The Monty Hall Problem

The Monty Hall Problem is a statistics problem that can be solved using your A-Level Maths knowledge of probability trees, but that didn't stop it being the subject of much debate amongst some very clever mathematicians. Even now, not everyone agrees on the answer...

The Problem

You're on a game show where you can win a car, or one of two goats. The car and goats are hidden behind a door each. You select a door from one of the three. The game show host, who knows what is behind every door opens one of the unpicked doors to reveal a goat. The host gives you the choice of sticking with the door you have already picked, or swapping to the other remaining open door.

What should you do to stand the best possible chance of winning the car?

Show Solution

The Solution

The answer is that you should swap to the other unopened door. This may sound a bit unusual, and often catches many people out, but you can actually double your chances of winning the car by switching doors.

There are many different ways to tackle The Monty Hall Problem, but perhaps the simplest way and one that you should be familiar with from any Statistics modules you are doing for A-Level Maths is that of using probability trees.

The figure below shows the full probability tree for this problem, assuming you pick Door 1 to start with:

The probability tree above shows the spread of outcomes possible if you pick Door 1. As can be seen there is a ??\frac{1}{3}?? chance that the car is behind Door 1, in which case switching doors, gives up the car and gives you the goat. However, there is ??2 \times \frac{1}{3}?? chance that if you pick Door 1 that the car is behind a different door, in which case, swapping doors gives you the car (as the host always opens the other door that hides a goat). Essentially ??\frac{1}{3}?? of the time sticking is beneficial and ??\frac{2}{3}?? of the time swapping doors is beneficial.

Furthermore, considering the cases where you pick Door 2 or Door 3 at the start, yield analagous situations, i.e. in each case switching is beneficial ??\frac{2}{3}?? of the time and sticking with the door you have is beneficial ??\frac{1}{3}?? of the time.

Therefore, it is, in general, twice as good to switch doors, when given the chance, than to stick with the door you have picked.

This may sound a little counter-intuitive, but one way to convince yourself is to imagine that the same game is played with, say, 100 doors. You pick a door, and then the host opens 98 doors revealing a goat behind each, and asks whether you wish to swap doors or stick with your original choice (which originally only had a ??\frac{1}{100}?? chance of being the door with the car behind it).