1. Sample Space and Probability Part I. Purdue University School of Electrical and Computer Engineering ECE 302, Spring 2012 Prof.

Transcription

1 1. Sample Space and Probability Part I Purdue University School of Electrical and Computer Engineering ECE 302, Spring 2012 Prof.

2 ProbabilisCc Model Sample space Ω = the set of all possible outcomes of an experiment E.g., {+$1, - $1} for a one- month stock movement in the opcon example; {$102, $100, $98} for the stock value aoer two months; {Obama, McCain, neither} for a voter s preference, etc. Probability law which assigns to a set A of possible outcomes (also called an event) a number P(A), called the probability of A.

3 Sets: basic terms and notacon, 1 A countably infinite (or countable) set is a set with infinitely many elements which can be enumerated in a list, e.g., the set of all integers { 0, 1,1, 2,2, } An example of an uncountable set is the set of all real numbers between 0 and 1, denoted [0,1]

4 Sets: basic terms and notacon, 2 The set of all x that have a certain property P is denoted by { x x satisfies P}, e.g., the interval [0,1] can alternatively be written as { x 0 x 1}.

5 Set operacons: complement

6 Set operacons: union More generally, S n = S 1 S 2 = x x S n for some n n =1 { }

7 Set operacons: interseccon More generally, S n = S 1 S 2 = x x S n for all n n =1 { }

8 Disjoint sets

9 ParCCon

10 CollecCvely exhauscve sets

11 Example: three coin tosses H HHH H T HHT H T H T HTH HTT T H H T THH THT T H TTH T TTT Sample space Ω = { HHH,HHT,HTH,HTT,THH,THT,TTH,TTT} Event "heads on the first and second toss" is the set { HHH,HHT}

12 Example: three coin tosses

13 Example: three coin tosses

14 Example: three coin tosses

15 Example: three coin tosses

16 Example: three coin tosses

17 Example: three coin tosses

18 Example: presidencal eleccon % for Obama 100% 100% % for McCain

19 Example: presidencal eleccon % for Obama 100% 100% % for McCain Note: Even though the sample space is discrete in reality (the quantum is 1 voter out of 130M, or %), it is more conveniently modeled as continuous.

20 Example: presidencal eleccon % for Obama 100% 100% % for McCain Is this sample space enough for answering these questions: (1) Will Obama win the popular vote? (2) Will Obama win the election?

21 Example: presidencal eleccon % for Obama 100% 100% % for McCain Is this sample space enough for answering these questions: (1) Will Obama win the popular vote? (2) Will Obama win the election? (1) No: need to include 3 rd -party candidates (2) No: need both 3 rd -party candidates and electoral (not popular) vote

22 Example: presidencal eleccon % for Obama 100% 100% % for McCain

23 Example: presidencal eleccon % for Obama 100% 100% % for McCain

24 Sample space Ω ProbabilisCc Model Probability law: assigns to an event A a number P(A) sacsfying the following probability axioms: 1. P(A) 0 2. If A B =, then P(A B) = P(A) + P(B) If A 1, A 2, are disjoint, then P 3. P(Ω) =1 n =1 A n = P(A n ) n =1

25 RelaConship between probability and relacve frequency of occurrence

26 Example: three coin tosses Suppose all outcomes are equally likely P(Ω) =1 by axiom 3 P P ({ HHH} ) + P( { HHT} ) + + P( { TTT} ) = P(Ω) by axiom 2 ({ HHH} ) = P( { HHT} ) = = P( { TTT} ) =1/8 ({ HTT,TTH,TTT} ) = 3/8 by axiom 2 P(S 4 ) = P("two tails in a row") = P Discrete uniform probability law : if Ω consists of N equally likely outcomes, then, for any event A, P(A) = number of elements in A N

27 Example: presidencal eleccon % for Obama 100% 100% % for McCain

28 Example: presidencal eleccon % for Obama 100% 100% % for McCain

29 Example: some properces of probability laws

30 Example: some properces of probability laws

31 Example: some properces of probability laws

32 Example: some properces of probability laws

33 Example: some properces of probability laws

34 Example: some properces of probability laws (a) If A B then P(A) P(B) Proof : Let C = A c B Then B = A C and C A = (since C A c ). Therefore, Axiom 2 Axiom 1 applied to P(C ) P(B) = P(A C) = P(A) + P(C) P(A)

35 Example Linda is 31 years old, single, outspoken and very bright. She majored in philosophy. As a student, she was deeply concerned with issues of discrimination and social justice, and also participated in anti-nuclear demonstrations.

