the record for defending a wrong solution for Monty Hall's dilemma with very strong words) and if you give me a new problem of that kind and ask me for my spontaneous solution (without using paper, pencil and Bayes' theorem) I'd not be surprised to be wrong. Wolfgang |
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Subject: RE: BS: Mathematical Probability Query From: Jon Freeman Date: 11 Oct 00 - 12:34 PM Andy, my model only had the house lifting once and are events that actually happen. What it failed to take into account was that in several situations, the house was forced into picking one of the 2 remaining cups and although invalid, they still need to be taken into account for the oveall probability. Jon
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Subject: RE: BS: Mathematical Probability Query From: Jon Freeman Date: 11 Oct 00 - 12:26 PM Yep OK, got it now. I had previously modeled the result and was getting roughly 1/3 to 2/3 then tried this other one. The invalid 1/3 work against the house. Jon |
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Subject: RE: BS: Mathematical Probability Query From: AndyG Date: 11 Oct 00 - 12:23 PM Jon,
As I said earlier in the ammended table, you, like Escamillo are modelling events that don't happen.
PRIZE IS IN CUP 1
PRIZE IS IN CUP 2
PRIZE IS IN CUP 3
(again only for sticking as in your example)
AndyG |
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Subject: RE: BS: Mathematical Probability Query From: Mark Clark Date: 11 Oct 00 - 12:13 PM Boy, I can see one has to hang in here a lot more than I can if one is to keep up. I see my race car puzzle was way to easy for this group. I remember Marilyn Vos Savant discussing the Monty Hall (cups) problem in the Parade Sunday roto section. A search for Marilyn turned up another good explanation, for the terminally curious, of why one should switch cups after one has been turned over.
Marion, I think Spaw was referring to three card monty, a common street shell game in large urban areas.
- Mark |
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Subject: RE: BS: Mathematical Probability Query From: Jon Freeman Date: 11 Oct 00 - 12:06 PM Andy, that was what I had thought until I tried to model every possibility on a spread sheet. It seems to me that 1/3 of the moves being considered are invalid as they would involve the house uncovering the winning cup. Bartholomew, I may have got that one right but I'm not higher minded - I am very confused here. Jon |
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Subject: RE: BS: Mathematical Probability Query From: AndyG Date: 11 Oct 00 - 11:58 AM Jon,
The sum of the probabilities must be 1.
The probability that your first choice is the winner is 1/3.
AndyG |
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Subject: RE: BS: Mathematical Probability Query From: Jim the Bart Date: 11 Oct 00 - 11:29 AM Sorry I doubted you, Jon. I yield the floor to the higher minded among us and will be quiet now. |
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Subject: RE: BS: Mathematical Probability Query From: Jon Freeman Date: 11 Oct 00 - 11:18 AM I must admit I am getting more and more confused with the cups one. I thought I could see and had modeled an answer and then tried to go through every possibility I can see occurring. Which led me to this:
I have only considered the stick here but the change would be the opposite. This yeilds a 6/12 result - what moves have I missed? Jon
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Subject: RE: BS: Mathematical Probability Query From: Mary in Kentucky Date: 11 Oct 00 - 10:55 AM Always profreed... Thanks for the links Pete. Here's The Monty Hall Problem. And I liked this link (referred to in the above link)---Obstinacy, Comprehension, and the Monty Hall Problem. This could be applied to our thread about the unbelieveable, but I won't go there. |
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Subject: RE: BS: Mathematical Probability Query From: Mary in Kentucky Date: 11 Oct 00 - 10:50 AM hanks for the links Pete. Here's The Monty Hall Problem. And I liked this link (referred to in the above link) Obstinacy, Comprehension, and the Monty Hall Problem. This could be applied to our thread about the unbelieveable, but I won't go there. |
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Subject: RE: BS: Mathematical Probability Query From: Jeri Date: 11 Oct 00 - 10:24 AM Slowly, the clouds part, and a ray of light penetrates the dank, dark place that is my brain. Thinking about what Jim has said, I see that the the probabilites are affected by the "house-knows-something-you-don't" factor. I had a pretty slim chance of being right when I picked at random from 1,000 cups. The house selected the other cup to not turn over, and the house KNOWS which cup hides the prize. Since my chances of being right in the first place were not so good, I switch to the house's pick. I don't know if this words as well with only three cups to start with - my chances of being right initially weren't so bad, but at least I've got the idea now. |
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Subject: RE: BS: Mathematical Probability Query From: GUEST,Pete Peterson Date: 11 Oct 00 - 10:04 AM first, thanks to Marion and also to Jim Dixon; between the two of you I finally have an explanation that makes sense to me. It's called the Monty Hall problem 9hence Catspaw's reference) because it is identical with the procedure on "let's make a deal"-- M. Hall, host. If you go into Yahoo/Science/Mathematics you can get quite a discussion of this problem, under that name, but none of the explanations are as clear as the two I have seen! thanks. |
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Subject: RE: BS: Mathematical Probability Query From: CamiSu Date: 11 Oct 00 - 09:36 AM Marion, if you have a basket with 10 apples in it and you take out 3, you have 10 apples. 7 in the basket and 3 in your hand! It's all in how you state the problem. Someday I must take a statistics course. I've tutored 3 adults through increasingly difficult ones and had so much fun that I want to do it for myself! |
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Subject: RE: BS: Mathematical Probability Query From: Jeri Date: 11 Oct 00 - 09:27 AM Well, I'll admit I'm mathamatically challenged. I've no problem with logic, just formulas, so I'm going to be difficult. When you choose the first cup, you have a one in three chance of picking the right cup. When you choose a second time, you have a one in two, or 50% chance of being right. Forget the third cup - you're making a choice between 2 cups, and it makes no difference if you pick the same cup again - it's still a choice. If it is the one with the prize, it will have survived a 1/2 chance because you made a second choice. I don't think adding up the probablilities makes sense. I think that taking one of the cups away just changes the number to pick from. You can't have a 2/3 chance of winning if there are only 2 cups. I wonder if anyone's done experiments... |
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Subject: RE: BS: Mathematical Probability Query From: Mary in Kentucky Date: 11 Oct 00 - 08:37 AM I think I mispoke on the words I used above when I said "empty." I'm still thinking about this one, but since I'm at work and should not be thinking... |