## 2011/08/31

### The Full Monty (Hall)

Warning: this post contains some mathematics. Please do not be afraid of it.

The so-called Monty Hall problem is fascinating in its simplicity coupled with a completely unobvious answer. It is a very good example of how people make probabilistic errors in making decisions. Even after confirming the maths, I still could not quite get it. My intuition is fighting with my reason and I cannot help but wonder if people routinely make these errors, whether I do.

Statement

The Monty Hall problem is based on a game show hosted by Monty Hall. The problem can be stated as follows (my own wording):

You are shown three boxes and must pick one. One of them contains a prize, the others are empty. The box containing the prize has been chosen randomly. After you have chosen a box the host will show you the contents of one of the other two boxes (it will be empty). You are then given the option of either changing your choice (and choose the remaining closed box) or sticking to your original choice. What is the probability of winning if you switch versus if you do not? Assume that if both of the unchosen boxes are empty one will be chosen at random to be opened.

The “obvious” answer, the one most people choose, is a half. That is, does not matter whether you switch or stay, the prize surely has an equal probability of being in either box.
Wrong. The probability of winning if you switch is two thirds and only one third if you do not. This is counter-intuitive. I myself, believing and having seen the maths, still find it hard. By showing the contestant one of the boxes Monty gives the contestant extra (useful) information. However, it would appear that this is information that our brains were not made to process very well.

Preparatory

Before we solve the problem, let us set up some things we will need. Let P be the box with the prize. P is 1,2 or 3 with probability one third. Let C be the box initially chosen (we will assume this is box 1). Let S be the box that Monty chooses to show to the contetant.

As I phrased the problem above, the solution is the probability of winning if you switch, given that box S was shown, which can be put in maths as (assuming box 2 is shown)
$Pr(P=3 |S=2).$

One thing to note is that the solution is NOT the probability of the prize being in box 3 given that is is not in box 2 or
$Pr(P=3 | P \neq 2)$
This is, in fact, a half, but we have been given more information than just that the prize is not in box 2 (even if we have trouble seeing it).

We could also decide to calculate the probability of winning given that the contestant will always switch. That is, we find the probability of winning before we know which box is shown to be empty, knowing the contestant will switch. This is the probability of winning with box 3 (if box 2 is shown) plus the probability of winning with box 2 (if box three is shown) Remember the contestant switches away from box 1. This can be shown in maths as
$Pr(P = 3 | S = 2) Pr(S = 2) + Pr(P = 2 | S = 3)Pr(S = 3).$

There is a subtle difference between these two formulations, which adds to the confusion. Lets call the first problem A and the second one B. Given the way I formulated the problem, most crucially the very last line, the two answers are the same. However, A’s answer will change if I modify the last line and B will remain at two thirds.

Solution 1

A very simple solution to problem B is given by drawing a table of the variables. We have P for prize and E for empty. Remember we assume the contestant chooses box 1 initially.

 Box 1 Box 2 Box 3 S Result on switch P E E 2 or 3 lose E P E 3 win E E P 2 win

Each of the three configurations of the boxes is equally likely and thus it must be that, if you switch (always) you win with probability two thirds.

Solution 2

A, perhaps, simpler way to solve B is to observe that the contestant loses if and only if they initially choose the car. This happens with probability one third. So the probability of winning must be two thirds.

Solution 3

An intuitive explanation for the result is to consider the two unchosen boxes together. These boxes must, together, contain the prize with probability two thirds. So switching will give the combined contents of the two boxes with probability two thirds (as one box will be revealed and the other can then be chosen).

Note that this does NOT mean that by revealing that one of the two boxes is empty, the other must therefore now contain the prize with the same (two thirds) probability. This explanation is appealing, but it is incorrect. Merely revealing which of the two remaining boxes is empty is not enough. We also need to know how the empty box was chosen (the last sentence of my problem statement). I will demonstrate this in the next solution.