36 Example Linda is 31 years old, single, outspoken and very bright. She majored in philosophy. As a student, she was deeply concerned with issues of discrimination and social justice, and also participated in anti-nuclear demonstrations. Let A = Linda is a bank teller Let B = Linda is a bank teller and is active in the feminist movement Which event is more likely?

37 Example Linda is 31 years old, single, outspoken and very bright. She majored in philosophy. As a student, she was deeply concerned with issues of discrimination and social justice, and also participated in anti-nuclear demonstrations. Let A = Linda is a bank teller Let B = Linda is a bank teller and is active in the feminist movement Which event is more likely? Since B A, we have P(B) P(A)

38 Example Linda is 31 years old, single, outspoken and very bright. She majored in philosophy. As a student, she was deeply concerned with issues of discrimination and social justice, and also participated in anti-nuclear demonstrations. Let A = Linda is a bank teller Let B = Linda is a bank teller and is active in the feminist movement Which event is more likely? Since B A, we have P(B) P(A) A. Tversky and D. Kahneman. Extensional Versus Intuitive Reasoning: The Conjunction Fallacy in Probability Judgment. Psychological Review, 90(4): , October 1983.

39 Example: some properces of probability laws

40 Example: some properces of probability laws

41 Example: some properces of probability laws

42 Example: some properces of probability laws

43 Example: some properces of probability laws

44 Example: some properces of probability laws

45 Example: some properces of probability laws (c) P(A B) P(A) + P(B) Proof : Use property (b) : P(A B) = P(A) + P(B) P(A B) 0, by Axiom 1 P(A) + P(B)

46 Example: some properces of probability laws (d) P(A B C) = P(A) + P(A c B) + P(A c B c C) Proof : Since A, A c B, A c B c C form a partition for A B C, the statement follows from Axiom 2.

47 CondiConal Probability NotaCon: P(A B) Probability of event A, given that event B occurred DefiniCon: assuming P(B) 0, CondiConal probabilices specify a probability law on the new universe B (exercise)

48 Example: Problem of Points Eight equally likely outcomes for three fair coin flips Best two out of three fair coin flips Helen bets on H, Tom bets on T (a) What s the probability that Helen wins 1 st round? (b) What s the probability that Helen wins overall? (c) The game is interrupted aoer Helen wins 1 st round. What s the condiconal probability that she would have won overall?

49 Example: Problem of Points Eight equally likely outcomes for three fair coin flips Best two out of three fair coin flips Helen bets on H, Tom bets on T (a) What s the probability that Helen wins 1 st round? (b) What s the probability that Helen wins overall? (c) The game is interrupted aoer Helen wins 1 st round. What s the condiconal probability that she would have won overall? SoluCon (a) ½ - - follows directly from the fact that the coin is fair. (b) ½ - - follows from the fact that all eight outcomes are equally likely.

50 Problem of Points, SoluCon H HHH H wins 1 st round H T HHT H wins overall H T H T HTH HTT T H H T THH THT T H TTH P(H wins 1 st round) = 1/2 P(H wins 1 st round and H wins overall) = 3/8 T TTT P(H wins overall H wins 1 st round) = P(H wins 1 st round and H wins overall)/ P(H wins 1 st round) = (3/8)/(1/2) = 3/4

51 AlternaCvely, View the blue set as the new sample space. Three out of four equally likely outcomes result in H s overall win. Therefore, P(H wins overall H wins 1 st round) = 3/4. HHH HHT HTH HTT H wins 1 st round H wins overall

52 Example: Intrusion DetecCon Event A: intrusion Event B: alarm Suppose that, from past experiences and measurements, we know that P(A) = 0.02; false alarm probability P(B A c ) = 0.05; and missed detection probability P(B c A) = Find P(B), P(A B), P(B c and A), and P(B and A c ).

53 Example: Intrusion DetecCon Event A: intrusion Event B: alarm Suppose that, from past experiences and measurements, we know that P(A) = 0.02; false alarm probability P(B A c ) = 0.05; and missed detection probability P(B c A) = Find P(B), P(A B), P(B c and A), and P(B and A c ). A B c A B A B A c A c B c A c

54 Example: Intrusion DetecCon Event A: intrusion Event B: alarm Suppose that, from past experiences and measurements, we know that P(A) = 0.02; false alarm probability P(B A c ) = 0.05; and missed detection probability P(B c A) = Find P(B), P(A B), P(B c and A), and P(B and A c ). P(A) = 0.02 A P(B c A)=0.01 B c A B A B A c A c B c A c P(B c A) = P(A)P(B c A) = =

55 Example: Intrusion DetecCon Event A: intrusion Event B: alarm Suppose that, from past experiences and measurements, we know that P(A) = 0.02; false alarm probability P(B A c ) = 0.05; and missed detection probability P(B c A) = Find P(B), P(A B), P(B c and A), and P(B and A c ). A B c A P(A c ) = 0.98 A c P(B A c )=0.05 B A B A c B c A c P(B A c ) = P(A c )P(B A c ) = = 0.049

