Monty Hall Problem: Probability of Winning When Switching Doors

[rbzima]

The Monty Hall Problem states that during a gameshow, a contestant can choose one of three doors. One of these three doors contains a car, whereas the other two doors contain a gag prize. After selecting your door of choice, the host will open one of the two gag prize doors. At this point, is it better to switch to the other door, or to stay.

The problem I'm looking at right now, and having a little bit of difficulty is the following. Suppose the contestant can only switch once, yet there are 6 different doors to choose from. After selecting the door, the host will randomly choose 3 of the gag prize doors to open and show you.

My question is this: What are the respective probabilities at each part of the tree?

 

[HallsofIvy]

In the original Monty Hall problem, the a-priori probability the prize is behind anyone door is 1/3. After you pick a door, Monty Hall opens one of the remaining two doors showing that the prize is NOT behind that door. Since he would not open the door you have already picked, in any case, the probability the prize is behind that door remains 1/3. Because Monty Hall's "superior knowledge" (he knows where the prize is) guarentees that he will not open the door the prize is behind, the probability the prize is behind the door he did not open is now 2/3. You would improve your odds by switching.

This new 6 door situation is pretty much the same. Assuming there are 5 gag doors and 1 with the real prize, a-priori, the chance of the prize being behind anyone of them is 1/6. After the host (who has additional information- he knows which doors do not have the prize) opens three doors to show to show they do NOT have the prize, since he would not, in any case, open the door you have picked, the probability the prize is behind the door you picked remains 1/6. However, the "5/6" probability that the prize was behind one of the doors you did NOT pick is now divided amoung the 2 remaining doors. The probability the prize is behind one of them is now 5/12 which is larger than 1/6. You would increase your odds of winning by switching.

The usual "error" is to think the doors remain "equally likely". They don't because you have additional information given you by the host.

Going back to the original, three door, problem, here is an interesting variation. Suppose that, after you make a choice, the game show host, NOT knowing which door the prize is behind, opens one of the two remaining doors at random. Show that, assuming the prize happens not to be behind that door, the probability the prize is behind either of the unopened doors is still 1/3 (and so they do not sum to 1 because this is a "conditional" probability).