The Monty Hall Problem: Why Switching Wins

You have probably been told that you should switch, and you have probably not believed it. So let us not argue about it. Let us fix a choice, write down every place the car could be, and count what happens in each one. Here are the rules, so that we are arguing about the same game. Three doors. Behind one of them is a car, and behind each of the other two is a goat. You choose a door, and it stays shut. Then the host opens one of the two doors you did not choose, and what he shows you is always a goat. He can always do that, because he knows where the car is. Two doors are left shut, and he offers you the swap. Two things get fixed now, and then nothing else moves. First the doors. Number them one, two and three. Second, your choice. You pick door one, and it stays picked for the whole lecture. Nothing is lost by fixing it, because before you choose, the three doors are identical to you. So the only thing still free is the car. It is behind door one, or door two, or door three, and each of those is equally likely. That is three cases, and I am going to walk all three of them. And here is where this ends up, so that you can watch it arrive rather than take it from me. Staying wins one time in three. Switching wins two times in three. You are not asked to believe those two lines yet. They are the thing being tested. So: the first case. The car is behind door one, which is the door you picked. The other two doors have goats behind them. Remember that we can see all three of them, and the player cannot. The host has to pick a door that is not yours and does not have the car behind it. Here both of the doors he is allowed to touch have goats, so he has a free choice. Say he opens door three. Now count the two strategies. If you stay on door one, you win the car. If you switch to door two, you get a goat. This is the case where staying is right, and it is the only one. And his free choice here changes nothing about the count. Whichever of the two he opens, you are staying on a car, or switching to a goat. Case one is the case that costs you the car if you switch. Case two. The car is behind door two, so your own door has a goat, and so does door three. Now watch what the host is allowed to do. He cannot touch your door, and he cannot touch door two, because the car is there. So he has exactly one legal move left. He must open door three. He is not choosing there. He is being pushed, by the car, onto the one door he is still allowed to open. And that push is the whole of this problem. So count again. Stay on door one and you have a goat. Switch to door two and you have the car. Switching wins this case. Case three, and it is the mirror of case two. The car is behind door three. Your door is a goat, and so is door two. He cannot touch your door, and he cannot touch door three, because that is where the car is. So he is forced onto door two. Forced again, for the second time out of three. Stay, and you have the goat behind door one. Switch, and you have the car. Switching wins again. There is the whole enumeration. Three cases, equally likely, nothing left out and nothing hidden. Read the staying column: one win and two losses. Now read the switching column: two wins and one loss. One in three against two in three, and it did not come out of a formula. It came out of writing down every case. And if you want to know where the fifty fifty feeling comes from, it comes from looking at two shut doors and treating them as interchangeable. They are not. One of them you picked blind out of three. The other one is whatever survived the host. So here is the sentence that matters. In two of the three cases the host had no choice at all. He was forced, and which door he was forced onto depended on where the car was. That is information, and switching is how you collect it. Which tells you exactly which assumption is doing the work, and gives us something to test. What if the host is not forced?

So change exactly one thing about the game. The host forgets where the car is. He still opens one of the two doors you did not pick, but now he chooses between them by flipping a coin. Which means that sometimes he opens the car himself. When that happens there is nothing left to decide, so we throw the round away and deal again. Everything else is untouched. Now count it again, the same way as before. Three places for the car, but each of those splits in two, because there are two doors his coin might send him to. Six branches. Same picture as before. Door one is yours, and the car is behind door one, door two or door three, exactly as it was. Start with the car behind door one. He flips, and opens door two: a goat, and staying wins. Or the coin lands the other way and he opens door three: also a goat, and staying wins again. Now the car is behind door two. If his coin sends him to door two, he has opened the car himself, and the round is void. If it sends him to door three, he shows a goat, and switching wins. And the car behind door three is the mirror of that. His coin sends him to door two, where he shows a goat, and switching wins. Or it sends him to door three, where he has shown the car, and the round is void again. Six branches, all equally likely, because the car is uniform and the coin is fair. Two of them end with the car standing there in the open, and those two get thrown out. So look at the four that survive. In two of them staying wins. In the other two switching wins. Two against two, and the advantage has gone. And look at what that does to your own door. In the first version it sat at one third whatever he did, because he could never fail to show a goat. Here, the fact that he happened to miss the car is itself evidence, and it lifts your door from one third to one half. Nothing else moved. Same doors, same first choice, same one in three that the car is behind door one at the start. The only thing that changed is whether the host knew where the car was. That is the real content of this puzzle, and it is why the first argument worked at all. A host who is forced two times in three is telling you something. A host who is flipping a coin is telling you nothing. So if the switching argument has always felt slippery, that is the joint it turns on. Now let me turn the numbers up until the thing is obvious.

Last picture, and this is the one that usually finishes the argument. A hundred doors. One car, ninety nine goats. You pick the door in the corner there, and it stays picked. The chance that you found the car with that pick is one in a hundred. You will not argue with me about that, because there was nothing at all to go on when you chose. Now the host, who knows exactly where the car is, opens ninety eight of the other doors, and every single one of them has a goat behind it. He is not lucky. He is stepping around the car, ninety eight times in a row. Two doors are left. Yours, and the one door out of ninety nine that he walked past every single time. So ask the question in this form. Did your one in a hundred just turn into one in two? Of course it did not. Your door is still the door you picked blind out of a hundred. All the probability that was spread thinly over the other ninety nine has been swept, by a host who was never allowed to touch the car, onto the single door he left standing. And the three door game is that same game with the numbers turned down. Your door holds a third. The other two hold two thirds between them, and the host pours all of it onto whichever door he does not open. So here is the whole thing on one board. If the host knows where the car is and has to avoid it, switching wins two times in three. If he is opening doors at random and merely happened to miss the car, it is one half either way, and you may do whatever you like. Three things to keep. Your first door keeps whatever chance it had at the moment you picked it. A host who must avoid the car sweeps all the rest onto a single door. And a host who is guessing sweeps nothing at all. Which is why this puzzle was never really about the doors. It was about the host, and whether he was filtering. When he is, switch.