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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.


marked as duplicate by Steven Stadnicki, Leucippus, Chris Godsil, JonMark Perry, Hans Engler May 11 '18 at 2:21

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ You have a $2/3$ chance of being wrong on your first guess. Each time you are wrong and switch to the remaining door, you will win. $\endgroup$ – zhw. May 10 '18 at 21:52
  • 1
    $\begingroup$ This question has been asked and answered countless times on this site, as well as a bunch of other places - if there's a specific place where you're confused then please give more details; otherwise, this question just seems to me like a duplicate of the various others. $\endgroup$ – Steven Stadnicki May 10 '18 at 22:03
  • $\begingroup$ It is essential that the problem state that the host (Monty Hall) knows which door hides the car and thus always picks a door that reveals a goat. $\endgroup$ – David G. Stork May 10 '18 at 23:34

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.

  • $\begingroup$ I get what you mean $\endgroup$ – Benny May 10 '18 at 22:45
  • $\begingroup$ Makes sense, cheers $\endgroup$ – Benny May 10 '18 at 22:45
  • $\begingroup$ Glad I could help. $\endgroup$ – TheLeogend May 10 '18 at 23:15

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...

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: $1-1/3 = 2/3$.


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 $1/3$.


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