It’s probability time again!

The discussions about the Tuesday’s Child problem were intense, so a break from probability puzzles was in order. But now that your brains are well rested, it is time to move on to the next puzzle from the Probability Paradoxes post: the Monty Hall problem. If you thought that the first puzzle was controversial, you ain’t seen nothing yet! Monty Hall has generated so much discussion (and, at times, acrimony), that there was enough material for someone to write a book on the subject.

In a possibly futile attempt to avoid some of that controversy, I will be taking a different approach to the discussion this time. I tried to come up with the “best” or “most reasonable” assumptions needed to solve the Tuesday’s Child puzzle, but in the case of Monty Hall I will side-step those sorts of questions and simply provide a framework for calculating solutions for a range of possible assumptions.

So here we go. Remember that the scenario we are presented with is as follows.

Suppose you’re on a game show, and you’re given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what’s behind the doors, opens another door, say No. 3, which has a goat. He then says to you, “Do you want to pick door No. 2?” Is it to your advantage to switch your choice?

I am going to jump right into the mathematics of conditional probability again. I will use “Car” to denote the possibility that door No. 1 (your door) has a car behind it and “Goat” to denote the possibility that it has a goat behind it. Without necessarily assuming Monty opens door No. 3, I will denote by R (for “reveal”) the possibility that Monty opens a door other than the one you have chosen and thereby reveals a goat.

With all of that in mind, I will compute the probabilities P(Car |  R) and P(Goat | R), that is the probability that you have picked a car or a goat respectively, given that Monty has shown you a goat behind another door. Even allowing for the goat being as good-looking as the one pictured here, I will also assume that you really want to win a car not a goat. On that basis, you should switch doors if P(Goat | R) is greater than P(Car | R) and stick with your original choice if P(Goat | R) is less than P(Car | R). If the two probabilities are the same, there is no advantage or disadvantage in switching.

Despite being a very easy theorem to prove, Bayes’ Theorem is one of the most fundamental results in probability and I will make use of it here.

The probability P(R) can be broken down in terms of the two mutually exclusive possibilities ‘Car’ and ‘Goat’:

Combining this with Bayes’ Theorem and the (fairly uncontroversial) facts P(Car) = 1/3 and P(Goat) = 2/3, we can conclude that

Tidying this up a bit gives us a formula that will be able to deal with many (but by no means all) possible assumptions for the problem.

Of course, once we have this, we also know P(Goat | R) as it is 1 – P(Car | R).

All we need to do to use this formula is come up with assumptions that can be formulated in terms of P(R | Car) and P(R | Goat). Here are some examples of what we can do.

Case 1 (Classical)

Here we assume that Monty knows everything in advance about where the goats and the car are and that he can and always will show you a goat behind another door. This means that

P(R | Car) = P(R | Goat) = 1

Plugging this into the formula gives P(Car | R) = 1/3 and P(Goat | R) = 2/3, which means that we are better off switching. This is the standard solution to the Monty Hall problem and may come as a surprise to many people as the most common response, even with the assumption of an omniscient Monty, is that it doesn’t matter whether you switch or not.

Case 2 (Nice Monty)

This time we will still assume that Monty knows everything, but now he really wants you to win the car. So, he will only open a door to show you a goat if you have made the wrong initial choice. If you picked the car in the first place, he will simply open your door and congratulate you on your win. This means that

P(R | Car) = 0 and P(R | Goat) = 1 and so P(Car | R) = 0 and P(Goat | R) = 1. Once again you should switch. This time this conclusion should be obvious.

Case 3 (Nasty Monty)

Now Monty still has all the facts, but would prefer you to pick a goat. Perhaps Monty loves goats or perhaps the show is losing money and he doesn’t want to fork out for another car. Either way, these assumptions mean that

P(R | Car) = 1 and P(R | Goat) = 0 and so P(Car | R) = 1 and P(Goat | R) = 0. Now you should stick to your guns. Again, this conclusion should be obvious.

Case 4 (Lucky Monty)

Now we will stop assuming that Monty knows everything. When he opens the second door, he has no idea whether he’s going to spoil everything by revealing the car. If he does that, the show’s segment would, of course, end a little sooner than he had hoped. If you have picked the car, he will be fine, but if you picked a goat, he has a 50/50 chance of revealing the car instead of a goat. Translating all of this into mathematics gives:

P(R | Car) = 1 and P(R | Goat) = 1/2 and this time P(Car | R) = 1/2 and P(Goat | R) = 1/2. So, if Monty just happened to be lucky and show you a goat, it will not make any difference if you switch or not. Your odds will be the same either way.

Many other possible assumptions about what Monty does or does not know or do can also be cast in terms of P(R | Car) and P(R | Goat) and for each of these possibilities, it is therefore straightforward to decide whether to stick or switch.

In the debate over the Monty Hall problem, some have argued that Case 4 is a “better” interpretation of the puzzle than Case 1. While I won’t comment on that, I am sure that my attempt to evade the issue will not be enough to keep everybody happy with this solution. It seems that the Monty Hall controversy will not die.

Update: for anyone not convinced by the formula for the classical case, this spreadsheet may help.

Possibly Related Posts (automatically generated):

• Probability Paradoxes (7 June 2010)
• Tuesday’s Child (8 June 2010)
• Eliminating the irrelevant (10 June 2010)
• Sleeping Beauty (26 August 2014)