johns@utcsrgv.UUCP (David Johns) (09/19/83)
You've made it as a contestant on Let's Make a Deal. Monty shows you three doors, only one of which has a good prize behind it. You make your choice and Monty opens one of the doors with a bad prize behind it (as he always does). He now asks you whether you would like to switch to the remaining door. The million dollar question is.... should you switch or does it matter??
pearlman@trw-unix.UUCP (Laura Pearlman) (09/22/83)
It doesn't matter which door you choose in the "Let's Make
a Deal" problem.
Proof: Let S be the set of prizes behind the three doors.
By a theorem proved earlier in this newsgroup, all elements
of S are equal. Thus all prizes are bad, and it doesn't
matter whether you switch or not.
-- Laura Pearlman
...{sdcrdcf,decvax}!trw-unix!pearlmanmcewan@uiucdcs.UUCP (mcewan ) (09/29/83)
#R:utcsrgv:-228700:uiucdcs:28200019:000:970 uiucdcs!mcewan Sep 28 11:22:00 1983 You've made it as a contestant on Let's Make a Deal. Monty shows you three doors, only one of which has a good prize behind it. You make your choice and Monty opens one of the doors with a bad prize behind it (as he always does). He now asks you whether you would like to switch to the remaining door. The million dollar question is.... should you switch or does it matter?? Switch. The third door has twice the probability of having the good prize as the door you chose. When you made your choice, each door had a 1/3 probability of having the good prize. Regardless of which door you chose, Monty will show you a door with a bad prize (if you chose a bad door, he shows you the other one; if you chose the good door, he chooses a door at random), so this tells you nothing about the probability of the door you chose - thus the probability stays 1/3. Therefore, the probability of the last door having the good prize is 1-1/3 = 2/3. Scott McEwan
leichter@yale-com.UUCP (Jerry Leichter) (10/17/83)
Here is an interesting argument, due to Dave Wittenberg, that shows that switching doors DOES gain: Suppose you follow a different algorithm: Choose a door at random. Monty shows you a "bad door". Flip a fair coin; if it is heads, keep the door you already have; if it is tails, switch. Now what are your chances? Well, the coin washes out any information from your first choice. It makes no difference what the original probabilities were. After Monty shows you a bad door, you are ALWAYS in the position of having two doors available to you, one good, one bad. The coin toss as described is indistinguishable from a fair coin toos to choose on of these two doors. Hence, this strategy gives you a 50% chance of winning, while always keeping the original door gives you a 1/3 chance. So switching CAN improve your chances...and a detailed argument for 2/3 if you always switch is plausible. The case analysis someone sent out is incorrect because it assume each of the four cases listed is equally probable, which is not the case. -- Jerry decvax!yale-comix!leichter leichter@yale