Chapter 7 Probability Basics

Transcription

1 Making Hard Decisions Chapter 7 Probability Basics Slide 1 of 62

2 Introduction A A,, 1 An Let be an event with possible outcomes: 1 A = Flipping a coin A = {Heads} A 2 = Ω {Tails} The total event (or sample space) of event is the collection of all possible outcomes of Ω= {Heads, Tails} A A Formally: Ω= A A A A = A 1 2 n 1 n i = 1 i n Slide 2 of 62

3 Probability Calculus Probability rules may be derived using VENN DIAGRAMS 1. Probabilities must be between 0 and 1 for all possible outcomes in the sample space : Ω A 1 Ω 0 Pr( A i ) 1, for all outcomes Ai that are in Ratio of the area of the oval and the area of the total rectangle can be interpreted as the probability of the event Ω Slide 3 of 62

4 Probability Calculus 2. Probabilities must add up if both events cannot occur at the same time: A 1 A 2 Ω A 1 A2 = φ Pr( A1 A2 ) = Pr( A1 ) + Pr( A2 ) Slide 4 of 62

5 Probability Calculus A A 3. If,, 1 n are all the possible outcomes and not two of these can occur at the same time, their Total Probability must sum up to 1: A1 A2 A 3 A,, 1 An Ω are said to be collectively exhaustive and mutually exclusive Slide 5 of 62

6 Probability Calculus 4. The probability of the complement of equals 1 minus the probability of A 1 A 1 A1 A1 Pr( A1 ) = 1 Pr( A1 ) Ω Slide 6 of 62

7 Probability Calculus 5. If two events can occur at the same time the probability of either of them happening or both equals the sum of their individual probability minus the probability of them both happening at the same time. A1 A2 Ω Pr( A1 A2 ) = Pr( A1 ) + Pr( A2 ) Pr( A1 A2 ) Slide 7 of 62

8 Probability Calculus 6. Conditional probability: Dow Jones Up { Ω New Total Event based on the condition that we know that the Dow Jones went up Stock Price Up Slide 8 of 62

9 Probability Calculus: Conditional Probability Pr( Stock Dow ) = Pr( Stock Dow Pr( Dow ) ) Intuition: If I know that the market as a whole will go up, the chances of the stock of an individual company going up will increase. Pr( A B) = Pr( A B) Pr( B) Informally: Conditioning on an event coincides with reducing the total event to the conditioning event Slide 9 of 62

10 Probability Calculus: Conditional Probability Example: The probability of drawing an ace of spades in a deck of 52 cards equals 1/52. However, if I tell you that I have an ace in my hands, the probability of it being the ace of spades equals ¼. Pr( Ace Space) 1/ 52 Pr( Spades Ace) = = = Pr( Ace) 4 / Note also that: Pr( B A) = Pr( A B) Pr( A) Slide 10 of 62

11 Probability Calculus 7. Multiplicative Rule: Calculating the probability of two events happening at the same time. Pr( Ai B) = Pr( B A) Pr( A) = Pr( A B) Pr( B) 8. Independence between two events: Informally, two events are independent if information about one does not provide you any information about the other and vice versa. Consider: A A,, 1 An B B,, 1 Bm Event with possible outcomes Event with possible outcomes Slide 11 of 62

12 Probability Calculus: Independence A Example: is the event of flipping a coin and is the event of throwing a dice. If you know the outcome of flipping the coin you do not learn anything about the outcome of throwing the dice (regardless of the outcome of flipping the coin). Hence, these two events are independent. Formal definition of independence between event and event : B B A Pr( Ai Bj) = Pr( Ai) A For all possible combinations and i Bj Slide 12 of 62

13 Probability Calculus: Independence Equivalent definitions of independence between event and event : 1. B Pr( Bj Ai) = Pr( Bj) A For all possible combinations and i 2. Pr( A B ) = Pr( A) Pr( B ) i j i j A For all possible combinations and i Bj Bj A Independence/dependence in influence diagrams: No arrow between two chance nodes implies independence between the uncertain events An arrow from a chance event A to a chance event B does not mean that "A causes B". It indicates that information about A helps in determining the likelihood of outcomes of B. Slide 13 of 62

