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How this issue be resolved mathematically?

First of all, the following facts are critical to a correct understanding of the problem, and need to be stated more explicitly:

- The host is not malicious. He doesn't just offer a chance to switch when the contestant's original guess is correct. Rather, he always offers the chance. (With a malicious host, it would always be better to stick to your original guess, since the very fact the host gave you a chance to change your mind would mean that your guess was correct!)
- The host knows where the car is. When he opens a door to reveal a goat, that wasn't an accident; he's never going to open the door that reveals the car!
- When the host has a choice of doors to open, he chooses randomly.

However, when the question is properly and clearly stated, with the above assumptions made, then Marilyn's answer is correct (although her reasoning was at fault, because she didn't make clear use of these assumptions).

Why is this answer correct?

One way to look at the problem is this. If you adopt the non-switching strategy, you will win whenever your original guess was correct (which has a 1/3 probability of happening), and lose otherwise. If you adopt the switching strategy, you will lose whenever your original guess was correct, but you will win whenever your original guess was wrong (which has a 2/3 probability of happening).

This is a good argument in favour of the 1/3, 2/3 theory, but it doesn't explain what's wrong with the 1/2, 1/2 theory. After all, it seems perfectly reasonable that, if a door is opened revealing a goat, there should now be a 50-50 chance to the car being behind one of the remaining two doors.

The key is this. That 1/2, 1/2 theory would be correct *if the
host opened a door completely at random, and it happened to reveal
a goat*. But (from item 2 above) we know that the host will never open
the door revealing the car. So there is additional information revealed
by the host's choice of which door to open, besides the obvious information
that that door revealed a goat.

Suppose you choose door 1 and the host opens door 2 (meaning the car
is either behind door 1 or door 3). Although it's true that the basic
probabilities of the car being behind door 1 or door 3 are equal,
*that's not the relevant issue here*. Instead, we are after the
*conditional probabilities* that car is behind door 1 or door 3,
*given that* the host opened door 2.

What this means is: think of playing the game many, many times. Obviously, on average the car will be behind door 1 1/3 of these times, behind door 2 1/3 of these times, and behind door 3 1/3 of these times. So, out of all the times you play the game, the proportion that have the car behind door 1 is equal to the proportion that have the car behind door 3.

But the question for us is: if we restrict our attention *only to those
cases in which you chose door 1 and the host opened door 2*, what proportion
of *those games* have the car behind door 1, and what proportion have the
car behind door 3? The answers are now no longer equal.

First let's think intuitively. On average, for every 6 times you play the game and choose door 1, there will be 2 times when the car is behind door 1 (in which case the host might open either door 2 or door 3, so that means 1 time out of the 6 the host will open door 2, and 1 time out of the 6 the host will open door 3). Also, on average, there will be 2 times when the car is behind door 2 (in which case the host must open door 3), and there will be 2 times when the car is behind door 3 (in which case the host must open door 2).

So on average, for every six times you play the game and choose door 1, there will be

- one time when the car is behind door 1 and the host opens door 2
- one time when the car is behind door 1 and the host opens door 3
- two times when the car is behind door 2 and the host opens door 3
- two times when the car is behind door 3 and the host opens door 2.

*Out of all the three times when the host opens door 2, one of
them has the car behind door 1 and two of them have the car behind
door 3*. So, out of all the times when you choose door 1 and the
host opens door 2, on average 1/3 of those times have the car behind
door 1 and 2/3 of those times have the car behind door 3. That's why
switching gives you a 2/3 chance of winning.

The way this is formalized mathematically is as follows. Suppose
door 1 is your choice. Let *A* be
the event that the car is behind door 1, and *B* the event that
the host opened door 2. What we want is the conditional probability
of *A*, given *B*. This is the probability of (*A* and *B*) divided by
the probability of *B*.

The probability of *A* is one third (the car has a 1/3 chance of being
behind door 1). The probability of (*A* and *B*) is one-half the
probability of *A* (since when the car is behind door 1 the host might
open either door 2 or door 3, with equal probability). So the probability
of (*A* and *B*) is 1/6.

The probability of *B* is one half; we'll leave the proof ot
that to you as an exercise.

Therefore, the conditional probability of *A* given *B* is (1/6)/(1/2) = 1/3.

Similarly, if *C* is the event that the car is begind door 3,
the probability of *C* is one third, and the probability of (*C* and *B*)
is also one-third (since when the car is behind door 3 the host has no
choice but to open door 2).

Therefore, the conditional probability of *C* given *B* is (1/3)/(1/2) = 2/3.

That's a mathematical justification of the fact that switching gives you a 2/3 chance of winning, while sticking with your original choice gives you only a 1/3 chance.

Notice what would be different if the host did not know where the car was and simply opened a door that just happened to reveal a goat. In that case, we're asking for the probability that the car is behind door 1 given that the host opened door 2 and given that the door the host opened revealed a goat. If you work out that probability, it turns out to be 1/2.

This page last updated: May 26, 1998

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