Probability … and how it can change your life Dan Simon Cleveland State University ESC Revised December 30, 2010

The Monty Hall Problem There are three closed doors. Behind two of them is a goat. Behind one of them is a milion dollars. You get to pick one door and keep whatever is behind it. Pick a door, but don’t open it. Monty opens a door and shows you that there is a goat behind it. Should you switch or stay with your original selection? 2

The Monty Hall Problem 87% of people stay with their original selection 13% of people switch 3

Conditional Probability Randomly choose a shape. What is the probability that you chose a circle? P(circle) = ? Randomly choose a white shape. What is the probability that you chose a circle? P(circle | white) = ? 4

Conditional Probability P(circle | white) = P(circle AND white) / P(white) = (2/8) / (3/8) = 2/3 General rule: P(A | B) = P(A, B) / P(B) 5

Conditional Probability P(A | B) = P(A, B) / P(B) But also, P(B | A) = P(B, A) / P(A) So P(A, B) = P(B, A) = P(B | A) P(A) Substitute this into the first equation to get P(A | B) = P(B | A) P(A) / P(B)  Bayes’ Theorem 6

The Monty Hall Problem You pick door #1, Monty opens door #3  $ 2 = {money is behind door #2} D 3 = {Monty opens door #3} Bayes’ Theorem: P(A | B) = P(B | A) P(A) / P(B) P($ 2 | D 3 ) = P(D 3 | $ 2 ) P($ 2 ) / P(D 3 ) 7

The Monty Hall Problem P($ 2 | D 3 ) = P(D 3 | $ 2 ) P($ 2 ) / P(D 3 ) 8 P($ 2 | D 3 ) = (1) (1/3) / (1/2) = 2/3  Switch! 11/31/2

Bayes’ Theorem and AIDS An AIDS test gives a false positive only 0.1% of the time. Therefore, if you test positive, you have a 99.9% chance of dying within 10 years. Right? 9

Bayes’ Theorem and AIDS Bayes’ Theorem: P(A | B) = P(B | A) P(A) / P(B) A = {I have AIDS}, B = {I test positive} P(A) = (i.e., 1 in 10,000 non-drug- abusing white male Americans have AIDS) P(B | A) = 1 (i.e., if I have AIDS, the test will certainly be positive) P(B) = P(B | ~A) P(~A) + P(B | A) P(A) = (0.001) P(0.9999) + (1) (0.0001) =

Bayes’ Theorem and AIDS Bayes’ Theorem: P(A | B) = P(B | A) P(A) / P(B) A = {I have AIDS}, B = {I test positive} P(A | B) = (1) (0.0001) / (0.0011) = 1/11  Don’t spend your retirement savings “The Drunkard’s Walk,” by Leonard Mlodinow 11

The Birthday Problem What is the probability that in a room of n people, at least 2 of them share a birthday? The probability that no one has the same birthday is equal to (364/365)(363/365)…((366-n)/365) 12

The Birthday Problem 13