The Monty Hall Problem

The Setup:
You're on a game show and are given the choice of three doors: Behind one door is a car; behind the other two, 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?"

Before you go off trying to solve this, there are a few things to remember:

• the player may initially choose any of the three doors (not just Door 1) and the host may open any different door revealing a goat (not necessarily Door 3).
• the host is constrained to always open a door revealing a goat.
• the host is constrained to always to make the offer to switch.
• the host opens one of the remaining two doors randomly if the player initially picked the car.

The Problem:
Is it to your advantage to switch your choice?

The Solution:
Yes, you should switch—doing so doubles the probability of winning the car from one in three to two in three.

The Explanation:
In contrast to Blue Eyes, the Monty Hall problem is explained by raising the numbers rather than reducing them. Imagine instead of three doors, there were three hundred doors but still only one contained a car. After you selected your door, let's say No. 1 again, the host opened two hundred and ninety-eight doors, all with goats behind them, leaving only the door you chose and one other, let's say No. 164. Now intuition tells us to switch because there's a 299/300 chance our initial guess was wrong and the other door holds the car.
The lower numbers in the Monty Hall problem conceal the fact that the host is not choosing one door to open but choosing one door to keep closed (the car). The only time you lose by switching is the 1 in 300 time when you actually guessed correctly to begin with and the host has selected a random door. Likewise, the only time switching loses in the Monty Hall problem is the 1 in 3 time when you guessed the correct door initially.