56 Example: Intrusion DetecCon Event A: intrusion Event B: alarm Suppose that, from past experiences and measurements, we know that P(A) = 0.02; false alarm probability P(B A c ) = 0.05; and missed detection probability P(B c A) = Find P(B), P(A B), P(B c and A), and P(B and A c ). A B c A B A B A c Event B A c B c A c P(B) = P(B A) + P(B A c )

57 Example: Intrusion DetecCon Event A: intrusion Event B: alarm Suppose that, from past experiences and measurements, we know that P(A) = 0.02; false alarm probability P(B A c ) = 0.05; and missed detection probability P(B c A) = Find P(B), P(A B), P(B c and A), and P(B and A c ). P(A) = 0.02 A B c A Event B P(B A)=0.99 B A B A c A c B c A c P(B A) = P(A)P(B A) = = P(B A c ) = P(B) = P(B A) + P(B A c )

58 Example: Intrusion DetecCon Event A: intrusion Event B: alarm Suppose that, from past experiences and measurements, we know that P(A) = 0.02; false alarm probability P(B A c ) = 0.05; and missed detection probability P(B c A) = Find P(B), P(A B), P(B c and A), and P(B and A c ). P(A) = 0.02 A B c A Event B P(B A)=0.99 B A B A c A c B c A c P(B A) = P(A)P(B A) = = P(B A c ) = P(B) = P(B A) + P(B A c ) = =

59 Example: Intrusion DetecCon Event A: intrusion Event B: alarm Suppose that, from past experiences and measurements, we know that P(A) = 0.02; false alarm probability P(B A c ) = 0.05; and missed detection probability P(B c A) = Find P(B), P(A B), P(B c and A), and P(B and A c ). P(B A) = P(B) = P(A B) = P(A B)/P(B) =0.0198/

60 Example: Intrusion DetecCon Event A: intrusion Event B: alarm Suppose that, from past experiences and measurements, we know that P(A) = 0.02; false alarm probability P(B A c ) = 0.05; and missed detection probability P(B c A) = Find P(B), P(A B), P(B c and A), and P(B and A c ). P(A) = 0.02 A P(B c A)=0.01 P(B A)=0.99 B c A B A Event B P(B A c )=0.05 B A c P(A c ) = 0.98 A c P(B c A c )=0.95 B c A c P(B c A) = P(A)P(B c A) = = P(B A) = P(A)P(B A) = = P(B A c ) = P(A c )P(B A c ) = = P(B) = P(B A) + P(B A c ) = = P(A B) = P(A B)/P(B) =0.0198/

61 Example 1.18: False- PosiCve Puzzle A crime commiked on an island, populacon 5000 A priori, all are equally likely to have commiked it Based on a forensic test, a suspect is arrested Apart from the test, no other evidence Accuracy of the test is 99.9%, i.e., P(test is posicve suspect is guilty) = and P(test is negacve suspect is innocent) = You are on the jury. Do you have reasonable doubt?

62 False- PosiCve Puzzle: SoluCon Proceed similar to the intruder example P(+) = P(+ and guilty) + P(+ and innocent)

63 False- PosiCve Puzzle: SoluCon Proceed similar to the intruder example P(+) = P(+ and guilty) + P(+ and innocent) = P(+ guilty) P(guilty) + P(+ innocent) P(innocent)

64 False- PosiCve Puzzle: SoluCon Proceed similar to the intruder example P(+) = P(+ and guilty) + P(+ and innocent) = P(+ guilty) P(guilty) + P(+ innocent) P(innocent) =

65 False- PosiCve Puzzle: SoluCon Proceed similar to the intruder example P(+) = P(+ and guilty) + P(+ and innocent) = P(+ guilty) P(guilty) + P(+ innocent) P(innocent) = =

66 False- PosiCve Puzzle: SoluCon Proceed similar to the intruder example P(+) = P(+ and guilty) + P(+ and innocent) = P(+ guilty) P(guilty) + P(+ innocent) P(innocent) = = P(guilty +) = P(+ and guilty) / P(+) = / !!!

67 False- PosiCve Puzzle: SoluCon Proceed similar to the intruder example P(+) = P(+ and guilty) + P(+ and innocent) = P(+ guilty) P(guilty) + P(+ innocent) P(innocent) = = P(guilty +) = P(+ and guilty) / P(+) = / !!! The seemingly reliable test is not very reliable at all!

