# The Monty Hall Deception

**Posted:**March 1, 2013

**Filed under:**Bayesian Games, Extensive-Form Games, Game theory blog, Incomplete information |

**Tags:**Bayesian statistics, game shows, Monty Hall Leave a comment

When I was in middle school, I consumed a lot of typical nerd literature like Richard Feynman’s “Surely You’re Joking, Mr. Feynman” and anthologies of mathematics puzzles from the Scientific American by Martin Gardner. In the latter, I first encountered the Monty Hall Problem, and it goes something like this:

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?

It turns out that, yes, it is always to your advantage to switch your choice. This is a solution that has been notoriously difficult for people to wrap their heads around. After all, when you picked a door, the probability of having picked the door with a car was still 1/3, and after a door was opened, there would still be a car and a goat behind the remaining two doors – it seems as through the probability of choosing the door with the car ought to be ½ regardless of the door chosen.

The Monty Hall Paradox is in fact not a paradox at all, but rather just some clever sleight of hand. The trick is that people are drawn to the fact that there are only two doors rather than three doors remaining, and assume that the host’s having opened a door is favorable to the player. People tend not to realize that the game has imperfect information – the player does not know where on the game tree he is, whereas the host does. Additionally, people assume that the host has no stake in the game (and this is not unreasonable, because the problem does not explicitly describe a parsimonious host! On the other hand, intuitively, we know that the host isn’t going to ruin the game by opening the door with the car.) So, if we assume that the host is profit maximizing and we model the problem as an extensive form game with imperfect information, then the conditional probabilities would be easy to see.

Now, just for fun, we’ll assign some utilities to the outcomes. What is a goat worth? According to a popular Passover song in Aramaic, a (small) goat is worth about 2 Zuz, and according to the traditional Jewish prenuptial document, a wife is worth about 200 Zuz. So, a goat is worth about 1/100^{th} of a wife. I asked my roommate, Anna, how many cars she thought a wife was worth, and she determined that a wife was worth three cars. By transitivity, then, a car is worth about 33 goats. (I think goats have become quite a bit less valuable since that song was written, or maybe goats back then were a lot better than our goats.) So, if the player wins the game, he will walk away with a utility of 33, and the host will walk away with the 2 goats.

In this game, the light gray branches are dominated because the host has no incentive to open the door that the player has already chosen, and the dark gray branches are dominated because, of the remaining two doors, the host would not open the door that has the car. We can tell that in the top branch, the host as 2 possible choices for doors to open, whereas in the lower two branches, the host is constrained to only one door (since, if the player has chosen a goat door, there is only one goat door left to open.)

So, since the player has no idea after the first stage what door has the car, we assume he picks door No.1 (as in the game). If he observes that the host opens door 3, he would know that there are two cases where the host opens door 2: in the world where the car is behind door 2, the host chooses door 3 100% of the time, and in the world where the car is behind door 1, the host chooses door 3 50% of the time. It’s actually twice as likely that we are on the 100% branch as that we are on the 50% branch – and that’s the branch where the car is hidden behind the other door.

What if we know that the host has opened a door, but we don’t know which one? Then, we can’t condition on a prior, because we don’t know what the prior is – we don’t get any new information by observing which door was opened, and switching doors would not help.