Posts Tagged ‘statistic’

Intuition, and the Monty Hall Problem

Thursday, May 22nd, 2008

Statistics is hard. It’s one of those fields, where your intuition seems to take you gently and seductively towards safety, and then you wake up in a bathtub full of ice, missing a kidney. Interface design is often the same way (but, for once, that’s not what I’m writing about). When we say that we find something “intuitive” it means that we have seen something like it before—we are saying that it is familiar. The thing with statistics is that similar looking situations can behave very differently.

The first time I had this point driven home was when I was presented the Monty Hall problem. Posit: You are on a game show and three doors stand before you. Behind two of the doors lies nothing, behind one door is a pile of cash. The game begins by you picking one of the doors. The host (Monty Hall) then opens one of the two remaining doors, revealing one that is empty. Then it’s your choice: Do you stick with the door you originally picked, or do you switch to the other door? Does it matter?

Intuition says is that it doesn’t matter. It’s 50%-50%. It’s the same as throwing a coin—the money is in either one door or the other. This kind of thinking will lose you a kidney. You’ve been tricked by “intuition” to not got suckered by the gamblers fallacy, only to get suckered by something new.

It’s best to switch doors. In fact, switching doors will make you win twice as often as staying. While I’ve been able to prove that for a long time—by enumerating all possible outcomes and counting&mash;it never really clicked in my head why it’s true. It never became familiar. It never became intuitive. Until a couple days ago. Here’s the explanation the occurred to me (in the shower, as always) that made it all make sense:

The chance that you pick the right door on the first choice is 1/3: there are empties and one prize. Conversely, that means that there is a 2/3 probability that prize is in one of the other two doors.

3 closed doors.

If you could open the other two doors, see which ones contains the money, and take it, you would have clearly do that. That’s twice as good as getting to open only one door! Well, when Monty opens one of those doors that doesn’t contain the prize, he’s effectively giving you that ability: there is still a 2/3 probability that the prize is behind the set of doors you didn’t pick.

It’s obvious, now, that switching will give you a 2/3 chance of winning. By opening a door that doesn’t contain a prize, Monty has effectively concentrated that 2/3rds probability into the one remaining door.

This way of looking at it makes it really easy to calculate similar problems. For example, take the 5 door case: you pick a door, Monty opens 3 empty doors, and you have to choose if you should switch. Your first pick has a 1/5 chance of getting the prize, which means that there is a 4/5 chance that it is in the set of doors you didn’t pick. Monty opens 3 of those 4 doors, which means that 4th unopened door takes that full 4/5 probability. Thus, switching is 4 times more likely to win than is staying.

What happens if Monty only opens two doors, instead of three? Should you switch? It’s easy to figure out. You still have 1/5 chance of getting the door on your first pick. The other doors still have a 4/5 chance of having the prize. He opens two, which means that 4/5 probability is spread equally over the remaining two doors. Thus switching will give you a probability of half of 4/5ths, which is a 4/10ths probability of winning. That’s better than 1/5 chance of winning, so switching is better.

What’s the moral of the story? That intuition is malleable and that statistics is hard, until you find the right way of framing the problem. And that’s a general rule.