Thursday, February 17, 2011

The Monty Hall Problem

Suppose you're on a game show, and you are given the choice of three doors: behind one door is a car; behind the other two are goats. Naturally your goal is to win the car, which is equally likely to be behind each door.

You pick a door (e.g. #1), and the host (who knows what's behind each door) opens another (e.g. #3) which has a goat. This means there are two doors left, one with a car and one with a goat. You are now given the opportunity to switch your choice (from #1 to #2). Is it to your advantage to do so?

Since you can't know which of the two remaining doors has the car, and since your initial pick had a chance of one-third, you might think that it does not matter. The chance is still 1/3 and you might as well stay with your original choice, right? Wrong! Switching actually doubles your chances to 2/3.

The Monty Hall problem is loosely based on the American television game show ‘Let's Make a Deal’. Monty Hall was the show's original host. So what's the deal here?

Let's make a list of the possible scenarios. There are three possible arrangements of the car and goats. The table below shows what happens when door 1 is initially picked — note that the result is the same when door 2 or 3 is the initial pick.

Door 1Door 2Door 3On switchOn stay

The table doesn't lie: when staying with your initial choice you win only one out of three times, while switching will result in a win two times out of three. Remember that Monty will never open the door with the car behind it. Him opening a door is thus an additional piece of information that skews the odds in favor of switching.

The easiest way of looking at it is that switching is only a disadvantage when you initially pick the car. That chance is 1/3, so that logically means that switching is an advantage in the remaining 2/3rd of the cases.

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