From the Exploring Data website - http://curriculum.qed.qld.gov.au/kla/eda/
© Education Queensland, 1997

Monty Hall Puzzle - Explanations of the Solution

One interesting aspect of this puzzle is that no one explanation seems to satisfy everybody. If you want to convince an entire class of skeptical students, you will need all of the solutions below, at least.

Explanation 1 - my favourite

The probability that the contestant chose the correct door initially is 1/3, since there are three doors each of which has an equal chance of concealing the prize. The probability that the door Monty Hall chooses conceals the prize is 0, since he never chooses the door that contains the prize. Since the sum of the three probabilities is 1, the probability that the prize is behind the other door is 1 - (1/3 + 0), which equals 2/3. Therefore the contestant will double the chance of winning by switching.

Explanation 2 - looking at an extreme case

Most people who get this puzzle wrong reason that after Monty reveals a losing door there are two doors left, one of which contains the prize, and therefore the probability of each concealing the prize is 1/2. This explanation dispels that line of reasoning.

Imagine that there were a million doors. Monty knows which door conceals the prize, so he then opens 999 998 losing doors. You are now confronted with two doors, the one you chose initially and the one Monty has left. Do they each have a 50% chance of concealing the prize?

Once someone is convinced by this argument, they are usually receptive to Explanation 1.

'Explanation' 3 - simulate the problem

This is not really an explanation, but is does seem to convince any remaining doubters. Simulate the game with some playing cards, and play the game, say 24 times, with the person not switching each time, and record the number of wins (about 8 wins on average). Then play the game the same number of times with the person switching each time, and record the number of wins (about 16 wins on average). That will dispel the notion that the odds are the same whether the contestant switches or doesn't switch, and give credence to the probabilities given in the solution.

You can also find simulations of the puzzle on the websites listed on the Monty Hall page.

Explanation 4 - conditional probability

Students studying conditional probability may prefer an explanation given in that genre. This explanation came from a web page maintained by Vinay Kashyap.

The a priori probability that the prize is behind door X, P(X) = 1/3

The probability that Monty Hall opens door B if the prize were behind A, P(Monty opens B|A) = 1/2

The probability that Monty Hall opens door B if the prize were behind B, P(Monty opens B|B) = 0

The probability that Monty Hall opens door B if the prize were behind C, P(Monty opens B|C) = 1

The probability that Monty Hall opens door B is then p(Monty opens B) = p(A)*p(M.o. B|A) + p(B)*p(M.o. B|B) + p(C)*p(M.o. B|C) = 1/6 + 0 + 1/3 = 1/2

Then, by Bayes' Theorem,

P(A|Monty opens B) = p(A)*p(Monty opens B|A)/p(Monty opens B) = (1/6)/(1/2) = 1/3 and P(C|Monty opens B) = p(C)*p(Monty opens B|C)/p(Monty opens B) = (1/3)/(1/2) = 2/3

In other words, the probability that the prize is behind door C is higher when Monty opens door B, and you SHOULD switch!

Extension Questions

The repeated simulation of this puzzle gives rise to some challenging problems involving the binomial distribution. For the following questions I am assuming that students have not yet come across the concept of confidence intervals and hence will use common sense in determining a probabilistic meaning for likely and almost certainly.

1. Assume I run the simulation 24 times, not switching each time, with P(not switching) = 1/3. Let the number of wins = w. What is the possible range for values of w? Within what range of values is w likely be? Explain your reasoning.

2. Repeat question 1, assuming that a switch is made each time, with P(switch) = 2/3.

3. Assume that the simulation is run n times, not switching each time, with P(not switching) = 1/3. What is the minimum value of n that will ensure that the proportion of wins will almost certainly be between .26 and .40? Justify your solution.