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:

- There are three doors on stage. A car is behind one of them and a goat is behind each of the other two doors.
- The contestant chooses a door; he will win whatever is behind that door (either the car, or one of the goats).
- 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. - 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.

## One thought on “The Monty Hall Problem”