Strategic Games: a Mini-Course Krzysztof R. Apt CWI, Amsterdam, the - PowerPoint PPT Presentation

IENBR: Example 1 X Y A 2 , 1 0 , 0 B 0 , 1 2 , 0 C 1 , 1 1 , 2 No strategy strictly or weakly dominates another one. C is never a best response. Eliminating it we get X Y A 2 , 1 0 , 0 B 0 , 1 2 , 0 from which in two steps we get X A 2 , 1 Strategic Games: a Mini-Course – p. 34/11

IENBR Theorem If G ′ is an outcome of IENBR starting from a finite G , then s is a Nash equilibrium of G ′ iff it is a Nash equilibrium of G . If G is finite and is solved by IENBR, then the resulting joint strategy is a unique Nash equilibrium of G . (Apt, ’05) Outcome of IENBR is unique (order independence). Strategic Games: a Mini-Course – p. 35/11

IENBR: Example 2 Location game on the open real interval (0 , 100) . s i + s 3 − i − s i  if s i < s 3 − i  2    100 − s i + s i − s 3 − i p i ( s i , s 3 − i ) := if s i > s 3 − i 2    50 if s i = s 3 − i  No strategy strictly or weakly dominates another one. Only 50 is a best response to some strategy (namely 50). So this game is solved by IENBR, in one step. Strategic Games: a Mini-Course – p. 36/11

More on Nash Equilibria Strategic Games: a Mini-Course – p. 37/11

Overview Best response dynamics. Potential games. Congestion games. Examples. Price of Stability. Mixed strategies. Nash Theorem. Strategic Games: a Mini-Course – p. 38/11

Best Response Dynamics Consider a game G := ( S 1 , . . ., S n , p 1 , . . ., p n ) . An algorithm to find a Nash equilibrium: choose s ∈ S 1 × . . . × S n ; while s is not a NE do choose i ∈ { 1 , . . ., n } such that s i is not a best response to s − i ; s i := a best response to s − i od Strategic Games: a Mini-Course – p. 39/11

Best Response Dynamics, ctd Best response dynamics may miss a Nash equilibrium. Example (Shoham and Leyton-Brown ’09) H T E H 1 , − 1 − 1 , 1 − 1 , − 1 T − 1 , 1 1 , − 1 − 1 , − 1 E − 1 , − 1 − 1 , − 1 − 1 , − 1 Strategic Games: a Mini-Course – p. 40/11

Potential Games (Monderer and Shapley ’96) Consider a game G := ( S 1 , . . ., S n , p 1 , . . ., p n ) . Function P : S 1 × . . .S n → R is a potential function for G if ∀ i ∈ { 1 , . . ., n } ∀ s − i ∈ S − i ∀ s i , s ′ i ∈ S i p i ( s i , s − i ) − p i ( s ′ i , s − i ) = P ( s i , s − i ) − P ( s ′ i , s − i ) . Intuition: P tracks the changes in the payoff when some player deviates. Potential game: a game that has a potential function. Strategic Games: a Mini-Course – p. 41/11

Example Prisoner’s dilemma for n players. � 2 � j � = i s j + 1 if s i = 0 p i ( s ) := 2 � j � = i s j if s i = 1 For i = 1 , 2 p i (0 , s − i ) − p i (1 , s − i ) = 1 . So P ( s ) := − � n j =1 s j is a potential function. Strategic Games: a Mini-Course – p. 42/11

Potential Games, ctd Theorem (Monderer and Shapley ’96) Every finite potential game has a Nash equilibrium. Proof 1. The games ( S 1 , . . ., S n , p 1 , . . ., p n ) and ( S 1 , . . ., S n , P, . . ., P ) have the same set of Nash equilibria. Take s for which P reaches maximum. Then s is a Nash equilibrium of ( S 1 , . . ., S n , P, . . ., P ) . Proof 2. For finite potential games the best response dynamics terminates. Strategic Games: a Mini-Course – p. 43/11