68 False- PosiCve Puzzle: Some IntuiCon Suppose this is repeated on 1000 islands Suppose we test all 5,000,000 people on 1000 islands

69 False- PosiCve Puzzle: Some IntuiCon Suppose this is repeated on 1000 islands Suppose we test all 5,000,000 people on 1000 islands On average, we expect roughly the following test results: ~1 tests - 5,000,000 people 1000 guilty 4,999,000 innocent ~999 test +

70 False- PosiCve Puzzle: Some IntuiCon Suppose this is repeated on 1000 islands Suppose we test all 5,000,000 people on 1000 islands On average, we expect roughly the following test results: ~1 tests - 5,000,000 people 1000 guilty 4,999,000 innocent ~999 test + ~4999 test + ~4,994,001 test -

71 False- PosiCve Puzzle: Some IntuiCon Suppose this is repeated on 1000 islands Suppose we test all 5,000,000 people on 1000 islands On average, we expect roughly the following test results: ~1 tests - 5,000,000 people 1000 guilty 4,999,000 innocent ~999 test + ~4999 test + ~4,994,001 test - ~999 criminals out of ~5998 who tested + i.e., roughly 1/6

72 False- PosiCve Puzzle: Some IntuiCon Suppose this is repeated on 1000 islands Suppose we test all 5,000,000 people on 1000 islands On average, we expect roughly the following test results: ~1 tests - 5,000,000 people 1000 guilty 4,999,000 innocent ~999 test + ~4999 test + ~4,994,001 test - ~999 criminals out of ~5998 who tested + i.e., roughly 1/6 I.e., there are so few criminals that the bulk of people who test posicve are innocent!

73 So

74 So PLEASE

75 So PLEASE never confuse P(A B) with P(B A)!

76 Example: P(A B) vs P(B A) Let A = On US ElecCon Day November 6, 2012, there is a hurricane in Florida and Louisiana, tornadoes in Midwest, a tsunami in New York, and an earthquake in California.

77 Example: P(A B) vs P(B A) Let A = On US ElecCon Day November 6, 2012, there is a hurricane in Florida and Louisiana, tornadoes in Midwest, a tsunami in New York, and an earthquake in California. Let B = On US ElecCon Day in November 6, 2012, the voter turnout is lower than it was on ElecCon Day November 4, 2008.

78 Example: P(A B) vs P(B A) Let A = On US ElecCon Day November 6, 2012, there is a hurricane in Florida and Louisiana, tornadoes in Midwest, a tsunami in New York, and an earthquake in California. Let B = On US ElecCon Day in November 6, 2012, the voter turnout is lower than it was on ElecCon Day November 4, P(B A) is huge. If A happens, a lot of people will be too preoccupied to vote.

79 Example: P(A B) vs P(B A) Let A = On US ElecCon Day November 6, 2012, there is a hurricane in Florida and Louisiana, tornadoes in Midwest, a tsunami in New York, and an earthquake in California. Let B = On US ElecCon Day in November 6, 2012, the voter turnout is lower than it was on ElecCon Day November 4, P(B A) is huge. If A happens, a lot of people will be too preoccupied to vote. P(A B) is Cny. Lower- than turnout could happen for many reasons besides natural distasters.

80 Total Probability Theorem A 1 B A 2 A 3

81 Total Probability Theorem A 1 B A 2 A 3 One way of compucng P(B): P(B) = P(B A 1 ) + P(B A 2 ) + P(B A 3 ) = P(B A 1 )P(A 1 ) + P(B A 2 )P(A 2 ) + P(B A 3 )P(A 3 )

82 Total Probability Theorem A 1 B A 2 A 3 One way of compucng P(B): P(B) = P(B A 1 ) + P(B A 2 ) + P(B A 3 ) = P(B A 1 )P(A 1 ) + P(B A 2 )P(A 2 ) + P(B A 3 )P(A 3 ) More generally, P(B) = Σ i P(B A i )P(A i ) if A i s are mutually exclusive and B is a subset of the union of A i s

83 Bayes Rule Prior model: probabilices P(A i )

84 Bayes Rule Prior model: probabilices P(A i ) Measurement model: P(B A i ) condiconal probability to observe data B given that the truth is A i

85 Bayes Rule Prior model: probabilices P(A i ) Measurement model: P(B A i ) condiconal probability to observe data B given that the truth is A i Want to compute posterior probabilides P(A i B) CondiConal probability that the truth is A i given that we observed data B

86 Bayes Rule Prior model: probabilices P(A i ) Measurement model: P(B A i ) condiconal probability to observe data B given that the truth is A i Want to compute posterior probabilides P(A i B) CondiConal probability that the truth is A i given that we observed data B P(A i B) = P(A i B) P(B) = P(B A i )P(A i ) P(B)

