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## Wireless Philosophy

### Course: Wireless Philosophy > Unit 2

Lesson 8: Probability# 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.

## Want to join the conversation?

- 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)

## Video transcript

(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