How does the Monty Hall Problem work?

Question

So I was reading a few maths problems and I stumbled upon the Monty Hall problem. If you are not familiar with the Monty Hall problem it goes something like this:

You are on a game show (with the objective of winning a car) and you have three doors A, B and C. Each door has one of two goats or a car behind it. You pick a door, let's say A, and before revealing what's behind A. Monty shows you what's behind C - a goat. Now do you switch your door or keep with your door?

I initially thought that it's 50-50 chance of winning if you switch, however, the answer is 2/3, and I have tried to research into this, but couldn't find an answer that explained this clearly. Please, can someone explain why the probability of winning the car is 2/3 rather than 50-50.

Answer 1

If none of these intuitive answers convince you, let's do some concrete probability solving.

With $2/3$ chance you pick a door with a goat at the beginning. If you switch, you have a 100% chance of winning because Monty will have to reveal the remaining door with the goat.

With 1/3 chance you pick the door with the car behind it. Now if you switch, then you will definitely lose, because the two remaining doors both have goats behind them.

So you can see you will win $2/3$ of the time if you switch.

Answer 2

It's hard to say what kind of explanation you will find compelling if you don't point to what you've already read. But at any rate...

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Because there are two goats and only one prize, Monty can always show you a goat in one of the two doors that you didn't choose. Therefore, revealing the goat gives you no additional information about the door you chose.

Since you initially had a $1/3$ chance of choosing correctly, with no increase in information, that's your chance of winning if you stay with the same door. Therefore, the probability of winning if you switch must be what remains: $1-1/3 = 2/3$.

Answer 3

The problem becomes more clear if we consider, for example, $100$ doors with one car behind.

In this case we have

• $99/100$ chance to be wrong
• $1/100$ chance to guess the door with the car behind

Then in $99\%$ of the cases changing the door will be a good strategy to win the car.

With three door the same explanation holds but the chances are respectively $2/3$ and $1/3$.