87 Denominator in Bayes Rule via Total Probability Theorem If B A j then we can use the total probability theorem j to compute P(B) :

88 Denominator in Bayes Rule via Total Probability Theorem If B A j then we can use the total probability theorem j to compute P(B) : P(A i B) = P(B A i )P(A i ) P(B) = P(B A i )P(A i ) P(B A j )P(A j ) j

89 MulCplicaCon Rule P n i=1 A i = P(A 1 )P(A 2 A 1 )P(A 3 A 1 A 2 ) P A n n 1 A i i=1 provided all the conditioning events have nonzero probability,

90 Example 1.10 Three cards are drawn from a deck of 52 cards, without replacement. At each step, each one of the remaining cards is equally likely to be picked. What s the probability that none of the three cards is a heart? Let A i ={i- th card is not a heart}, i=1,2,3

91 Example Three cards are drawn from a deck of 52 cards, without replacement. At each step, each one of the remaining cards is equally likely to be picked. What s the probability that none of the three cards is a heart? Let A i ={i- th card is not a heart}, i=1,2,3 P(A 1 ) = A 1 P(A 1 c ) = A 1 c

92 Example Three cards are drawn from a deck of 52 cards, without replacement. At each step, each one of the remaining cards is equally likely to be picked. What s the probability that none of the three cards is a heart? Let A i ={i- th card is not a heart}, i=1,2,3 P(A 1 ) = P(A 1 c ) = A 1 A 1 c P(A 2 A 1 ) = A 1 A 2 A 1 A 2 c

93 Example Three cards are drawn from a deck of 52 cards, without replacement. At each step, each one of the remaining cards is equally likely to be picked. What s the probability that none of the three cards is a heart? Let A i ={i- th card is not a heart}, i=1,2,3 P(A 3 A 1 A 2 ) = A 1 A 2 A 3 P(A 2 A 1 ) = A 1 A 2 P(A 1 ) = A A 1 A 2 A 3 c A 1 A 2 c P(A 1 c ) = A 1 c

94 Example Three cards are drawn from a deck of 52 cards, without replacement. At each step, each one of the remaining cards is equally likely to be picked. What s the probability that none of the three cards is a heart? Let A i ={i- th card is not a heart}, i=1,2,3 P(A 3 A 1 A 2 ) = A 1 A 2 A 3 P(A 2 A 1 ) = A 1 A 2 P(A 1 ) = A A 1 A 2 A 3 c A 1 A 2 c P(A 1 c ) = A 1 c P(A 1 A 2 A 3 ) = P(A 1 )P(A 2 A 1 )P(A 3 A 1 A 2 ) =

95 Example: Prisoner s Dilemma (p. 58) Three prisoners: A, B, and C. One will be executed next day, two released. Prisoner A asks the guard to tell him the name of one of the other two who will be released. Guard says that B will be released. (Assume that the guard is telling the truth.) A argues: before my chances to be executed were 1/3, now they are 1/2 since I know it s either me or C. What s wrong with his reasoning?

96 Prisoner s Dilemma: A few remarks The prisoners only know that one of them will be executed, but do not know which one. Thus, from their point of view, a reasonable model is a discrete uniform probability law that assigns each of them probability 1/3 to be executed. The guard knows which one of them will be executed. Since A already knows that either B or C will be released, it does not seem like the guard s response should influence A s chances in any way. In fact, under a reasonable assumpcon, A s probability to be executed is scll 1/3 (from A s point of view), as shown in the next few slides.

97 Prisoner s Dilemma: Problem Statement and AddiConal Modeling We know P(A to be executed) = 1/3. P(A to be executed Guard s response is B) =?

98 Prisoner s Dilemma: Problem Statement and AddiConal Modeling We know P(A to be executed) = 1/3. P(A to be executed Guard s response is B) =? But we do not yet have enough informacon to compute this condiconal probability. We need P(Guard s response is B A to be executed).

99 Prisoner s Dilemma: Problem Statement and AddiConal Modeling We know P(A to be executed) = 1/3. P(A to be executed Guard s response is B) =? But we do not yet have enough informacon to compute this condiconal probability. We need P(Guard s response is B A to be executed). It s reasonable for A to assume that if he is to be executed, the guard has a chance to respond B or C : P(Guard s response is B A to be executed) = 1/2.