Congestion Games n > 1 players, Finite set E of facilities (road segments, primary production factors, . . . ), each strategy is a non-empty subset of E , each player has a possibly different set of strategies, d j : { 1 , . . ., n } → R is the delay function for using j ∈ E , d j ( k ) is the delay for using j when there are k users of j , x j ( s ) := |{ r ∈ { 1 , . . ., n } | j ∈ s r }| is the number of users of facility j given s , c i ( s ) := � j ∈ s i d j ( x j ( s )) , (We use here cost functions c i instead of payoff functions p i .) Strategic Games: a Mini-Course – p. 44/11

Example 5 drivers. Each driver chooses a road from Katowice to Gliwice, More drivers choose the same road: more delays. KATOWICE 1/2/3 1/4/5 1/5/6 GLIWICE Strategic Games: a Mini-Course – p. 45/11

Example as a Congestion Game 5 players, 3 facilities (roads), each strategy: (a singleton set consisting of) a road, cost function:  1 if s i = 1 and |{ j | s j = 1 }| = 1    2 if s i = 1 and |{ j | s j = 1 }| = 2      3 if s i = 1 and |{ j | s j = 1 }| ≥ 3  c i ( s ) := 1 if s i = 2 and |{ j | s j = 2 }| = 1    . . .      6 if s i = 3 and |{ j | s j = 3 }}| ≥ 3  Strategic Games: a Mini-Course – p. 46/11

Possible Evolution (1) KATOWICE 1/2/3 1/4/5 1/5/6 GLIWICE Strategic Games: a Mini-Course – p. 47/11

Possible Evolution (2) KATOWICE 1/2/3 1/4/5 1/5/6 GLIWICE Strategic Games: a Mini-Course – p. 48/11

Possible Evolution (3) KATOWICE 1/2/3 1/4/5 1/5/6 GLIWICE Strategic Games: a Mini-Course – p. 49/11

Possible Evolution (4) KATOWICE 1/2/3 1/4/5 1/5/6 GLIWICE We reached a Nash equilibrium using the best response dynamics. Strategic Games: a Mini-Course – p. 50/11

Congestion Games, ctd Theorem (Rosenthal, ’73) Every congestion game is a potential game. Proof. x j ( s ) � � P ( s ) := d j ( k ) , j ∈ s 1 ∪ . . . ∪ s n k =1 where (recall) x j ( s ) = |{ r ∈ { 1 , . . ., n } | j ∈ s r }| , is a potential function. Conclusion Every congestion game has a Nash equilibrium. Strategic Games: a Mini-Course – p. 51/11

Another Example Assumptions: 4000 drivers drive from A to B. Each driver has 2 options (strategies). U T/100 45 A B 45 T/100 R Problem: Find a Nash equilibrium (T = number of drivers). Strategic Games: a Mini-Course – p. 52/11

Nash Equilibrium U T/100 45 A B 45 T/100 R Answer: 2000/2000. Travel time: 2000/100 + 45 = 45 + 2000/100 = 65. Strategic Games: a Mini-Course – p. 53/11

Braess Paradox Add a fast road from U to R. Each driver has now 3 options (strategies): A - U - B, A - R - B, A - U - R - B. U T/100 45 0 A B 45 T/100 R Problem: Find a Nash equilibrium. Strategic Games: a Mini-Course – p. 54/11

Nash Equilibrium U T/100 45 0 A B 45 T/100 R Answer: Every driver will choose the road A - U - R - B. Why?: The road A - U - R - B is always a best response. Strategic Games: a Mini-Course – p. 55/11

Small Complication U T/100 45 0 A B 45 T/100 R Travel time: 4000/100 + 4000/100 = 80! Strategic Games: a Mini-Course – p. 56/11

