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The Monty Hall Problem

| May 16th, 2008

I assume that most of you have watched the movie ‘21‘. Well if you haven’t, I recommend that you watch it. Around 20 minutes into the movie there is a scene where the professor poses a problem in front of the protagonist:

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? [via Whitaker, Craig F. (1990). [Letter]. “Ask Marilyn” column, Parade Magazine p. 16]

The protagonist chooses to switch. It is not very intuitive as to why one should deviate from one’s original choice. It seems as if the probability of winning the car just bumped up to 1/2 from 1/3 after the host opens a door with a goat behind it. But it can be empirically proven that switching the choice is the rational strategy for the player. This ubiquitous game show dilemma is the Monty Hall problem.

Let me try to prove why is it so.

The first column represents the probable contents of the door originally chosen by the player (assume door 1). The second column represents the contents of the door opened by the host (assume door 3). The last column shows the contents of the other door, which the player chooses if he makes a switch. It can be clearly seen in the table above that probability of winning the car changes from 1/3 (as in the first column) to 2/3 (as in the last column). This probability can be observed if do some kind of random sampling.
The above is valid only for scenarios like the problem above where a player is given some choices and some choices are revealed based on which he has to make a decision.

Posted on Friday, May 16th, 2008 at 1:34 pm in Algorithm | rss feed for comments| Leave a response, or trackback