In his Coding Horror Blog, Jeff Atwood writes about the Monty Hall Problem and some variants. The classic problem presents the situation in which the game show host allows a contestant to choose one of three doors, one of which opens to reveal a prize while the other two reveals goats. The host then opens one of the other doors, reliably choosing one that has a goat, and invites the contestant to switch to the remaining unopened door. The problem is to determine the probability of winning the prize by switching and staying. The variants deal with cases in which the host does not reliably choose a door with a goat, but happens to do so.

Jeff cites Monty Hall, Monty Fall, Monty Crawl (PDF) by Jeff Rosenthal, which explains why the variants have different probabilities in terms of the "Proportionality Principle", which the appendix acknowledges to be a special case of Bayes' Theorem.

One of Jeff's anonymous commenters presented the Monty Maul Problem:

Hypothetical Situation:

The Monty Maul problem. There are 1 million doors. You pick one, and the shows host goes on a bloodrage fueled binge of insane violence, knocking open doors at random with no knowledge of which door has the car. He knocks open 999,998 doors, leaving your door and one unopened door. None of the opened doors contains the car.

Are your odds of winning if you switch still 50/50, as outlined by the linked Rosenthal paper? It seems counter-intuitive even for people who've wrapped their head around the original problem.

If you take as absolute the problem's statement the host is randomly knocking doors open, then yes, the fact that only goats were revealed is strong evidence that only goats were available because you picked the door with the prize, which, when combined with the low prior probability that you picked the door with the prize, gives equal probability to either of the unopened doors having the prize.

However, the fact that only goats were revealed is also strong evidence that the host deliberately avoided opening the door with the prize, and therefor switching is a winning strategy. After all, the probability of this happening if the host really is choosing doors randomly is 2 in a million, but it is guaranteed if the host deliberately opened only doors with goats.

Note that this principal still applies in variants with fewer doors. Unless there is an actual penalty for switching doors (which could happen if the host only sometimes offers the opportunity to switch, and is more likely to do so when the contestant chooses the winning door), any uncertainty about the host choosing doors randomly implies that it is a good strategy to switch.

I'm told that game theorists often have arguments over a similar sort of situation:

In a competitive game with another person, it's often necessary to assume that they'll behave rationally in order to determine what your best strategy is. (If they'll choose randomly and erratically, the nature of the rules might make it impossible to determine which of the options is better.)

But what about the cases where there are, say, two strategies which are equally good if the opponent is rational but only one of which is better if they act irrationally? Is it consis... (read more)