Brian Hayes recently provided some evidence that there are still many out there who are confounded by the Monty Hall problem.

The Monty Hall problem, perhaps the best-known counter-intuitive probability problem, gets a nice treatment in Jeffery Rosenthal’s *Struck by Lightning: The Curious World of Probabilities*, and is also explained well (perhaps better) in Mark Haddon’s novel *The Curious Incident of the Dog in the Night-time*. Professor Rosenthal has some further Monty Hall explanations here.

I just found an alternate version of the problem in Martin Gardner’s *The Second Scientific American Book of Mathematical Puzzles and Diversions *(published also by Penguin in a slightly different form as *More Mathematical Puzzles and Diversions*). In the chapter “Probability and Ambiguity” (chapter 19 in both versions of the book), Gardner describes the problem of the three prisoners. Here is a condensed description of the problem:

Three prisoners, A, B, and C are in separate cells and sentenced to death. The governor has selected one of them at random to be pardoned. Finding out that one is to be released, prisoner A begs the warden to let him know the identity of one of the others who is going to be executed. “If B is to be pardoned, give me C’s name. If C is to be pardoned, give me B’s name. And if I’m to be pardoned, flip a coin to decide whether to name B or C.”

The warden tells A that B is to be executed. Prisoner A is pleased because he believes that his probability of surviving has gone up from 1/3 to 1/2. Prisoner A secretly tells C the news, who is also happy to hear it, believing that his chance of survival has also risen to 1/2.

Are A and C correct? No. Prisoner A’s probability of surviving is still 1/3, but prisoner C’s probability of receiving the pardon is 2/3.

It is reasonably easy to see that the 3 prisoners problem is the same as the Monty Hall problem. Seeing the problem in this different formulation might help those who continue to struggle with it.

It is a nice activity to simulate both the 3 prisoners problem and the Monty Hall problem in Fathom – try it and confirm the surprising results that the “second prisoner” is pardoned 2/3 of the time, and that 2/3 of the time, the winning curtain is not the one you selected first.

There are many, many, ways to write these simulations. Here are the attributes and formulas for a Fathom implementation of the three prisoners:

The table below (click on it to see a larger version) shows a separate simulation for the Monty Hall problem. Here we are assuming three curtains “1”, “2”, and “3”, one of which has a prize behind it. You pick one, and then Monty reveals the contents behind one of the other curtains (the curtain with the prize behind it is not shown). In the game, you have the option of switching your choice for the curtain that has not been revealed.

After creating the attributes, you can “run the simulation” by adding data to the collection (Collection->New Cases…), the more the better.

Incidentally, Gardner’s use of A, B, and C reminds me of Stephen Leacock’s “A, B, and C: The Human Element in Mathematics.”

**Addendum**

A quick search shows that the connection between the Monty Hall problem and the Three Prisoners is well known (see the wikipedia entries on Monty Hall and the Three Prisoners), and that both are alternate formulations of an older problem, known as Bertrand’s box paradox.

For more such insights, log into www.international-maths-challenge.com.

*Credit for article given to dan.mackinnon*****