14 Probability Calculus: Conditional Independence Example: The performance of a person on any IQ test is uncertain and may range anywhere from 0% to 100%. However, if you to know that the person in question is highly intelligent it is expected his\her score will be high, e.g. ranging anywhere from 90% to 100%. On the other hand, the person s IQ does not explain this remaining uncertainty, and it may be considered measurement error affected by other conditions. For example, having a good night sleep during the previous night. On any two IQ tests, these measurement errors may be reasonably modeled as independent, if we know the IQ of the person. Slide 14 of 62

15 Probability Calculus: Conditional Independence A A,, 1 An B B,, 1 Bm C C,, 1 Cp Event with possible outcomes Event with possible outcomes Event with possible outcomes A B C Formal definition: Event and event are conditionally independent given event if and only if Pr( A B, C ) = Pr( A C ) i j k i k A, B For all possible combinations and i j Ck Informally: If I already know C, any information or knowledge about B does not tell me anything more about A Slide 15 of 62

16 Probability Calculus: Conditional Independence Equivalent definitions: Event and event are conditionally independent given event if and only if A C B 1. Pr( B A, C ) = Pr( B C ) j i k j k A, B For all possible combinations and i j Ck 2. Pr( A B C ) = Pr( A C ) Pr( B C ) i j k i k j k A, B For all possible combinations and i j C k Slide 16 of 62

17 Probability Calculus: Conditional Independence Equivalent definitions: Event and event are conditionally independent given event if and only if A C B 1. Pr( B A, C ) = Pr( B C ) j i k j k A, B For all possible combinations and i j Ck 2. Pr( A B C ) = Pr( A C ) Pr( B C ) i j k i k j k A, B For all possible combinations and i j C k Slide 17 of 62

18 Probability Calculus: Conditional Independence Conditional independence in influence diagrams: C C A B A B Slide 18 of 62

19 Probability Calculus: Law of Total Probability Let B,, B 1 3 be mutually exclusive, collectively exhaustive: B 1 A B 3 B 2 Pr( A) = Pr( A B) + Pr( A B ) + Pr( A B ) Pr( A) = Pr( A B) Pr( B) + Pr( A B ) Pr( B ) + Pr( A B ) Pr( B ) Slide 19 of 62

20 Probability Calculus: Law of Total Probability Example: X = System fails SYSTEM: X, X = failure, X = No Failure A = Component A fails, B = Component B fails, C = Component C fails A B C Assume that components A, B and C operate independently. Slide 20 of 62

21 Probability Calculus: Law of Total Probability Task: Write the probability of failure Pr(X) as a function of the component failure probabilities Pr(A), Pr(B) and Pr(C). 1.Pr( X ) = Pr( X A) Pr( A) + Pr( X A) Pr( A) = = 1 Pr( A) + Pr( X A)Pr( A) 2.Pr( X A) = Pr( X B, A) Pr( B A) + Pr( X BA, ) Pr( B A) = Pr( X BA, ) Pr( B) + 0 Pr( B) = Pr( X BA, ) Pr( B) 3.Pr( X ) = Pr( A) + Pr( X B, A) Pr( B) Pr( A) Substitute result 2 into 3 Slide 21 of 62

22 Probability Calculus: Law of Total Probability Intermediate conclusion: Hence we need to further develop Pr( X B, A) 4.Pr( X BA, ) = Pr( X CBA,, )Pr( C BA, ) + Pr( X CBA,, ) Pr( C BA, ) = 1 Pr( C) + 0 Pr( C) = Pr( C) Substitute result 4 into 3 5.Pr( X ) = Pr( A) + Pr( C) Pr( B) Pr( A) 6.Pr( A) = 1 Pr( A) Substitute result 6 into 5 7.Pr( X) = Pr( A) + Pr( C) Pr( B) Pr( C) Pr( B) Pr( A) Slide 22 of 62

23 Probability Calculus: Law of Total Probability Example: Oil Wildcatter Problem Dry (?) Max Profit -100K Drill at Site 1 Low (?) High (?) 150K 500K Drill at Site 2 Dry (0.2) -200K Low (0.8) 50K Payoff at site 1 is uncertain. Dominating factor in eventual payoff at Site 1 is the presence of a dome or not. Slide 23 of 62

24 Probability Calculus: Law of Total Probability DOME Pr(Dome) Pr(No Dome) Outcome Pr(Outcome Dome) Outcome Pr(Outcome No Dome) Dry Dry Low Low High High Slide 24 of 62

