The Monty Hall Problem

I love statistics problems even though they frustrate me more often they entertain me; perhaps the elation from my rare successes overwhelms the despair from my myriad failures. In any case, the Monty Hall problem is my favourite statistical frustration!

Digression: I hope that my kids aren’t reading this blog post. They would mock me mercilessly for thinking that someone could have a favorite statistics problem.

It’s pretty simple:

  1. There are three doors on stage. A car is behind one of them and a goat is behind each of the other two doors.
  2. The contestant chooses a door; he will win whatever is behind that door (either the car, or one of the goats).
  3. After the contestant chooses a door and before it is opened, Monty will open one of the unchosen doors and show him that it was hiding a goat. Important: Monty will always open a door that has a goat behind it. He will never open the door with the car.
  4. Monty will then offer the contestant a chance to change his guess and choose the other closed door.

Should the contestant change his choice of doors?

The obvious answer is that it doesn’t matter since the probability of the car being behind any door is 1/3, so switching doors wouldn’t change his chances.

Spoiler: This is a frustrating statistics problem and so by definition the obvious answer is wrong!

The correct answer is that the contestant should change doors; his odds of winning will double! From 1/3 to 2/3!

Why? Ahhh, there’s the rub! It’s not obvious. Here’s the explanation that makes the most sense for me.

  • Assume the contestant picks door A.
    • The odds that the car is behind door A is 1/3.
    • The odds that the car is behind one of the other two doors (i.e. B or C) is 2/3.
  • When Monty shows the contestant that there is a goat behind either door B or door C, the odds that the car is behind one of the other two doors (i.e. B or C) is still 2/3. The door with the car doesn’t change just because Monty opened another door.
  • But now there is only one door left in that pair of doors that the contestant didn’t choose, so there must be a 2/3 chance that the car is behind that door!

Historical note: I watched this show as a kid and recall thinking that Monty was cheating, and would only open the door to show a goat if the contestant had already picked the door with a car. Not True! He would always open one of the doors. I belatedly apologize for thinking ill of him.

Another perspective that some people find useful is to consider a different contest. Suppose there are a million doors and only one of them hides a car. When the contestant chooses a door, there is only one chance in a million that he picked the door with the car. The odds that the car is behind one of the other doors is 999,999/1,000,000. In this contest, Monty opens 999,998 doors of the unchosen 999,999 doors, leaving one other door closed.

Should the contestant change his choice of doors?

Of course he should! There is only one chance in a million that the contestant guessed right the first time; changing his choice will increase the probability of winning a car from 1/1,000,000 to 999,999/1,000,000.

Confession: I don’t find this perspective particularly illuminating even though I understand the logic and would change my choice in this contest in a heartbeat. For some reason I don’t think that the two contests are similar enough to generalize from one to the other.

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