Does it happen? from Wikipedia (‘Braess Paradox’): In Seoul, South Korea, a speeding-up in traffic around the city was seen when a motorway was removed as part of the Cheonggyecheon restoration project. In Stuttgart, Germany after investments into the road network in 1969, the traffic situation did not improve until a section of newly-built road was closed for traffic again. In 1990 the closing of 42nd street in New York City reduced the amount of congestion in the area. In 2008 Youn, Gastner and Jeong demonstrated specific routes in Boston, New York City and London where this might actually occur and pointed out roads that could be closed to reduce predicted travel times. Strategic Games: a Mini-Course – p. 57/11

Price of Stability Definition PoS: social welfare of the best Nash equilibrium social welfare of the social optimum Question: What is the price of stability for the congestion games? Strategic Games: a Mini-Course – p. 58/11

Example n A B x n - (even) number of players. x - number of drivers on the bottom road. Two Nash equilibria 1 / ( n − 1) , with the social cost n + ( n − 1) 2 . 0 /n , with the social cost n 2 . Social optimum Take f ( x ) = x · x + ( n − x ) · n = x 2 − n · x + n 2 . We want to find a minimum of f . f ′ ( x ) = 2 x − n , so f ′ ( x ) = 0 if x = n 2 . Strategic Games: a Mini-Course – p. 59/11

Example n A B x Best Nash equilibrium 1 / ( n − 1) , with social cost n + ( n − 1) 2 . Social optimum f ( x ) = x 2 − n · x + n 2 . 2 ) = 3 Social optimum = f ( n 4 n 2 . n +( n − 1) 2 4 n 2 = 4 PoS = ( n + ( n − 1) 2 ) / 3 . 3 n 2 lim n →∞ PoS = 4 3 . Strategic Games: a Mini-Course – p. 60/11

Price of Stability Theorem (Roughgarden and Tárdos, 2002) Suppose delay functions (e.g., T/ 100 ) are linear. Then the price of stability for the congestion games is ≤ 4 3 . Strategic Games: a Mini-Course – p. 61/11

Mixed Extension of a Finite Game Probability distribution over a finite non-empty set A : π : A → [0 , 1] such that � a ∈ A π ( a ) = 1 . Notation: ∆ A . Fix a finite strategic game G := ( S 1 , . . ., S n , p 1 , . . ., p n ) . Mixed strategy of player i in G : m i ∈ ∆ S i . Joint mixed strategy: m = ( m 1 , . . ., m n ) . Strategic Games: a Mini-Course – p. 62/11

Mixed Extension of a Finite Game (2) Mixed extension of G : (∆ S 1 , . . ., ∆ S n , p 1 , . . ., p n ) , where m ( s ) := m 1 ( s 1 ) · . . . · m n ( s n ) and � p i ( m ) := m ( s ) · p i ( s ) . s ∈ S Theorem (Nash ’50) Every mixed extension of a finite strategic game has a Nash equilibrium. Strategic Games: a Mini-Course – p. 63/11

Kakutani’s Fixed Point Theorem Theorem (Kakutani ’41) Suppose A is a compact and convex subset of R n and Φ : A → P ( A ) is such that Φ( x ) is non-empty and convex for all x ∈ A , for all sequences ( x i , y i ) converging to ( x, y ) y i ∈ Φ( x i ) for all i ≥ 0 , implies that y ∈ Φ( x ) . Then x ∗ ∈ A exists such that x ∗ ∈ Φ( x ∗ ) . Strategic Games: a Mini-Course – p. 64/11

Proof of Nash Theorem Fix ( S 1 , . . ., S n , p 1 , . . ., p n ) . Define best i : Π j � = i ∆ S j → P (∆ S i ) by best i ( m − i ) := { m ′ i ∈ ∆ S i | p i ( m ′ i , m − i ) attains the maximum } . Then define best : ∆ S 1 × . . . ∆ S n → P (∆ S 1 × . . . × ∆ S n ) by best ( m ) := best 1 ( m − 1 ) × . . . × best 1 ( m − n ) . Note m is a Nash equilibrium iff m ∈ best ( m ) . best ( · ) satisfies the conditions of Kakutani’s Theorem. Strategic Games: a Mini-Course – p. 65/11

