The Monty Hall Problem
The Monty Hall problem is a strange result arising from a very simple situation. In this video, Bryce Gessell explains why it seems so counterintuitive and why the solution isn't counterintuitive at all.
Speaker: Bryce Gessell, Duke University.
Speaker: Bryce Gessell, Duke University.
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- Let's say the prize is behind A.
If you choose A, then either B or C will be opened. Stay with your choice -> win. Switch -> lose.
If you choose B, then C will be opened. A can't be opened because that is where the prize is. Stay with your choice -> lose. Switch -> win.
If you choose C, then B will be opened. Once again, A can't be opened because that is where the prize is. Stay with your choice -> lose. Switch -> win.
In two of the equally probable situations you win by switching, i.e. the probability of winning if you switch is 2/3.
P(win | switch) = P(first choice is A) x P(second choice is A | first choice was A) + P(first choice not A) x P(second choice is A | first choice not A)
The probability that your first choice is A is 1/3. The probability that your first choice is not A is 2/3.
If you are switching, then your second choice can't be A if A was your first choice, so P(second choice is A | first choice is A) = 0. If your first choice wasn't A, then you will definitely choose A on your second choice if you switch, i.e. P(second choice is A | first choice not A) = 1
P(win | switch) = 1/3 x 0 + 2/3 x 1 = 2/3
Similarly, P(win | stay) = 1/3 x 1 + 2/3 x 0 = 1/3
The whole thing is based on the fact that door A will definitely not be opened if you
initially choose B or C. That is valuable knowledge.
It doesn't matter how many doors there are. If there are 4 doors, then the probability of winning if you switch is 1/4 x 0 + 3/4 x 1/2 = 3/8. If you don't switch then the probability of winning is 1/4 x 1 + 3/4 x 0 = 1/4, i.e. still not as good as switching.
If there are N doors, then the probability of winning if you switch is 1/n x 0 + (n-1)/n x 1/(n-2) = 1/n x (n-1)/(n-2). If you don't switch the probability of winning is 1/n x 1 + (n-1)/n x 0 = 1/n, which is less than 1/n x (n-1)/(n-2).
So it pays to switch no matter how many doors there are.(1 vote)
(intro music) Hi! My name is Bryce Gessell, and I'm a philosophy graduate student at Duke University. In this video, I'll be explaining the Monty Hall Problem. Imagine that you're in the final round of a game show, and you're just one step away from winning the grand prize. The prize is behind one of three different doors. All you have to do to win is pick the right door. But there's a wrinkle. After you select one of the doors, but before you see what's behind it, the host to the game show opens one of the other two doors. The second door doesn't have the prize behind it. The host then gives you a choice. You can stick with the door you've already picked, or you can switch to the third door. Once you've decided, all the doors will open, and you'll see if you've won. The question is, should you switch? This is called the Monty Hall problem, named after an actual game show host on an old television show. The first answer of most people is that switching doesn't matter. At the beginning, you had a one-in-three chance of winning. Now that there are only two doors, you seem to have a fifty-fifty chance of getting the prize, whether you switch or not. You might as well just flip a coin. But that isn't right. In fact, if you switch, your chance to win doubles. How can that be? The best way to understand why switching doubles your chance to win is to actually try the switching strategy. Let's name the three doors "A," "B," and "C." For our simulation, we'll say that the prize is behind door A. Now suppose that you're the contestant, and you choose door A. The host opens door B to show you that there's nothing behind it. Now you get the choice: should you switch, or stay with door A? We'll go with the switching strategy. Not knowing where the prize is, you choose to switch to door C. Unfortunately, there's nothing behind door C, and you lose. But what if you had chosen one of the other two doors to start? Let's say you choose door B. The host opens door C, and again there's nothing there. Now your choice is between switching to door A or staying with door B. Since we're testing the switching strategy, you choose to switch to A and you've won. Now suppose you choose door C to begin. The host opens door B, it's empty, and you can choose to switch from door C to door A. Our strategy is to switch, so you choose to switch from door C to door A. Hey, there it is again! You won. But take a moment to think about what we've just seen. If you adopt the switching strategy, the only way you can lose the game is if you choose the door with the prize behind it on the first try. But the odds are choosing the door with the prize behind it on the first try are just one-out-of- three. This means that the odds of winning the game by switching are two-out-of-three. When you switch, your chances of winning double. Even after seeing the explanation, you still might be stumped. After all, when the host has opened the door without the prize, there are only two doors remaining. Since the prize could be behind either door, why aren't your chances of winning just fifty-fifty? To see the answer, let's divide the three doors into two different groups: group Yours and group Others. The only door in group Yours is the one you choose originally, and the other two doors are in Others. Before any doors are opened, it's easy to see that group Yours has a one-in-three chance of winning, while group Others has a two-in-three chance, divided equally between the two doors in that group. But when the host opens the second door, you learn something. And what you've learned is that the prize isn't behind one of the two doors in Others. But what happens to the one-in-three chance that belonged to the open door? It has to go somewhere, because the total chances at winning a prize always have to add to one. That one-in-three chance from the open door can't go to the door in the Yours group, because you haven't learned anything about the door in Yours. You only learn something about the doors in the Others group. So the entire one-in-three chance from the open door gets reassigned to the remaining closed door in Others. The chance of winning with the door in the Yours group is still only one-in-three. Even though there are two remaining doors, we can split the chances evenly, because we have more information about the Others group than we do about the Yours group. And that's the Monty Hall Problem. Hope you enjoyed it! Subtitles by the Amara.org community