## How 21 got the Monty Hall problem wrong

Thursday, May 7th, 2009 by NikiLast night I watched the movie 21, based on the book *Bringing Down the House*, about the MIT blackjack team. There is a scene in the movie in which math professor Mickey Rosa poses the Monty Hall problem to a student, Ben Campbell. Predictably enough Ben is able to figure out the correct strategy and realizes that switching doors yields a 2/3s probability of winning the car (which is roughly the same chance you’ll predict the entire plot of the movie within five minutes). The problem is, he’s wrong.

At first the error is subtle. There is not enough information about the behavior of the host to determine whether the best strategy is to switch or not. The problem is not constrained enough. This is made glaringly obvious when Mickey Rosa asks “how do you know he’s not trying to play a trick on you?” If there is the possibility that the host is trying to play a trick on you, then there is the possibility that the host only allows you to switch if you have already chosen the car. In that case, switching doors would result in a 0% chance of winning the car.

Only when you constrain the problem does it make sense to switch doors. If (and only if) the host always opens a door, the door always reveals a goat, and the host always offers the player the option of switching does switching result in winning the car 2/3s of the time. Wikipedia has a good analysis of this constrained version using Bayesian products. (The short answer, the host is essentially giving you the option of switching from your initial choice of one door, to the two remaining doors and if the car is behind one of those two, you win).

Of course if the host only gives you the option to switch when you’ve chosen the car it won’t take long for people to catch on, and everyone will win all the time. It makes more sense for the host to offer a switch when you’ve selected the car with probability p, and offer a switch when you’ve selected a goat with probability q. Thus the host opens the door with probability: .

Borrowing from the analysis on the wikipedia page, we see that:

We can now use this new piecewise function to determine the probability of winning the car by switching:

So:

If we let p = 1 and q = 1 (in other words, the host always opens a door) then the probability of winning the car by switching is 2/3s, which is what we should expect.

**Update**:

It makes the most sense for the host to always offer the switch when you’ve chosen the car (p=1), and to offer the switch when you’ve selected the goat with probability q. By solving the inequality:

We find that we should switch when q > 1/2.