PHPE 400 Individual and Group Decision Making Eric Pacuit - PowerPoint PPT Presentation

PHPE 400 Individual and Group Decision Making Eric Pacuit University of Maryland 1 / 24

Allais Paradox Red (1) White (89) Blue (10) S 1 A 1 M 1 M 1 M B 1 M 5 M 0 S 2 C 1 M 0 1 M D 0 0 5 M A � B iff C � B 2 / 24

U ( 5 M ) U ( 5 M ) U ( 1 M ) U ( 1 M ) U ( 0 ) U ( 0 ) 0 . 1 0 . 9 1 0 . 1 0 . 9 1 [ 1 M : 0 . 01 , 1 M : 0 . 89 , 1 M : 0 . 01 ] [ 1 M : 0 . 01 , 0 : 0 . 89 , 1 M : 0 . 01 ] [ 0 : 0 . 01 , 1 M : 0 . 89 , 5 M : 0 . 01 ] [ 0 : 0 . 01 , 0 : 0 . 89 , 5 M : 0 . 01 ] 3 / 24

U ( 5 M ) U ( 5 M ) U ( 1 M ) U ( 1 M ) U ( 0 ) U ( 0 ) 0 . 1 0 . 9 1 0 . 1 0 . 9 1 [ 1 M : 0 . 01 , 1 M : 0 . 89 , 1 M : 0 . 01 ] [ 1 M : 0 . 01 , 0 : 0 . 89 , 1 M : 0 . 01 ] [ 0 : 0 . 01 , 1 M : 0 . 89 , 5 M : 0 . 01 ] [ 0 : 0 . 01 , 0 : 0 . 89 , 5 M : 0 . 01 ] 3 / 24

U ( 5 M ) U ( 5 M ) U ( 1 M ) U ( 1 M ) U ( 0 ) U ( 0 ) 0 . 1 0 . 9 1 0 . 1 0 . 9 1 [ 1 M : 0 . 01 , 1 M : 0 . 89 , 1 M : 0 . 01 ] [ 1 M : 0 . 01 , 0 : 0 . 89 , 1 M : 0 . 01 ] [ 0 : 0 . 01 , 1 M : 0 . 89 , 5 M : 0 . 01 ] [ 0 : 0 . 01 , 0 : 0 . 89 , 5 M : 0 . 01 ] 3 / 24

Red (1) White (89) Blue (10) S 1 A 1 M 1 M 1 M B 0 1 M 5 M S 2 C 1 M 0 1 M D 0 0 5 M A � B iff C � D 3 / 24

Independence Independence For all L 1 , L 2 , L 3 ∈ L and a ∈ ( 0 , 1 ] , L 1 ≻ L 2 if, and only if, [ L 1 : a , L 3 : ( 1 − a )] ≻ [ L 2 : a , L 3 : ( 1 − a )] . L 1 ∼ L 2 if, and only if, [ L 1 : a , L 3 : ( 1 − a )] ∼ [ L 2 : a , L 3 : ( 1 − a )] . 4 / 24

≺ 0.8 1 0.2 0 $4,000 0 $3,000 0 ≻ 0.2 0.8 0.25 0.75 $4,000 $3,000 0 0 5 / 24

≺ 0.8 1 0.2 0 $4,000 $3,000 0 0 0.25 0.75 0.25 0.75 0 0 ≺ 0.8 1 0.2 0 $4,000 $3,000 0 0 5 / 24

0.25 0.75 0.25 0.75 0 0 ≺ 0.8 1 0.2 0 $4,000 $3,000 0 0 ≻ 0.2 0.25 0.75 0.8 $4,000 $3,000 0 0 0 . 25 ∗ 0 . 8 = 0 . 2 0 . 25 ∗ 1 = 0 . 25 5 / 24

≺ 0.8 0.2 1 0 $4,000 0 $3,000 0 sure thing 0.25 0.75 0.25 0.75 0 0 ≻ 0.8 0.2 1 0 $4,000 $3,000 0 0 gamble 5 / 24

Allais Paradox We should not conclude either 6 / 24

Allais Paradox We should not conclude either (a) The axioms of cardinal utility fail to adequately capture our understanding of rational choice, or 6 / 24