25 Probability Calculus: Law of Total Probability Pr( Dry) = Pr( Dry Dome)Pr( Dome) + Pr( Dry No Dome)Pr( No Dome) = *0.400 = Pr( Low) = Pr( Low Dome)Pr( Dome) + Pr( Low No Dome)Pr( No Dome) = *0.400 = Pr( High) = Pr( High Dome)Pr( Dome) + Pr( High No Dome)Pr( No Dome) = *0.400 = Slide 25 of 62

26 Probability Calculus: Law of Total Probability Dome (0.600) Dry (0.600) Low (0.250) High (0.150) Dry (0.850) -100K 150K 500K -100K No Dome (0.400) Low (0.125) 150K LAW OF TOTAL PROBABILITY High (0.025) 500K Dry ( = 0.70) -100K Low ( = 0.20) 150K High ( = 0.10) 500K Informally: when we apply LOTP we are collapsing a probability tree Slide 26 of 62

27 Probability Calculus: Bayes Theorem Let B,, B 1 3 be mutually exclusive, collectively exhaustive: B 1 A B 3 B 2 1. From the multiplicative rule it follows that: Pr( A B ) = Pr( B A) Pr( A) = Pr( A B ) Pr( B ) j j j j Slide 27 of 62

28 Probability Calculus: Bayes Theorem 2. Dividing the LHS and RHS of 1. by Pr(A) yields: Pr( B A) = j Pr( A B ) Pr( B ) j Pr( A) 3. We may rewrite Pr(A) using the Law of Total Probability, yielding Pr( A) = Pr( A B) Pr( B) + Pr( A B ) Pr( B ) + Pr( A B ) Pr( B ) j 4. Substituting the result of 3. into 2. gives perhaps the most well known theorem in probability theory: Bayes Theorem. Slide 28 of 62

29 Probability Calculus: Bayes Theorem Pr( B A) j = Pr( A B ) Pr( B ) j Pr( A B ) Pr( B ) + Pr( A B ) Pr( B ) + Pr( A B ) Pr( B ) j Thomas Bayes (1702 to 1761): Bayes theory of probability was published in His conclusions were accepted by Laplace in 1781, rediscovered by Condorcet, and remained unchallenged until Boole questioned them. Since then Bayes' techniques have been subject to controversy. Source: Slide 29 of 62

30 Probability Calculus: Bayes Theorem Oil Wildcatter Problem Example Continued: We drilled at site 1 and the well is a high producer. Given this new information what are the chances that a dome exists? (Perhaps that information is important when attracting additional investors.) 1. From the rule for conditional probability it follows that: Pr( Dome High ) = 2. From the LOTP it follows that: Pr( High Dome)Pr( Dome) Pr( High) Pr( High) = Pr( High Dome) Pr( Dome) + Pr( High NoDome) Pr( NoDome) Slide 30 of 62

31 Probability Calculus: Bayes Theorem Oil Wildcatter Problem Example Continued: 3. Substitution of 2 in 1 yields: Pr( Dome High ) = Pr( High Dome)Pr( Dome) = Pr( High Dome)Pr( Dome) + Pr( High No Dome)Pr( No Dome) 0.150* * * = Pr(Dome) The Prior Probability Pr(Dome Data) The Posterior Probability Data = The well is a high produces Slide 31 of 62

32 Probability Calculus: Bayes Theorem Oil Wildcatter Problem Example Continued: Dome (0.600) Dry (0.600) Low (0.250) -100K 150K Notice that Pr(Dry),Pr(Low) and Pr(High) have been inserted in the tree.these were calculated using LOTP. No Dome (0.400) High (0.150) Dry (0.850) Low (0.125) High (0.025) -100K 150K 500K Notice that Pr(Dome High) has been inserted as well This one was calculated using Bayes Theorem. BAYES THEOREM Dry (0.7) Low (0.2) Dome (?) No Dome (?) Dome (?) -100K -100K 150K We need to fill out the Remainder of the question Marks. High (0.10) No Dome (?) Dome (0.90) No Dome (?) 150K 500K 500K Slide 32 of 62

