The Monty Hall Problem (12/10/2009): See, there's this set of logic problems, or maybe it's one problem with many instances. Anyway, it's come to be known as the Monty Hall problem. There's a really good book about it called, reasonably enough, The Monty Hall Problem, by Jason Rosenhouse. That's Dr. Rosenhouse to you and me. Allow me to summarize:

You go on the old TV show, "Let's Make a Deal" hosted by Monty Hall. Monty plucks you from the audience (maybe because you were dressed like a chicken) and invites you to select one of three closed doors on stage. They are labelled "1," "2," and "3" but are otherwise indistinguishable. You get to keep whatever is behind whichever door you choose. Behind one door is a car; behind the other two are worthless "prizes." Probably goats. Monty knows where the car is, and you know Monty knows where the car is. You make your choice (let's say you picked door number "2"). Monty will then open another door that conceals a goat and ask if you'd like to trade your door for the remaining unopened door. So the game begins: you pick a door, then Monty opens another door to reveal a goat. And then comes the crux of the game: Monty invites you to either stick with your choice and keep whatever is behind the door you originally picked or switch your choice to the remaining unopened door. What should you do -- stick with your first choice or switch?

What the hell do you mean, "What should you do?" How can it possibly matter? I mean, you knew all along that whether you picked the winning door or a losing door Monty could always open a door with a goat behind it -- Monty knows. So he did just that; so what? It may be good fun for the TV audience, but it doesn't have any effect at all on what's behind your door, or ipso facto, hi diddly ho, on whether you should stick with your door or switch. Right? I mean, c'mon! The car is either behind "your" door or the remaining closed door. It's 50-50, right? How can switching make any difference?

Except it does. You'd better switch. You're twice as likely to win the car if you switch. Play the game a thousand times, you'll be ass-deep in Cadillacs if you always switch.

So a mathermatician, Dr. Rosenhouse, writes a book about this and explains all about the problem and its correct resolution, writes about the controversies about it when lay and not-so-lay mathematicians go public with their reasoning about it, relates it to other less trivial issues (like false positives in medical testing). Good for him. Here's the rest of the story.

A philosophy grad student in Tucson, Arizona, runs across a version of this problem. Sets out to defend his initial certainty ("Switch or stick, it just can't matter!"), writes a monte carlo simulation in which he has a computer play the game a hundred thousand times, becomes convinced he's wrong ("Son of a bitch! It does matter") and relates his experiences to another grad student. Fellow grad student becomes livid ("Why would you change your mind based on that? What does that computer simulation have to do with anything?"). So the first student, convinced that the the code and its execution are actually a species of proof, realizes that the concept of "proof" is up for grabs and sets out to write his master's thesis (or qualifying paper, whatever you want to call it), examining just what it means to say that something is a proof. Never gets it done, for better or for worse, and so instead of claiming his degrees and pinballing from one one-year appointment to another, from one university to another, and finally throwing in the towel and taking up computers again, skips all that and just takes up computers again and hello there, have we met?

Incidentally, Dr. Rosenhouse relates that none other than Paul Erdos followed exactly the same trajectory -- certainty, simulation, revision of conviction -- except, being a mathematician, Erdos was not happy about not having a conventional proof of why switching is the proper strategy.

Nevermind! The most amazing thing (to me) about the Monty Hall problem right now has nothing to do with doors and probabilities, or even with epistemic pursuasion and the compulsion of belief. No: it is that I started to write that paper over twenty (twenty!) years ago. It can't possibly have been twenty years! Right? I mean, c'mon!

New project: Just how succinctly can I explain the correctness of the claim that switching is the proper strategy? William F. Buckley, Jr., once wrote that two things were almost impossible to write about clearly (without pictures): celestial navigation and how to tie a shoelace. Then he proceeded to write about celestial navigation. Let's add "the correct solution to the Monty Hall problem&quot to the list. So here goes:

At the start, you have only one chance in three of selecting the door that conceals the car. You have two chances in three of selecting a door that conceals a goat. Suppose you did select the winning door. In that case, and only in that case, switching will result in a loss. Suppose, as is twice as likely, that you selected a losing door. Whichever losing door you selected, switching will result in your selecting the winning door (because Monty will not have opened the door with the car, and we've assumed that you picked a losing door; the only door remaining, the only door you can switch to, must conceal the car). Not switching produces a win only if you picked the one winning door at the outset. Switching produces a win if you picked either of the two losing doors. Therefore, you are twice as likely to win if you switch.

:: back to the slow blog ::