Allais Paradox We should not conclude either (a) The axioms of cardinal utility fail to adequately capture our understanding of rational choice, or (b) those who choose A in S 1 and D is S 2 are irrational. 6 / 24

Allais Paradox We should not conclude either (a) The axioms of cardinal utility fail to adequately capture our understanding of rational choice, or (b) those who choose A in S 1 and D is S 2 are irrational. Rather, people’s utility functions ( their rankings over outcomes ) are often far more complicated than the monetary bets would indicate.... 6 / 24

L. Buchak. Risk and Rationality . Oxford University Press, 2013. 7 / 24

Ellsberg Paradox 30 60 Lotteries Blue Yellow Green L 1 1 M 0 0 L 2 0 1 M 0 L 1 � L 2 iff L 3 � L 4 8 / 24

Ellsberg Paradox 30 60 Lotteries Blue Yellow Green L 3 1 M 0 1 M L 4 0 1 M 1 M L 1 � L 2 iff L 3 � L 4 8 / 24

Ellsberg Paradox 30 60 Lotteries Blue Yellow Green L 1 1 M 0 0 L 2 0 1 M 0 L 3 1 M 0 1 M L 4 0 1 M 1 M L 1 � L 2 iff L 3 � L 4 8 / 24

Ambiguity Aversion I. Gilboa and M. Marinacci. Ambiguity and the Bayesian Paradigm . Advances in Economics and Econometrics: Theory and Applications, Tenth World Congress of the Econometric Society. D. Acemoglu, M. Arellano, and E. Dekel (Eds.). New York: Cambridge University Press, 2013. 9 / 24

Flipping a fair coin vs. flipping a coin of unknown bias: “The probability is 50-50”... 10 / 24

Flipping a fair coin vs. flipping a coin of unknown bias: “The probability is 50-50”... ◮ Imprecise probabilities ◮ Non-additive probabilities ◮ Qualitative probability 10 / 24

Newcomb’s Paradox A very powerful being, who has been invariably accurate in his predictions about your behavior in the past, has already acted in the following way: 1. If he has predicted that you will open just box B , he has in addition put $1,000,000 in box B 2. If he has predicted you will open both boxes, he has put nothing in box B . What should you do? R. Nozick. Newcomb’s Problem and Two Principles of Choice . 1969. 11 / 24

Dominance Reasoning w 1 w 2 A 1 3 B 2 4 12 / 24

Dominance Reasoning w 1 w 2 A 1 3 B 2 4 12 / 24

Dominance Reasoning Dominance reasoning is appropriate only when probability of outcome is independent of choice . (A nasty nephew wants inheritance from his rich Aunt. The nephew wants the inheritance, but other things being equal, does not want to apologize. Does dominance give the nephew a reason to not apologize? Whether or not the nephew is cut from the will may depend on whether or not he apologizes .) 12 / 24

$ 1 , 000 , 000 $ 1000 A B Choice: one-box: choose box B two-box: choose box A and B 13 / 24

$1 million in A $0 in closed box closed box one-box A $1,000,000 $0 two- A $1,001,000 $1,000 box 14 / 24

$1 million in A $0 in closed box closed box one-box A $1,000,000 $0 two- A $1,001,000 $1,000 box act-state dependence: P ( s ) � = P ( s | A ) 14 / 24

Newcomb’s Paradox B = 1M B = 0 1 Box 1M 0 2 Boxes 1M + 1000 1000 15 / 24

Newcomb’s Paradox B = 1M B = 0 B = 1M B = 0 1 Box 1M 0 1 Box h 1 − h 2 Boxes 1M + 1000 1000 2 Boxes 1 − h h 15 / 24

Newcomb’s Paradox J. Collins. Newcomb’s Problem . International Encyclopedia of Social and Behavorial Sciences, 1999. 16 / 24

Newcomb’s Paradox There is a conflict between maximizing your expected value (1-box choice) and dominance reasoning (2-box choice). 17 / 24

Newcomb’s Paradox There is a conflict between maximizing your expected value (1-box choice) and dominance reasoning (2-box choice). What the Predictor did yesterday is probabilistically dependent on the choice today, but causally independent of today’s choice. 17 / 24

