1. Probability … and how it can change your life Dan Simon Cleveland State University ESC 120 1 Revised December 30, 2010

2. 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

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

4. 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

5. 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

6. 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

7. 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

8. 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

9. 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

10. 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) = 0.0001 (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) = 0.0011 10

11. 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

12. 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

13. The Birthday Problem 13