100 Prisoner s Dilemma: SoluCon Let E i ={prisoner i will be executed} for i=a,b,c Let G j ={guard names prisoner j} for j=b,c

101 Prisoner s Dilemma: SoluCon Let E i ={prisoner i will be executed} for i=a,b,c Let G j ={guard names prisoner j} for j=b,c 1/3 1/3 1/3 E A E B E C

102 Prisoner s Dilemma: SoluCon Let E i ={prisoner i will be executed} for i=a,b,c Let G j ={guard names prisoner j} for j=b,c 1/2 E A G B 1/3 E A 1/2 E A G C 1/3 E B 1/3 E C

103 Prisoner s Dilemma: SoluCon Let E i ={prisoner i will be executed} for i=a,b,c Let G j ={guard names prisoner j} for j=b,c 1/2 E A G B 1/3 1/3 1/3 E A E B 1/2 0 1 E A G C E B G B E B G C E C

104 Prisoner s Dilemma: SoluCon Let E i ={prisoner i will be executed} for i=a,b,c Let G j ={guard names prisoner j} for j=b,c 1/2 E A G B 1/3 1/3 1/3 E A E B E C 1/ E A G C E B G B E B G C E C G B E C G C

105 Prisoner s Dilemma: SoluCon Let E i ={prisoner i will be executed} for i=a,b,c Let G j ={guard names prisoner j} for j=b,c 1/2 E A G B 1/6 1/3 1/3 1/3 E A E B E C 1/ E A G C E B G B E B G C E C G B E C G C 1/6 0 1/3 1/3 0

106 Prisoner s Dilemma: SoluCon Let E i ={prisoner i will be executed} for i=a,b,c Let G j ={guard names prisoner j} for j=b,c G B 1/2 E A G B 1/6 1/3 1/3 1/3 E A E B E C 1/ E A G C E B G B E B G C E C G B E C G C 1/6 0 1/3 1/3 0

107 Prisoner s Dilemma: SoluCon Let E i ={prisoner i will be executed} for i=a,b,c Let G j ={guard names prisoner j} for j=b,c G B 1/2 E A G B 1/6 1/3 1/3 1/3 E A E B E C 1/ E A G C E B G B E B G C E C G B E C G C 1/6 0 1/3 1/3 0 P(E A G B ) = P(E A G B )/P(G B )

108 Prisoner s Dilemma: SoluCon Let E i ={prisoner i will be executed} for i=a,b,c Let G j ={guard names prisoner j} for j=b,c G B 1/2 E A G B 1/6 1/3 1/3 1/3 E A E B E C 1/ E A G C E B G B E B G C E C G B E C G C 1/6 0 1/3 1/3 0 P(E A G B ) = P(E A G B )/P(G B ) = (1/6)/(1/ /3) = 1/3

109 Prisoner s Dilemma Problem 13 in FiFy Challenging Problems in Probability with SoluDons by F. Mosteller, Dover, 1965.

110 The Monty Hall Puzzle (Ex. 1.12) Prize behind one of three doors. Contestant picks a door. Host (who knows where the prize is) opens one of the remaining two doors which does not have the prize. Contestant is offered an opportunity to stay with his door, or to switch to another door. Stay or switch?

111 The Monty Hall Puzzle: SoluCon If stay, P(win) = 1/3

112 The Monty Hall Puzzle: SoluCon If stay, P(win) = 1/3 If switch, the only way to lose is if inically pointed to the door with prize: P(lose) = 1/3 and so P(win) = 2/3

113 The Monty Hall Puzzle: SoluCon If stay, P(win) = 1/3 If switch, the only way to lose is if inically pointed to the door with prize: P(lose) = 1/3 and so P(win) = 2/3 Conclusion: must switch! Switching is advantageous because the host s accon tells you something. If you inically picked a door with no prize, he is forced to open the other door with no prize.

114 The Monty Hall Puzzle: Discussion Crucial parts of the problem statement: the host knows where the prize is he must open the door with no prize

115 The Monty Hall Puzzle: Discussion Crucial parts of the problem statement: the host knows where the prize is he must open the door with no prize Thus, if you have chosen a door with no prize, you are forcing him to open the only other door with no prize and thus show you where the prize is.

116 The Monty Hall Puzzle: Another SoluCon Use the total probability theorem to evaluate the probabilices of winning under the two strategies.

117 The Monty Hall Puzzle: Another SoluCon Use the total probability theorem to evaluate the probabilices of winning under the two strategies. Let W = win and N = originally point to a door with no prize.

118 The Monty Hall Puzzle: Another SoluCon Use the total probability theorem to evaluate the probabilices of winning under the two strategies. Let W = win and N = originally point to a door with no prize. P(N) = 2/3; P(N c ) = 1/3.

119 The Monty Hall Puzzle: Another SoluCon Use the total probability theorem to evaluate the probabilices of winning under the two strategies. Let W = win and N = originally point to a door with no prize. P(N) = 2/3; P(N c ) = 1/3. If you do not switch, P(W N) = 0 and P(W N c ) = 1.

