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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.

## 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.
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