V ( A ) = � w V ( w ) · P A ( w ) (the expected value of act A is a probability weighted average of the values of the ways w in which A might turn out to be true) 18 / 24

V ( A ) = � w V ( w ) · P A ( w ) (the expected value of act A is a probability weighted average of the values of the ways w in which A might turn out to be true) EDT: P A ( w ) := P ( w | A ) (Probability of w given A is chosen) CDT: P A ( w ) = P ( A � → w ) (Probability of if A were chosen then w would be true ) 18 / 24

Suppose 99% confidence in predictors reliability. B 1 : one-box (open box B ) B 2 : two-box choice (open both A and B ) N : receive nothing K : receive $1,000 M : receive $1,000,000 L : receive $1,001,000 19 / 24

Suppose 99% confidence in predictors reliability. B 1 : one-box (open box B ) B 2 : two-box choice (open both A and B ) N : receive nothing K : receive $1,000 M : receive $1,000,000 L : receive $1,001,000 V ( B 1 ) = V ( M ) P ( M | B 1 ) + V ( N ) P ( N | B 1 ) 19 / 24

Suppose 99% confidence in predictors reliability. B 1 : one-box (open box B ) B 2 : two-box choice (open both A and B ) N : receive nothing K : receive $1,000 M : receive $1,000,000 L : receive $1,001,000 V ( B 1 ) = V ( M ) P ( M | B 1 ) + V ( N ) P ( N | B 1 ) = 1000000 · 0 . 99 + 0 · 0 . 01 19 / 24

Suppose 99% confidence in predictors reliability. B 1 : one-box (open box B ) B 2 : two-box choice (open both A and B ) N : receive nothing K : receive $1,000 M : receive $1,000,000 L : receive $1,001,000 V ( B 1 ) = V ( M ) P ( M | B 1 ) + V ( N ) P ( N | B 1 ) = 1000000 · 0 . 99 + 0 · 0 . 01 = 990 , 000 19 / 24

Suppose 99% confidence in predictors reliability. B 1 : one-box (open box B ) B 2 : two-box choice (open both A and B ) N : receive nothing K : receive $1,000 M : receive $1,000,000 L : receive $1,001,000 V ( B 1 ) = V ( M ) P ( M | B 1 ) + V ( N ) P ( N | B 1 ) = 1000000 · 0 . 99 + 0 · 0 . 01 = 990 , 000 V ( B 2 ) = V ( L ) P ( L | B 2 ) + V ( K ) P ( K | B 2 ) 19 / 24

Suppose 99% confidence in predictors reliability. B 1 : one-box (open box B ) B 2 : two-box choice (open both A and B ) N : receive nothing K : receive $1,000 M : receive $1,000,000 L : receive $1,001,000 V ( B 1 ) = V ( M ) P ( M | B 1 ) + V ( N ) P ( N | B 1 ) = 1000000 · 0 . 99 + 0 · 0 . 01 = 990 , 000 V ( B 2 ) = V ( L ) P ( L | B 2 ) + V ( K ) P ( K | B 2 ) = 1001000 · 0 . 01 + 1000 · 0 . 99 19 / 24

Suppose 99% confidence in predictors reliability. B 1 : one-box (open box B ) B 2 : two-box choice (open both A and B ) N : receive nothing K : receive $1,000 M : receive $1,000,000 L : receive $1,001,000 V ( B 1 ) = V ( M ) P ( M | B 1 ) + V ( N ) P ( N | B 1 ) = 1000000 · 0 . 99 + 0 · 0 . 01 = 990 , 000 V ( B 2 ) = V ( L ) P ( L | B 2 ) + V ( K ) P ( K | B 2 ) = 1001000 · 0 . 01 + 1000 · 0 . 99 = 11 , 000 19 / 24

Let µ be the assigned to the conditional B 1 � → M (and B 2 � → L ) (both conditionals are true iff the Predictor put $1,000,000 in box B yesterday). B 1 : one-box (open box B ) B 2 : two-box choice (open both A and B ) N : receive nothing K : receive $1,000 M : receive $1,000,000 L : receive $1,001,000 20 / 24