The Monty hall problem is easy and simple, and does not deserve the deep confusion it has engendered. I am going to explain the problem, its two solutions, and the situations where they are true, such that a child could understand, in the span of 700 words (two pages of print). Starting… Now.
You are on a game show. Monty Hall presents to you three doors. Behind two are goats, behind the third a chic Italian convertible. At Monty’s behest, you mark one door. He then opens one of the other two, revealing a goat. The revealed goat wanders off. You must now choose between two remaining doors. Whatever door you choose, you get what’s behind it, and you are not hungry for goat. So which do you choose?
More precisely, what is the probability that the car is behind the marked door?
Some argue 1/2. There are two doors left. Each might have the goat. Fifty-fifty. Sure. Others argue there is a two-thirds chance that unmarked door hides the convertible because [complicated math].
Actually, neither answer is correct. It depends on whether Monty would have ever risked showing you the car.
He might not (that would kind of ruin the show). Let’s explore that case. You mark a door. Let’s name the door you marked A, and name the other doors B and C. There’s a 1/3 chance the convertible is behind any of the doors.
He then opens whichever unmarked door doesn’t have a car (let’s say he opens B). So, you have to choose between A and C.
This is the point where people get confused. Don’t.
Instead, rewind to before Monty opened door B. Suppose C hid a goat. Then, either A or B would have the car. If A had it, then maybe Monty would have opened B, maybe C, fifty fifty chance. If B had the car, then Monty would definitely have opened C. So, when C has a goat, three fourths of the time, Monty would open C.
Suppose, instead, the car were behind C. Then Monty would not have opened it.
Flashforward to now. C hasn’t been opened. B was. Originally, there was a 1/3 chance C hid a car and a 2/3 chance it hid a goat. But, if it held a goat, 3/4 of the time, it would have been opened. You know you’re not in one of those times, so you could only be in one fourth of the cases where C hides a goat, but you could be in any of the cases where C hides a car.
So, instead of C hiding a goat being twice as likely as C hiding a car, it is half as likely, because one fourth of twice is half. That is to say, there is a two-thirds chance that C hides a car. So you should choose C.
This is the standard solution.
But it’s not always true.
Suppose instead that Monty were a bad host, and opens an unmarked door totally at random. Maybe he could have showed you the convertible, let the game be ruined, who cares?
Again, assume you marked A, and he opened B, revealing goat.
Rewind to before Monty opened B.
Suppose C had the car. Fifty fifty chance he opened B instead of C.
Suppose C instead hid a goat. Fifty-fifty chance he opened B, but half of that time, B would have had a car behind it.
Flashforward again. Half the time C had a car, the world would be different from how it actually is (Monty would have opened C). Three fourths of the time C had a goat, the world would look different from how it actually is (Monty would have opened C or revealed a car when opening B).
Originally, C had a goat twice as often as it had a car, but only one fourth of those cases remain possible (the case where Monty chose B and it revealed a goat), compared to half of cases where C hides a car (the case where Monty chose B). One fourth of twice as often is the same as half of regular often (every other fortnightly meeting happens as often as every fourth weekly meeting). So, in your position now, seeing B open and its goat revealed, C hides a car equally often to C hiding a goat.
So it doesn’t matter if you choose A or B.
It all comes down to whether Monty is good at his job.
695 words.
(Actually, there’s a third, funkier possibility, where Monty doesn’t risk showing you the car but the car is equally likely to be behind either unopened door. We’ll get into that later.)
Indefensible Expectations 