mgroves

The Monty Hall problem seems very unintuitive–like a paradox.

There are 3 doors–behind two of them are goats, and behind one is a brand new car. You choose 1 door. Monty Hall then opens one of the other doors and shows you one of the two goats. You can now switch to the other door or stay with the current one. Which one do you choose?

Intuitively, you might think it doesn’t matter. There’s two doors left, one has a goat. 50/50 chance, right?

Wrong. Monty has revealed some information to you, so if you make the switch, you will have a better chance of winning the car. Think about it as if there were 3 initial choices, and 1 secondary choice. Consider the case where the car is behind door 1, goat 1 is behind door 2, and goat 2 is behind door 3:

• You pick door 1 and stay with door 1: You win a car.
• You pick door 2 and stay with door 2: Goat.
• You pick door 3 and stay with door 3: Goat.

Now, if you switch:

• You pick door 1, switch to door 3: Goat.
• You pick door 2, switch to door 1: You win a car.
• You pick door 3, switch to door 1: You win a car.

So, if you don’t switch, you have a 1 in 3 chance of the winning the car. If you do switch, you now have a 2 in 3 chance of winning the car. Crazy, isn’t it?

Need more proof? Here is a simulation of the Monty Hall game. As of this writing, it has been played 6109 times. Those who switched have won 66% of the time and lost 33% of the time exactly. Those who didn’t switch have won 34% of the time.

Now, here’s where it gets a little weird. Suppose the game was simply 2 doors–one a goat, one a car. Then you would have 50/50 odds no matter what. This shows that the additional probability of winning comes from Monty giving us extra information about what was behind door 3.

This doesn’t mean that you will win the car every time. What it does mean is that on average, you will win the car every 2 out of 3 times you play the game as long as you make the switch every time.