33 Probability Calculus: Bayes Theorem Oil Wildcatter Problem Example Continued: When we reverse the order of the chance nodes in a decision tree we need to apply Bayes Theorem Pr(Dome) Pr(No Dome) X Pr(X Dome) Pr(X No Dome) Pr(X Dome) Pr(X No Dome) Pr(X) Pr(Dome X) Pr(No Dome X) Check Dry Low High Check Next, we allocate the probabilities from the table at their appropriate locations in the tree Slide 33 of 62

34 Probability Calculus: Bayes Theorem Oil Wildcatter Problem Example Continued: AFTER BAYES THEOREM Dome (0.514) -100K Dry (0.7) No Dome (0.486) -100K Low (0.2) Dome (0.750) 150K No Dome (0.250) 150K High (0.10) Dome (0.900) No Dome (0.100) 500K 500K Slide 34 of 62

35 Probability Calculus: Bayes Theorem The Game Show Example: Suppose we have a game show host and you. There are three doors and one of them contains a prize. The game show host knows the door containing the prize but of course does not convey this information to you. He asks you to pick a door. You picked Door 1 and are walking up to door 1 to open it when the game show host screams: STOP!. You stop and the game show host shows Door 3 which appears to be empty. Next, the game show asks: "DO YOU WANT TO SWITCH TO DOOR 2?" WHAT SHOULD YOU DO? Slide 35 of 62

36 Probability Calculus: Bayes Theorem The Game Show Example: Assumption 1: The game show host will never show the door with the prize. Assumption 2: The game show will never show the door that you picked. Define: D i ={Prize is behind door i }, i=1,,3 H i ={Host shows Door i containing no prize after you selected Door 1}, i=1,,3 1. It seems reasonable to set prior probabilities: Pr( D i ) = 1 3 Slide 36 of 62

37 Probability Calculus: Bayes Theorem 2. Apply LOTP to calculate Pr(H 3 ): 3 Pr( H3) = Pr( H3 Di) Pr( Di) = = * + 1* + 0* = Calculate Pr(D 1 H 3 ): 1 1 * Pr( H 3 D1 ) Pr( D1 ) Pr( D H ) = = Pr( H ) Apply the complement rule: 1 Pr( D H 3 ) = 1 Pr( D1 H 3 ) = 1 3 i= 1 1 = 2 = So YES, you should SWITCH since you would increase your chances of winning! Slide 37 of 62

38 Probability Calculus: Uncertain Quantities Example: When a student attempts to log on to a computer time-sharing system, either all ports are busy (B) in which case the student will fail to obtain access, or else there is at least one port free (F), in which case the student will be successful in accessing the system. Total Event: Ω= { B, F} Definition: For a given total event Ω, a random variable (rv) is any rule that associates a number with each Ω outcome in. In mathematical language, a random variable is a function whose domain is the total event and whose range is the real numbers. Slide 38 of 62

39 Probability Calculus: Uncertain Quantities B F Ω X( B ) = 0 X( F) = X( B) = 1 Example: Consider the experiment in which batteries are examined until a good (G) is obtained. Let B denote a bad Battery. Total Event: Ω= { G, BG, BBG, BBBG,...} Slide 39 of 62

40 Probability Calculus: Uncertain Quantities Define a rv X as follows: X = The number of batteries examined before the experiment terminates. Then: X ( G) = 1, X ( BG) = 2, X ( BBG) = 3, etc The argument of the random variable function is typically omitted. Hence, one writes Pr( X = 2) = Pr( Second Battery Works) Note that the above statement only has meaning with the above definition of the random variable. It is good practice to always include the definition of a random variable in words. Slide 40 of 62

41 Probability Calculus: Uncertain Quantities The nature of random variables can be discrete and continuous. Definition: A discrete random variable is an rv whose possible values either constitute a finite set or else can be listed in an infinite sequence in which there is a first element, a second element, and so on. (Think of the previous batter example). A random variable is continuous if its set of possible values consists of an entire interval on the number line. (For example, the failure time of a component). Slide 41 of 62

42 Discrete Probability Distributions The nature of random variables can be discrete and continuous. Definition: A discrete random variable is an rv whose possible values either constitute a finite set or else can be listed in an infinite sequence in which there is a first element, a second element, and so on. (Think of the previous batter example). A random variable is continuous if its set of possible values consists of an entire interval on the number line. (For example, the failure time of a component). Slide 42 of 62

