Twist on Monty HallDate: 03/31/2001 at 17:27:39 From: rick singh Subject: Twist on Monty Hall Here is a twist on the classic Monty Hall problem. There are three doors. Each door contains a prize. Your goal is to maximize the probability of getting the best of the three prizes. You can open as many of the doors as you like, but you have to stick with the last door that you open. When you open a door, you do not know whether or not it is the best. Should you open only one door, or should you open one and then open the second and stick with the second - or should you open all three and stick with the third? I understand Monty Hall, but this one has elements I can't figure out. Thanks for your help. Date: 04/01/2001 at 16:41:37 From: Doctor Schwa Subject: Re: Twist on Monty Hall Hi Rick, The classic problem can be found in the Dr. Math FAQ: The Monty Hall Problem http://mathforum.org/dr.math/faq/faq.monty.hall.html Yours is a really interesting type of problem! I first heard it in the context of a princess choosing among a hundred suitors. At any time she can accept the proposal of the suitor she's seeing, but once she rejects one, he won't come back. How can she maximize her chances of getting the best one? The answer, of course, depends on the relative quality of the suitors. I think it turns out that the best strategy for her is to let the first 37 suitors go by, and then if she ever sees a suitor better than the best she's seen so far, keep him. Why 37? Well, that's a long story. Luckily, your problem has only three doors instead of a hundred suitors, so it's a bit easier to analyze. If you keep the first door, you have a 1/3 chance of getting the best prize. If you go to the second door, and just keep it no matter what, it still has a 1/3 chance of being best. Same with the third door. BUT, when you open the second door, you already have a bit of information: is the second door better than the first? If it is better than the first, keep it. If it's worse than the first, go on to the third and take whatever's there. This strategy gives you a 50% chance of getting the best prize. Let's let A stand for the best prize, and B the second best, and C the worst. Then the possibilities are ABC ACB BAC BCA CAB CBA With the strategy described above, you win whenever the second prize is best (BAC and CAB), but you also win in the case BCA, because with the second prize worse than the first, you go on to the third and win. So, half the time you get the best prize; in two cases out of six, ACB and CBA, you end up with the middle-quality prize (in ACB, since the second prize is worse than the first, you try the third; in CBA, since the second prize was better than the first, you stay with it). Only in case ABC will you get stuck with the worst prize. - Doctor Schwa, The Math Forum http://mathforum.org/dr.math/ |
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