The Monty Hall Problem

05.08.10

The Monty Hall problem is a problem in probability theory which puzzled very highly educated mathematicians as well as the man in the street when it first was presented in a letter to the American Statistician in 1975. The reason for the confusion lies in the fact that the answer is somewhat counterintuitive. When first presented with the problem, most people go through a quick (and wrong) line of reasoning and promptly arrive at the wrong conclusion. Think about this when reading on. The problem can be formulated as such:

Imagine that you are on a game show called 'Guess where the car is' where the objective is to win a car. The game is played in the following way. You are presented with three equal doors. You know that a car is placed behind one of the doors while goats are placed behind the two others. You, of course, want to end up with the door which has the car behind it. A key point here is that the host of the game show knows where the car is.



Next you select a door. Then, without opening this door, the game show host opens one of the doors with the goat behind it. You are now given a choice. Do you still want to open the door you initially chose, or do you now want to switch to the other remaining door?

The question is then, should you switch to the other door, or not? Does it even matter?

It is tempting to think that it makes no difference whether you change door or not; the probability is 50/50 since there is a car behind one door and a goat behind the other, but the situation is not quite as simple. As this line of reasoning will show, changing door will significantly increase your chance of winning the car (66.67% change, actually). Now if you are into probability theory you can calculate this relatively easy, but I will try to convince the non-mathematician with a non-mathematical justification. Let us refer to the doors as door A, B and C and let the car be behind door B.

When playing this game there are essentially three different outcomes.

- First outcome
You pick door A, the game show host reveals door C. If you change to door B you win.

- Second outcome
You pick door B, the game show host reveals door A or C. If you change to door C or A, you lose.

- Third outcome
You pick door C, the game show host reveals door A. If you change to door B you win.

As you can see there are two outcomes where you win the car and one where you don't, which means that you, by always changing the door, have a 2/3 (or 66.67% percentage) chance of winning.

Now for the mathematical formulation:
Consider following discrete random variables:
$$C \in \lbrace 1,2,3 \rbrace, \,\, S \in \lbrace 1,2,3 \rbrace, \,\, H \in \lbrace 1,2,3 \rbrace$$
where C is the door-number hiding the car, S is the door-number which is selected and H is the door-number which the game show host opens. Right from the start we can say that
$$P(C) = \frac{1}{3}, \,\,\, P(C|S) = P(C)$$
since the placement of the car is random and the player has no knowledge of where the car is. Furthermore we have that
$$P(H|C,S) = \begin{cases}0 \, \mathrm{if} \, H = S \\ 0 \, \mathrm{if} \, H = C \\ 1/2 \, \mathrm{if} \, S = C \\ 1 \, \mathrm{if} \, H \neq S \, \mathrm{and} \, S \neq C \end{cases}$$
We can then apply Bayes' rule to calculate the probability as such
$$P(C|H,S) = \frac{P(H|C,S)P(C|S)}{P(H|S)}$$
where
$$P(H|S) = \sum_{C = 1} ^3 P(H,C|S) = \sum_{C = 1} ^3 P(H|C,S)P(C|S)$$
So if the player selects door 1 and the host opens door 3 the chance of winning (by switching) is:
$$P(C=2|H=3,S=1) = \frac{1 \cdot 1/3}{1/2 \cdot 1/3 + 1 \cdot 1/3} = \frac{2}{3}$$

It really is strange how our intuition can lead us astray sometimes. :)