43 Discrete Probability Distributions Example: Y = Number of Raisins in an Oatmeal Cookie Assume possible outcomes: Y=1,,5 Definition: The probability mass function (PMF) of Y is the collection of probabilities such that Pr(Y=i)=p i Y Pr(Y=i) Note that: Pr( Y = 1) + Pr( Y = 2) + Pr( Y = 3) + Pr( Y = 4) + Pr( Y = 5) = 5 i= 1 Pr( Y = i) = 1 Slide 43 of 62

44 Discrete Probability Distributions Graphical Depictions of PMF s p i 0.2 p i y y A Histogram An Line Graph Definition: The cumulative distribution function (CDF) of Y at y is the sum of the probabilities such that Y y: F ( y) = Pr( Y y) = Pr( Y = i) ii : y Slide 44 of 62

45 Discrete Probability Distributions Graphical Depictions of CDF: F(y) In Decision Analysis a CDF is referred to as a CUMMULATIVE RISK PROFILE and a PMF is referred to as a RISK PROFILE 0.15 y Slide 45 of 62

46 Probability Calculus: Expected Value We know the random variable Y has many possible outcomes. However, if you were forced to give a BEST GUESS for Y, what number would you give? (Managers, CEO s, Senators etc. typically like POINT ESTIMATES, unfortunately). Why not use the expected value of Y? n n E [ Y] y Pr( Y y ) y p = = = n i i i i i= 1 i= 1 Interpretation: If you were able to observe many outcomes of Y, the calculated average of all the outcomes would be close to E[Y]. If Z = g(y): = = = E[ Z] g( y ) Pr( Y y ) g( y ) p i i i i i= 1 i= 1 n Slide 46 of 62

47 Probability Calculus: Expected Value If Z=aY+b, a,b constants, Y a rv: E[ Z] = ae[ Y] + b If Z=aX+bY, a,b const., X,Y a rv: E[ Z] = ae[ X] + be[ Y] Oatmeal Cookie Example: Raisin (0.10) 2 Raisins (0.15) 3 Raisins (0.30) 4 Raisins (0.35) 5 Raisins (0.10) #Raisins # Raisins*Pr(Y = # Raisins) 1 1*0.10= *0.15= *0.30= *0.35= *0.10= On average an oatmeal cookie has 3.2 Raisins Slide 47 of 62

48 Variance and Standard Deviation We know the random variable Y has many possible outcomes. If you were forced to give a BEST GUESS for the uncertainty in Y, what number would you give? Some people prefer to give the range of the outcomes of Y, i.e. the MAX VALUE of Y minus the MIN VALUE of Y. However, this completely ignores that some values of Y may be more likely than others. SUGGESTION: Calculate the BEST GUESS for the DISTANCE from the MEAN. The standard deviation can be thought of such a guess. The standard deviation of Y is the square root of the variance of Y. Slide 48 of 62

49 Variance and Standard Deviation Variance: Var Y E Y E Y 2 ( ) = σy = [ [ ] ] ( 2 2 ) = E[ Y 2 Y E[ Y] + E [ Y] ] = E Y E Y E Y + E Y 2 2 [ ] 2 [ ] [ ] [ ] = E Y E Y 2 2 [ ] [ ] Standard Deviation: ( ) 2 σ Y = σy = E Y E Y [ ] [ ] Slide 49 of 62

50 Variance and Standard Deviation If Z=aY+b, a,b constants, Y a rv: Var Z 2 ( ) = a Var( Y ) If Z=aX+bY, a,b const., X,Y indepent rv s: Var Z a Var X b Var Y 2 2 ( ) = ( ) + ( ) Example: A $35.75 (0.24) (0.47) Max Profit $20 $35 Note that: (0.29) $50 E[ A] = E[ B] B $35.75 (0.25) (0.35) (0.40) -$9 $0 $95 Pr(Profit 0 A) = 0; Pr(Profit 0 B) = 0.6 Slide 50 of 62

51 Variance and Standard Deviation Alternative A Prob Profit Profit^2 Prob*Profit Prob*(Profit^2) Variance St. Dev E[ Y ] = = E 2 [ Y ] = E[ Y 2 ] = E[ Y 2 ] E 2 [ Y ] σ Y Alternative B Prob Profit Profit^2 Prob*Profit Prob*(Profit^2) Variance St. Dev Slide 51 of 62

