Which is easier to analyze: 3 things or 100 things? Hang on, don’t answer yet.

The Monty Hall problem is a probability puzzle that’s notorious for messing with people’s intuition. If you haven’t heard it before, it goes like this. On the TV game show Let’s Make a Deal, a contestant is trying to win a car, which is hidden behind one of three closed doors. The contestant picks a door. Then the host of the show, Monty Hall, gives her one chance to switch to another door. He doesn’t tell her whether or not the car is behind the door she picked, but he does narrow down her choice by revealing what’s behind one of the doors she *didn’t* pick — not the one with the car, of course. For example:

Contestant picks Door 2.

Monty opens Door 1, showing the car isn’t there.

Monty asks: Do you want to keep Door 2, or switch to Door 3?

Think about it for a little while if you haven’t heard it before. OK, a little longer. Assuming the game always works this way (Monty always gives you the choice to switch doors, and he always narrows your choice down to just two doors after you’ve picked an initial one), *would you stay with your original door or switch?*

The simplest intuition is that it doesn’t matter: two identical doors left, there’s no information to help you decide where the car is, it’s 50-50. This is wrong, but it’s very powerful. It’s hard to find the flaw in the logic, and *really* hard to explain the flaw in a way that convinces somebody else.

But now let’s play the game with 100 doors instead of three. Same basic rules: you pick a door, and Monty gives you an option to switch *after narrowing the number of choices down to two*. For example:

You pick Door 12 (let’s say), because your oldest kid just turned 12.

Monty starts opening doors, one by one. He opens Door 1, no car. Door 2, no car. He keeps opening doors, no car behind any of them. He skips Door 12, because that’s the one you picked. Door 13, no car. Door 14, no car. On and on: Door 51, no car. Door 52, no car. Door 54, no car. (Yes, he did skip Door 53, didn’t he.) 55, no car. He opens the remaining doors, all the way through Door 100, all empty.

So now we’re down to Door 12, which you picked as a 1-in-100 shot, and Door 53, which Monty just happened to skip when he was opening every other door. Still think it’s 50-50?

Meanwhile, in the next studio over, a MontyClone™ is running the show, and your friend Bob is playing the same game. Bob picks Door 77 out of 100, because his mother is 77. MontyClone starts opening doors in order: 1, 2, 3, all the way up through 67. He skips Door 68. Then he opens 69, 70, all the way through 100 (he skips 77 because that’s the door that Bob picked). Then he asks Bob to choose between the mysteriously skipped 68, and the original choice, 77. Again: do you *really* think it’s 50-50?

And in yet another studio… But you get the point.

What’s a little clearer in the 100-door version, I hope, is that the two doors — the one you picked and the one Monty skipped — aren’t symmetric. Because, unless your 1-in-100 shot at Door 12 was right to begin with, *Monty always has to skip the door with the car behind it *when he’s opening all the other doors*.* If the car is behind Door 53, he skips 53. If it’s behind 68, he skips 68. Put another way, you end up choosing between two doors at the end: yours and Monty’s. You picked yours at random, a 1-in-100 shot. But *unless your door was right,* i.e., 99 times out of 100, Monty’s door isn’t random at all. He knows where the car is, and by the rules of the game, he points you right to it. You just have to take the hint.

It’s the same with just three doors, just a little harder to see. Put aside the door you picked and *focus on the single additional door that Monty skips*. Unless your initial guess was right (a 1-in-3 chance), the skipped door is the one with the car. If you can keep your door and Monty’s door separate in your head, then the 50-50 intuition might go away. But *things* *are a lot clearer with 100 doors than with just three.*

The approach here — replacing a small example with a larger one, and finding it actually simplifies things — may seem like a trick, but it’s actually a very common thing to do in math and physics. *Look at the behavior in the large limit. *

Here’s a more down to earth example. I started to follow the Giants when I lived in the Bay Area as a student, and I’ve been rooting for them in the World Series. My seven year old son is rooting for the Royals, because he saw them in person last year and likes their uniforms. Family conflict! But I’ve found it really relaxing to watch the Series together: someone will be happy no matter who wins! At first I was skeptical that I really felt this way: I’ve suffered through a *lot* of painful Red Sox defeats, and I figured it would hurt if my team lost, even if my kid was happy that his team won. But then I thought about the large limit. If I had 30 kids, one rooting for each major league team, then someone would be happy after every game, and I’d always have one really happy kid at the end of the season. Which actually sounds great! Looking at the large limit was the thing that convinced me. Financial theory tells you to diversify your investments; in my family, we’re diversifying our rooting interests as well.

Hey Eugene, been enjoying your new blog so far. Great MH “proof”, reminds me of a betting game: you must bet (even money) that the top card of a deck is red. However, you can first turn over one or more cards, so long as you eventually bet on a not-yet-disclosed top card. Best strategy?

Diversification can be wrong when there’s synergy. Red Sox fans share a bond. Not sure which way this goes on net.

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