The Monty Hall "pick a door...should you switch" problem

Many of you will have heard of this classic problem - known as the Monty_Hall_problem.

Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice?

There are two slightly different interpretations of this problem, with (unsurprisingly) different answers. (*Note: updated since my original post*)

1. IF we assume that Monty always has to pick a goat when he shows you a door, then you will do best to switch (there is a 2/3 probability of winning if you switch, 1/3 probability of winning if you stay). Your overall chance of winning a car if you plan to switch in this scenario is 2/3. See more discussion below.
2. IF Monty picks a door randomly, then switching and staying both give you the same result! He's either going to:
1. show you the car (in which case you've lost)...this will occur 1/3 of the time
2. show you a goat. You now have a 50/50 proposition, since Monty was not *required* to show you a goat. He randomly picked a door and now there is a 50/50 chance that the car is behind your door, and a 50/50 chance that it is behind the other not-yet-selected door. So whether you stay or switch you will get it right 1/2 of the time. If' you're the type of person who kicks themselves more if you switch and are then found to be wrong to switch, then by all means stay.

So in Case #2, your overall chance of winning a car is only 1/3! If you stay with your pick we already know that you have a 1/3 probability of winning. But if you were planning to switch (which would give you the 2/3 probability of #1) you get KO'd 1/2 the time when Monty actually reveals a car, so your net probability of wining is 1/2 * 2/3 = 1/3.

So why do so many people not believe the answer to #1? There are a couple of simple explanations to prove that it is true...maybe people don't have the attention span to actually go through the thinking. Two ways to bring yourself to a new understanding of probability:

a) Just try the combinations! (See also the simple table in the link I gave to wikipedia in the first sentence). If you do this, you will find that out of all possibilities (i.e. 3 possible initial guesses by you x 3 possible positions for the car) you will win 6 times out of 9 if you switch.

• For any time that you chose correctly at first (3/9 of the possibilities above) you will lose when you switch
• For any time that you chose incorrectly at first (6/9 of the possibilities above) you will win when you switch

b) What I have found compelling for many is to imagine a situation where there are many more doors, and Monty picks all but one of those doors and shows goats behind all of them. Now would you switch? Here it seems to be more reasonable to people that if there were, say, 100 doors and they pick 1 that they will have a 1/100 chance of winning (and that there is a 99/100 chance that the car is in one of the other 99 doors). So if Monty shows 98 of those doors have goats, you should switch. Only 1/100 of the time will you have selected the right door, but 99/100 of the time Monty will have found it for you out of all the other doors!

Here's a simple table (using the $ to indicate a good prize, and 0 to indicate a bad one). If the prize was originally behind door 1 you win by staying (in blue below). If the prize was originally behind ANY OTHER DOOR you win by switching (in green below).