his is a favorite probability problem of mine and I wanted to see who could anwser it correctly:
This guy dies and goes to heaven but before he can enter he is first confronted by St. Peter who reads through the man’s sins and says the following to him: “Due to some minor sins that you have committed, I can’t let you pass directly into heaven, I will leave your fate for eternity up to chance”
He shows the man three identical doors and explains to him: “one of these doors goes to heaven and the other two go to hell, you must choose one” The man reluctantly chooses door number 2.
St. Peter, knowing which is the door that leads to heaven says to the man: “ok, I’ll give you a little help. I know which one goes to heaven so I am sure that door number 1 goes to hell.” He opens door number 1 and demons and flames come out of it affirming that it did go to hell.
He then asks the man again: “Now there are two doors left, do you stick to your original choice, number 2 or do you change to door number 3”
Assuming that St. peter would have applied this same procedure no matter what door the man originally chooses (he opens one of the doors that the man did not choose to show that it went to hell), What is the probability on each door that is left to be the one that goes to heaven? In other words, which one should the man choose?
Here’s a simple way to think of it that helped me when I first heard this:
Let’s say you choose dor #1. You’ve just divided the doors into two groups, [1] and [2 & 3]. Obviously there’s a 2/3 chance that the second group is the winner, 1/3 that your group is the winner.
Now, you eliminate one choice from [2 & 3], but the probability remains 2/3 that the winner was in that group. Therefore it’s 2/3 for the other group (i.e., switching), and 1/3 for your original choice.
Here’s the way to think of the problem that works best for me:
You have three choices for which door to select first. St. Peter (or the game show host) has three choices for which door to make the winner. Thus, there are nine total possible situations. Of those nine situations, there are three where your first choice of door is correct, and six where it is incorrect. Thus, you’re twice as likely to win if you assume that your choice was incorrect.
Right, but if one of the choices from the larger group is removed, then your chances are half as good (since you’re not very well going to pick the door you know is bad), i.e. 1/2 * 2/3, or 1/3. So you don’t have much of a difference.
It’s beneficial for the man to stay if and only if he picked the correct door in the first place. Since these are equivalent events, and the probability that he picked the correct door is 1/3, the probability that it’s beneficial to stay is 1/3.
Oh, and iampunha: When you switch, you’re not picking one door. You’re picking both doors that you didn’t pick already.
Well Im glad to see that you strightdope surfers are smart guys, tha correct anwser is to swich, with a probability of 2/3 for going to heaven. (for anyone that has any doubts, imagine there are a hundred doors, the guy chooses door number 34 and st. peter opens all but door number 34, obviusly, and door number 72. If you don’t see it now, you won’t see it.)
Of course, your figure of 2/3 assumes that either a) the supplicant does not favor any particular door, or b) no one door is most likely to lead to heaven. If both a) and b) are false, all bets are off. But you knew that, right?
