Sunday, January 10, 2010

Monty Hall Problem

Here is the Wikipedia statement of the Monty Hall Problem along with its explanation for why people are puzzled by it.
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? (Whitaker 1990)
As the player cannot be certain which of the two remaining unopened doors is the winning door, most people assume that each of these doors has an equal probability and conclude that switching does not matter. In fact, in the usual interpretation of the problem the player should switch—doing so doubles the probability of winning the car, from 1/3 to 2/3.
The name of the problem is derived from "Let's Make a Deal," an American TV game show once hosted—and made popular—by Monty Hall. It placed people in the position described quite frequently.

Many people find it counter-intuitive to say that switching is advantageous. Here's my explanation for why it is.
You pick a door. Monty then say that you can either keep your selection or switch to both of the other doors. That is, you get to keep either what is behind your current door or what is behind the other two doors. The probability of winning the car is 1/3 if you don't switch and 2/3 if you switch. The fact that Monty is also willing to show you that one of the two other doors has a goat becomes meaningless. You already knew that.
I added this explanation to the Wikipedia page. It was erased by someone who claimed that a similar explanation was already on the page. I don't think the other explanation was as good mine, but I didn't bother complaining about it.

The other explanation did point, however, to a 1990 column by Cecil Adams that included my explanation. (And I thought I was original.) Here's his explanation, which is about 3/5 of the way down the (rather long) page and is essentially the same as mine.
Suppose we have the three doors again, one concealing the prize. You pick door #1. Now you're offered this choice: open door #1, or open door #2 and door #3. In the latter case you keep the prize if it's behind either door. You'd rather have a two-in-three shot at the prize than one-in-three, wouldn't you? If you think about it, the original problem offers you basically the same choice. Monty is saying in effect: you can keep your one door or you can have the other two doors, one of which (a non-prize door) I'll open for you.
So I guess I wasn't the first one to think of this explanation. But I still like it.

This video explains it well also.

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