# What is the Nash Equilibrium of the Monty Hall Problem?

The Monty Hall problem or paradox is famous and well-studied. But what confused me about the description was an unstated assumption.

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?

The assumption is that the host of the show does not have a choice whether to offer the switch. In fact, Monty Hall himself, in response to Steve Selvin's original formulation of the problem, pointed out that as the host he did not always offer the switch.

Because the host knows what's behind the doors, it would be possible and to his advantage to offer a switch more often to contestants who guess correctly. If he only offered the switch to contestants who guess correctly, all contestants who accept the offer would lose. However, if he did this consistently, the public would learn not to accept the offer and soon all contestants who first guess correctly would win.

If, for instance, he gave the offer to one third of incorrect guessers and two thirds of correct guessers, 2/9 contestants would be given the offer and should not switch and 2/9 contestants would be given the offer and should, which would bring the chances of winning back to 1/2 whether one accepts the offer or not, instead of 1/3 or 2/3.

Is this a Nash equilibrium for the Monty Hall problem (or the iterated Monty Hall problem) as a two-player zero-sum game? And if not, what is it, or aren't there any?

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The car probably doesn't come out of the host's salary, so he probably doesn't really want to minimize the payoff, he wants to maximize the show's ratings. But OK, let's suppose he did want to minimize the payoff, making this a zero-sum game. Then the optimal value of the game (in terms of the probability of winning the car) would be $1/3$. An optimal strategy for the contestant is to always refuse to switch, ensuring that the expected payoff is $1/3$. An optimal strategy for the host is never to offer a switch unless the contestant's first guess is correct, ensuring that the expected payoff is no more than $1/3$.

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Awesome, now what about the iterated version? –  Eric Mickelsen Feb 18 '13 at 2:55
The "iterated version" of a zero-sum game always has the same equilibrium as the "single-play" version. –  Robert Israel Feb 18 '13 at 6:14