52 Expected Value, Variance and Standard Deviation 10K Dry (0.7) Max Profit -100K Drill at Site 1 Low (0.2) 150K Example: Oil Wildcatter Problem Continued Drill at Site 2 0K High (0.1) Dry (0.2) Low (0.8) 500K -200K 50K Drill at Site 1 Prob Profit Profit^2 Prob*Profit Prob*(Profit^2) Variance St. Dev = E[Y ] = E 2 [ Y ] 2 = E[ Y ] = E[ Y 2 ] E 2 [ Y ] σ Y Slide 52 of 62

53 Expected Value, Variance and Standard Deviation 10K Dry (0.7) Max Profit -100K Drill at Site 1 Low (0.2) 150K Example: Oil Wildcatter Problem Continued Drill at Site 2 0K High (0.1) Dry (0.2) Low (0.8) 500K -200K 50K Drill at Site 2 Prob Profit Profit^2 Prob*Profit Prob*(Profit^2) Variance St. Dev = E[Y ] = E 2 [ Y ] 2 = E[ Y ] = E[ Y 2 ] E 2 [ Y ] σ Y Slide 53 of 62

54 Expected Value, Variance and Standard Deviation Max Profit EMV=10K Drill at Site 1 Dome (0.6) EMV=52.50K Dry (0.60) Low (0.25) High (0.15) -100K 150K 500K Prob Profit Prob*Profit EMV= 10K EMV=0K Drill at Site 2 No Dome (0.4) Prob Profit Prob*Profit EMV=-53.75K Dry (0.2) Low (0.8) Dry (0.850) Low (0.125) High (0.025) -100K 150K 500K -200K 50K Prob Profit Prob*Profit Prob Profit Prob*Profit Expected Values can be calculated by folding back the tree Slide 54 of 62

55 Continuous Probability Distributions Let: X = The failure time of a component Definition: Let X be a continuous rv. Then a probability density function (pdf) of X is a function f(x) such that for any two numbers a and b with a < b: b Pr( X [ ab, ]) f( xdx ) f(x) = a For f(x) to be a legitamate pdf we must have: f( x) x, for all possible values x a b x f( x) dx = 1 Slide 55 of 62

56 Continuous Probability Distributions Cumulative distribution function: f(u) F( x) = Pr( X x) = f( u) du x The p-th quantile x p : x u f(x) F( x ) = Pr( X x ) = p p p F(x) 1 p x p x x p x Slide 56 of 62

57 Continuous Probability Distributions Let: X = The failure time of a component f(x) Pr( a X b) a b x f(x) F( x) = Pr( X a) f(x) F( x) = Pr( X b) a x b Conclusion: Pr( a X b) = Pr( X b) Pr( X a) = F( b) F( a) x Slide 57 of 62

58 Continuous Probability Distributions Examples theoretical density functions: ( x µ ) Normal: f ( x) = exp( ), x R 2 σ 2π 2 σ Exponential: f ( x) = λ exp( λ x), x > 0 Beta: More in Chapter 9 Expected Value: Variance: Γ ( α + β) f x x x x Γ( α) Γ( β) α 1 β 1 ( ) = (1 ), [0,1] E[ X] = xf( x) dx 2 Var( X ) ( x E[ X ]) f ( x) dx = Formulas carry over from the discrete to the continuous case Slide 58 of 62

59 Dominance Revisited DETERMINISTIC DOMINANCE Assume random Variable X Uniformly Distributed on [A,B] Assume random Variable Y Uniformly Distributed on [C,D] PDF X Y 0 CDF 1 0 A B C D X Y A B C D Slide 59 of 62

60 Dominance Revisited STOCHASTIC DOMINANCE Assume random Variable X Uniformly Distributed on [A,B] Assume random Variable Y Uniformly Distributed on [C,D] PDF X Y Note: Pr(Y<z) < Pr(X< z) for all z 0 CDF 1 A X C B Y D 0 A C B D Slide 60 of 62

61 Dominance Revisited CHOOSE ALTERNATIVE WITH BEST EMV Assume random Variable X Uniformly Distributed on [A,B] Assume random Variable Y Uniformly Distributed on [C,D] E(X) E(Y) X PDF Y 0 CDF 1 0 C A C A X Y B B D D Slide 61 of 62

62 Making Decisions under Uncertainty Deterministic Dominance Present Stochastic Dominance Present Chances of an unlucky outcome increase Making Decisions based on EMV Slide 62 of 62