120 The Monty Hall Puzzle: Another SoluCon Use the total probability theorem to evaluate the probabilices of winning under the two strategies. Let W = win and N = originally point to a door with no prize. P(N) = 2/3; P(N c ) = 1/3. If you do not switch, P(W N) = 0 and P(W N c ) = 1. So, if you do not switch, P(W) = P(W N)P(N) + P(W N c )P(N c ) = 1/3.

121 The Monty Hall Puzzle: Another SoluCon Use the total probability theorem to evaluate the probabilices of winning under the two strategies. Let W = win and N = originally point to a door with no prize. P(N) = 2/3; P(N c ) = 1/3. If you do not switch, P(W N) = 0 and P(W N c ) = 1. So, if you do not switch, P(W) = P(W N)P(N) + P(W N c )P(N c ) = 1/3. If you switch, P(W N) = 1 because if you originally choose a door with no prize the host is forced to open the only other door with no prize.

122 The Monty Hall Puzzle: Another SoluCon Use the total probability theorem to evaluate the probabilices of winning under the two strategies. Let W = win and N = originally point to a door with no prize. P(N) = 2/3; P(N c ) = 1/3. If you do not switch, P(W N) = 0 and P(W N c ) = 1. So, if you do not switch, P(W) = P(W N)P(N) + P(W N c )P(N c ) = 1/3. If you switch, P(W N) = 1 because if you originally choose a door with no prize the host is forced to open the only other door with no prize. If you switch, P(W N c ) = 0 because if you originally choose the door with the prize, you will switch out of it and lose.

123 The Monty Hall Puzzle: Another SoluCon Use the total probability theorem to evaluate the probabilices of winning under the two strategies. Let W = win and N = originally point to a door with no prize. P(N) = 2/3; P(N c ) = 1/3. If you do not switch, P(W N) = 0 and P(W N c ) = 1. So, if you do not switch, P(W) = P(W N)P(N) + P(W N c )P(N c ) = 1/3. If you switch, P(W N) = 1 because if you originally choose a door with no prize the host is forced to open the only other door with no prize. If you switch, P(W N c ) = 0 because if you originally choose the door with the prize, you will switch out of it and lose. So, if you switch, P(W) = P(W N)P(N) + P(W N c )P(N c ) = 2/3.

124 Two- Envelopes Puzzle (p. 58) You are handed two envelopes, each containing an posicve integer number of dollars, unknown to you. The two amounts are different. You select at random one envelope and look inside. You can either scck with this envelope or take the other envelope. Your objeccve is to get the larger amount. Does it maker what you do?

125 Two- Envelopes Puzzle Surprisingly, the answer is yes. There exist strategies with a strictly beker than chance of picking the envelope with the larger amount.

126 Two- Envelopes Puzzle: Strategy Make independent flips of a fair coin uncl heads come up for the first Cme. Let X = 1/2 + number of tosses to get first H. If the amount in your envelope > X, stay If the amount in your envelope < X, switch

127 Two- Envelopes Puzzle: Analysis Strategy: If the amount in your envelope > X, stay If the amount in your envelope < X, switch where X = 1/2 + # tosses of a fair coin to get first H Let s show that this strategy has a larger than 1/2 probability to get the larger amount.

128 Two- Envelopes Puzzle: Analysis Strategy: If the amount in your envelope > X, stay If the amount in your envelope < X, switch where X = 1/2 + # tosses of a fair coin to get first H Let s show that this strategy has a larger than 1/2 probability to get the larger amount. Denote the two amounts d and D, d < D

129 Two- Envelopes Puzzle: Analysis Strategy: If the amount in your envelope > X, stay If the amount in your envelope < X, switch where X = 1/2 + # tosses of a fair coin to get first H Let s show that this strategy has a larger than 1/2 probability to get the larger amount. Denote the two amounts d and D, d < D Case 1: X < d always stay win with probability ½

130 Two- Envelopes Puzzle: Analysis Strategy: If the amount in your envelope > X, stay If the amount in your envelope < X, switch where X = 1/2 + # tosses of a fair coin to get first H Let s show that this strategy has a larger than 1/2 probability to get the larger amount. Denote the two amounts d and D, d < D Case 1: X < d always stay win with probability 1/2 Case 2: X > D always switch win with probability ½

131 Two- Envelopes Puzzle: Analysis Strategy: If the amount in your envelope > X, stay If the amount in your envelope < X, switch where X = 1/2 + # tosses of a fair coin to get first H Let s show that this strategy has a larger than 1/2 probability to get the larger amount. Denote the two amounts d and D, d < D Case 1: X < d always stay win with probability 1/2 Case 2: X > D always switch win with probability 1/2 Case 3: d < X < D stay if pick D, switch if pick d win!