Solution 4

Let us try a somewhat longer and comprehensive algebraic solution that will allow us to fully distinguish between A and B. First let us solve problem A.

By Bayes Theorem:
$Pr(P = 3 | S= 2) = \frac{Pr(S = 2 | P = 3)Pr(P = 3)}{Pr(S = 2)}.$

We know
$Pr(S = 2 | P = 3) = 1$
as Monty must show the empty box if there is a prize in three and
$Pr(P = 3) = \frac{1}{3}$
by definition.

Now to get the probability that box 2 is shown we can add the probability that it is shown and the prize is in box 1, 2 and 3 respectively.
$P(S = 2) = P(S =2 | P =1)Pr(P=1) + Pr(S=2|P=2)Pr(P=2)+P(S = 2| P =3)Pr(P=3).$

The middle term is zero as Monty would not show us the box with the prize and the second factor of each term is one third. This gives
$P(S =2) = \frac{1}{3}(Pr(S = 2| P =1) + 1).$

The key term is
$Pr(S = 2 | P =1).$
Let us call it q for now. This term explains how Monty chooses which empty box to show us. In the standard formulation q is 0.5 , meaning Monty chooses randomly. If q is 1 then Monty will always show the second box, if it is open (we could say Monty prefers to open the leftmost box)1.

The answer we are looking for is then
$Pr(P=3|S=2) = \frac{1}{q+1}.$

This ranges from ½ when q = 1 to 1 when q = 0. The value is two thirds in the standard case where q = ½.

When q =1 we get what is the intuitive answer. Each of the remaining two boxes is equally likely to have the prize. When q = 0, we know that Monty would not open box 2 if given a choice. So, the fact that box 2 is open means the prize must be in box 3.The fact that the probability of winning is always at least a half, means that it never hurts the contestant to switch.

What’s interesting is that the answer to problem B does not change, no matter what q is. This means that for a person playing the game, the probability of winning given which box is shown may vary, but on average, a person who always switches, will win two thirds of the time. We can easily show this by calculating
$Pr(P = 3 | S = 2) Pr(S = 2) + Pr(P = 2 | S = 3)Pr(S = 3).$

Let us do a simple numerical example. Suppose q is 1. So we know that if box 2 is opened there is probability 0.5 of winning. However, if box 3 were opened instead we would KNOW that the prize is in box 2. Monty would not have opened box three if box 2 were open. So we win with probability 1. However, before we know which box is opened, the probability of winning is
$\frac{1}{2}\times\frac{1}{3}(1+1) + 1 \times\frac{1}{3 \times 1} = \frac{2}{3}.$

In general both terms of the expression for B are easily seen to always be one third 1/3. By similar calculations to that I already did Pr(P = 2 | S = 3) can be found and the expression calculated. This gives the solution.

Why do we get it wrong?

The answer, probably, has something to do with our brains. Perhaps we are just bad at intuiting probabilities. As few as 13% of people choose to switch. People appear to have a tendency to think probability is evenly distributed across any possible outcomes.

The status quo bias may also partially explain it. People tend not to change something unless there is a very compelling reason to do so. People may also tend to value something more highly once it is their property or they have a right to it. This is the endowment effect and is a form of the status quo bias. Here it is that people value the box they already have more than the others.

I think, possibly, we have evolved (or been taught) to stick to things. So if we start something, we have to finish it. If we make a decision now, and later change our minds, we’re seen as fickle, undependable. And so, perhaps, from this we have a tendency to believe that our initial choice is correct. So we don’t switch.

Markets

I find the Monty Hall problem interesting because similar reasoning in the markets (which are so much more complex) means traders might make irrational decisions. The market will not be efficient. The irrationality of the traders can (potentially) be exploited by others (possibly computers) who notice it and who do not make the same errors. However, whether these kinds of errors are easy to detect is not clear and they may not be nearly as simple as this little problem.

Some references

On the problem