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The Infamous Monty Hall

A (Brief) History:

The Monty Hall problem is arguably the most infamous probability puzzle in recent history. It was originally proposed in its current form in 1975, but only really surged into the public spotlight in 1990 when in appeared in a Parade column written by Marilyn vos Savant. For more details on the history of this problem see Wikipedia: Monty Hall problem.

The Original:

As given in Marilyn's column the problem read:
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 [but the door is not opened], 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 phrasing here is ambiguous. Does Monty always show you a goat? Does he only show you a goat when switching would lead to a win? Does he only show you a goat when switching would lead to a loss? In fact, Monty could assign you any probability \( p \) of winning when switching that he wanted. All Monty needs to do is with probability \( p \) only show you a goat if switching wins; otherwise, he'd only show you a goat if switching loses. Some of the unstated assumptions undoubtedly led to additional confusion in what's already a conceptually difficult problem for many people.

Unambiguous Phrasing (Krauss and Wang 2003):
Suppose you're on a game show and you're given the choice of three doors [and will win what is behind the chosen door]. Behind one door is a car; behind the others, goats [unwanted booby prizes]. The car and the goats were placed randomly behind the doors before the show. The rules of the game show are as follows: After you have chosen a door, the door remains closed for the time being. The game show host, Monty Hall, who knows what is behind the doors, now has to open one of the two remaining doors, and the door he opens must have a goat behind it. If both remaining doors have goats behind them, he chooses one [uniformly] at random. After Monty Hall opens a door with a goat, he will ask you to decide whether you want to stay with your first choice or to switch to the last remaining door. Imagine that you chose Door 1 and the host opens Door 3, which has a goat. He then asks you "Do you want to switch to Door Number 2?" Is it to your advantage to change your choice?

Standing, of course, gives us a \( 1/3 \) probability of winning. But what if we switch? Well, the car is behind one of the doors, and if it's behind Door 1 with probability \( 1/3 \) it must be behind one of the remaining two doors with probability \( 2/3 \). But it's not behind Door 3 because Monty just showed us a goat. Aha, so the probability that the prize is behind Door 2 must be \( 2/3 \) now, so we should switch! Another way of seeing that this is correct is to pretend that you close your eyes and cover your ears right before Monty Hall reveals the goat. Now by switching you'll win the prize if it's behind either Door 2 or Door 3, so your probability of winning increases from \( 1/3 \) to \( 2/3 \) by switching.

My next entry will be a solution using Bayes' Theorem.


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Standard notation:

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