132 Two- Envelopes Puzzle: Analysis Strategy: If the amount in your envelope > X, stay If the amount in your envelope < X, switch where X = 1/2 + # tosses of a fair coin to get first H Let s show that this strategy has a larger than 1/2 probability to get the larger amount. Denote the two amounts d and D, d < D Case 1: X < d always stay win with probability 1/2 Case 2: X > D always switch win with probability 1/2 Case 3: d < X < D stay if pick D, switch if pick d win! Since P(Case 3) > 0 for any two amounts d and D, we have that the probability to get the larger amount is bigger than 1/2.

133 Two- Envelopes Puzzle: P(win) > 1/2 Case 1: X < d always stay win with probability 1/2 Case 2: X > D always switch win with probability 1/2 Case 3: d < X < D stay if pick D, switch if pick d win! Using the total probability theorem, we have: P(win) = P(win X<d) P(X<d) + P(win X>D) P(X>D) + P(win d<x<d) P(d<X<D)

134 Two- Envelopes Puzzle: P(win) > 1/2 Case 1: X < d always stay win with probability 1/2 Case 2: X > D always switch win with probability 1/2 Case 3: d < X < D stay if pick D, switch if pick d win! Using the total probability theorem, we have: P(win) = P(win X<d) P(X<d) + P(win X>D) P(X>D) + P(win d<x<d) P(d<X<D) = 1/2 P(X<d) + 1/2 P(X>D) + P(d<X<D)

135 Two- Envelopes Puzzle: P(win) > 1/2 Case 1: X < d always stay win with probability 1/2 Case 2: X > D always switch win with probability 1/2 Case 3: d < X < D stay if pick D, switch if pick d win! Using the total probability theorem, we have: P(win) = P(win X<d) P(X<d) + P(win X>D) P(X>D) + P(win d<x<d) P(d<X<D) = 1/2 P(X<d) + 1/2 P(X>D) + P(d<X<D) = 1/2 P(X<d) + 1/2 P(X>D) + 1/2 P(d<X<D) + 1/2 P(d<X<D)

136 Two- Envelopes Puzzle: P(win) > 1/2 Case 1: X < d always stay win with probability 1/2 Case 2: X > D always switch win with probability 1/2 Case 3: d < X < D stay if pick D, switch if pick d win! Using the total probability theorem, we have: P(win) = P(win X<d) P(X<d) + P(win X>D) P(X>D) + P(win d<x<d) P(d<X<D) = 1/2 P(X<d) + 1/2 P(X>D) + P(d<X<D) = 1/2 P(X<d) + 1/2 P(X>D) + 1/2 P(d<X<D) + 1/2 P(d<X<D) = 1/2 (P(X<d) + P(X>D) + P(d<X<D)) + 1/2 P(d<X<D)

137 Two- Envelopes Puzzle: P(win) > 1/2 Case 1: X < d always stay win with probability 1/2 Case 2: X > D always switch win with probability 1/2 Case 3: d < X < D stay if pick D, switch if pick d win! Using the total probability theorem, we have: P(win) = P(win X<d) P(X<d) + P(win X>D) P(X>D) + P(win d<x<d) P(d<X<D) = 1/2 P(X<d) + 1/2 P(X>D) + P(d<X<D) = 1/2 P(X<d) + 1/2 P(X>D) + 1/2 P(d<X<D) + 1/2 P(d<X<D) = 1/2 (P(X<d) + P(X>D) + P(d<X<D)) + 1/2 P(d<X<D) = 1/2 + 1/2 P(d<X<D) > 1/2

138 Two- Envelopes Puzzle: Comments Note that X can be any random variable having non- zero values at 3/2, 5/2, 7/2 (or, in fact, at any point(s) on ]1,2[, at any point(s) on ]2,3[, etc.)

139 Two- Envelopes Puzzle: Comments Note that X can be any random variable having non- zero values at 3/2, 5/2, 7/2 (or, in fact, at any point(s) on ]1,2[, at any point(s) on ]2,3[, etc.) According to the strategy, you will always switch if the amount in your inical envelope is $1.

140 Two- Envelopes Puzzle: Comments Note that X can be any random variable having non- zero values at 3/2, 5/2, 7/2 (or, in fact, at any point(s) on ]1,2[, at any point(s) on ]2,3[, etc.) According to the strategy, you will always switch if the amount in your inical envelope is $1. If it is known that the largest possible number of dollars in both envelopes is some integer L, then the strategy will scll result in a larger than 50% probability of success. However, it can be improved because having X > L is wasteful in this case. To adapt the strategy to this case, make X not take any values above L 1/2. For example, X can be taken to be discrete uniform over {3/2, 5/2,, L 1/2}.