Friday, May 18, 2012

A Monty Hall reductio ad absurdum

I wonder if most Internet users, by this time, are familiar with the Monty Hall problem.

I'm going to assume that whoever reading this is. So…


… the answer for the classical version is that the intuitive answer, "Might as well stay", is wrong.

A simple argument as to why your win rate from staying is 1/3 is: you only had a 1/3 chance to be right in the first place, so if you always stay, you'll win only 1/3 of the time. Naysayers will respond with something like "Yes, but that was before the second door was opened. Monty opening another door changes the game, and the chances on each door are now 1/2."

Here's a reductio ad absurdum argument I came up with for why it can't be fifty-fifty just because there are two doors left.

1. Suppose, for the sake of argument, that in the classical Monty Hall puzzle, the probability of containing a car is in fact 1/2 for both the contestant's original door and the remaining closed door. (The standard intuitive answer.)

2. Given that, then a player who always stays should expect to win about half of his games, not much more or fewer.

(If someone hedges on this, then they're possibly confused. They may think that it's "1/3 either way", but can't explain what's going on with the other third. Now, in some variants of the problem, that other third refers to situations in which Monty randomly reveals a car. It may also refer to situations in which Monty chooses not to open another door at all, instead saying "Congratulations, you won a goat!" But in the classical version, both of these outcomes are forbidden.

Another possible source of confusion is a sense in which committing beforehand to staying, instead of choosing after the goat has been revealed, will somehow make it a fundamentally different game, odds-wise. I'm not sure how to address that.)

3. Meet John. John is going to play the game 300 times and he will always stay. He expects to win about 150 games. (John has the same views as most newcomers to this puzzle.)

4. Meet Mary. Mary is going to play a less interesting version of the game, in which no second door is opened. She will simply choose one of three doors, and if her chosen door has the car, she wins; otherwise she loses. She will play 300 times and naturally expects to win 100 games.

(If someone disagrees with this, then they're probably under a very stange apprehension that the odds are fifty-fifty because "You either win the car or you don't." Some of these same people apply this sad logic to lottery tickets, and I have no idea how to persuade them out of it.)

5. So, John has just played his 300 games, and Mary played her 300 games. (I won't tell you how often they won because that's very close to the key point.) John and Mary now join forces to become a team, as is sometimes done on certain game shows. They will play 300 games together. Here's the way it will work: First, they discuss until they agree on a door, and tell the host which one they choose. Then, Mary will be blindfolded, and the host will open one of the other doors hiding a goat, revealing it to John. (The door and the goat are both silent, so Mary has no way of knowing which door was opened.) The host closes the door, and Mary's blindfold is removed. John is not allowed to indicate to Mary which door had been opened. Finally, the big reveal: John and Mary's original door is opened. (They don't get the chance to switch, because we only want to test the sticking strategy.)

6. So, how often should they win? Looking at it from John's perspective the game is basically the same as the one he had played before, so if the intuitive answer is correct, Team John-and-Mary should win about 150 games. But from Mary's perspective, it's the same as the games she had played before, so they should only win 100!

If someone doesn't see the light from this, it should prove very fascinating to try to isolate which condition(s) of the game they think are responsible for improving Mary's original 100 wins to John's original (actually very unlikely) 150 wins.

Is it the knowledge that at least one of the other doors hides a goat? But Mary already had that knowledge, based on the simple fact that there are two goats and only one "chosen door". She still only wins 100 of the 300 games she plays by herself, where no second door is opened.

So is it the knowledge of which door the host opened? But John holds that information while Mary does not! Which player's mental model "wins"? If it's John's, because the knowledge trumps the ignorance, then what if he plays with 99 blindfolded people; will the effect of their ignorance on the win rate overcome the effect of his knowledge? (And the same question in reverse if one thinks that Mary's ignorance determines the win rate.)

Perhaps the win rate will fall in between the two, at 125 out of 300. But that doesn't make much sense. We again have to ask what happens if you add more blindfolded players, or more un-blindfolded ones.

Maybe what matters is the degree of knowledge held by a person who has an actual stake in the game. For example, if John were blindfolded while the host shows a goat to the audience, then closes a door, John will still only win 1/3 of the time (it's just like a trivia show in which the correct answer flashes on the screen while the contestant is thinking) whereas if John gets to see the goat first, then he'll win 1/2 the time. But to suggest that that's how it works is magical thinking. John makes no decisions after picking the door initially, so informing him and/or the audience before opening his door won't help him play the game any better.

Maybe what makes a difference is whether or not the goat door is still physically open when the contestant's door is opened? So if the door is closed again, then the game "resets". But this, too, is magical thinking; if I know a door is empty then it can't possibly matter if it's currently closed or open. Such thinking approaches what babies supposedly believe about peek-a-boo, that when Daddy's face isn't visible it means Daddy has literally vanished. (I myself doubt it's that simple with babies, but that's another story.)

Finally, it's possible that a naysayer perceives the existence of the option to switch as being crucial to the fifty-fifty odds. Without that choice, one would only win 1/3 of the time, but if you are given the option and decide to stay, then you'll win 1/2 of the time.* This, again, is nonsense; the car can't move around just because the host says "Would you like this door instead?" Also, you can just re-do the whole above argument but with both John and Mary always switching instead. (Though it may be harder to work out what's going on given then the act of switching to the third door should give away to Mary which one had been opened as the second door. Does she just stay blindfolded the whole time? Do they both have to agree on the switch? If yes and yes, then it's just like the regular version, and switching will be the superior strategy.)

Ultimately, I hope I've make it clear that the rigmarole of opening another door cannot affect your odds of having picked correctly to begin with. If you get this, you should feel like the problem isn't so weird after all.

* I have a private view that part of what is going on with people's refusal to switch is a reluctance to "let go of" a car. Switching from a door that hides a car to one that does not feels like losing a car in a way that sticking with a vacant door, when switching would have won, does not. People then justify this desire-based intuition by arguing about probabilities; they're really talking about what things feel like, almost thinking in terms of Murphy's law ("With my luck, switching means I'll lose every time!"). Supporting this claim is the fact that nearly everyone chooses to stay while still arguing that it's fifty-fifty.

Interestingly, we can model people's loss aversion by creating a version of the game in which, if the contestant switches away from the car, she actually loses an entire car of her own. (Say she starts out with 100 cars, so she has something to lose.) In this case, the expected value from switching-versus-staying really is identical. If you always switch, then every three games, you win an average of two cars but lose one car, but if you always stay, you simply win one car per three games — a net of one car either way, plus the "staying" option is less nerve-wracking.