Comments First special case of Nash theorem: Cournot (1838). Nash theorem generalizes von Neumann’s Minimax Theorem (’28). An alternative proof (also by Nash) uses Brouwer’s Fixed Point Theorem. Search for conditions ensuring existence of Nash equilibrium. Strategic Games: a Mini-Course – p. 66/11

2 Examples Matching Pennies H T H 1 , − 1 − 1 , 1 T − 1 , 1 1 , − 1 ( 1 2 · H + 1 2 · T, 1 2 · H + 1 2 · T ) is a Nash equilibrium. The payoff to each player in the Nash equilibrium: 0. The Battle of the Sexes F B F 2 , 1 0 , 0 B 0 , 0 1 , 2 (2 / 3 · F + 1 / 3 · B, 1 / 3 · F + 1 / 3 · B ) is a Nash equilibrium. The payoff to each player in the Nash equilibrium: 2 / 3 . Strategic Games: a Mini-Course – p. 67/11

Mechanism Design Strategic Games: a Mini-Course – p. 68/11

Overview Decision problems. Direct mechanisms. Groves mechanisms. Examples. Optimality results. Strategic Games: a Mini-Course – p. 69/11

Intelligent Design Strategic Games: a Mini-Course – p. 70/11

Intelligent Design A theory of an intelligently guided invisible hand wins the Nobel prize WHAT on earth is mechanism design? was the typical reaction to this year’s Nobel prize in economics, announced on October 15th. [...] In fact, despite its dreary name, mechanism design is a hugely important area of economics, and underpins much of what dismal scientists do today. It goes to the heart of one of the biggest challenges in economics: how to arrange our economic interactions so that, when everyone behaves in a self-interested manner, the result is something we all like. (The Economist, Oct. 18th, 2007) Strategic Games: a Mini-Course – p. 71/11

Decision Problems Decision problem for n players: set D of decisions, for each player i a set of (private) types Θ i and a utility function v i : D × Θ i → R . Intuitions Type is some private information known only to the player (e.g., player’s valuation of the item for sale), v i ( d, θ i ) represents the benefit to player i of type θ i from the decision d ∈ D . Assume the individual types are θ 1 , . . ., θ n . Then � n i =1 v i ( d, θ i ) is the social welfare from d ∈ D . Strategic Games: a Mini-Course – p. 72/11

Decision Rules Decision rule is a function f : Θ 1 × . . . × Θ n → D. Decision rule f is efficient if n n � � v i ( f ( θ ) , θ i ) ≥ v i ( d, θ i ) i =1 i =1 for all θ ∈ Θ and d ∈ D . Intuition f is efficient if it always maximizes the social welfare. Strategic Games: a Mini-Course – p. 73/11

Set up Each player i receives/has a type θ i , each player i submits to the central authority a type θ ′ i , the central authority computes decision d := f ( θ ′ 1 , . . ., θ ′ n ) , and communicates it to each player i . Basic problem How to ensure that θ ′ i = θ i . Strategic Games: a Mini-Course – p. 74/11

Example 1: Sealed-Bid Auction Set up There is a single object for sale. Each player is a buyer. The decision is taken by means of a sealed-bid auction. The object is sold to the highest bidder. D = { 1 , . . . , n } , each Θ i is R + , � θ i if d = i v i ( d, θ i ) := 0 otherwise Let argsmax θ := µi ( θ i = max j ∈{ 1 ,...,n } θ j ) . f ( θ ) := argsmax θ. Note f is efficient. Payments will be treated later. Strategic Games: a Mini-Course – p. 75/11

Example 2: Public Project Problem Each person is asked to report his or her willingness to pay for the project, and the project is undertaken if and only if the aggregate reported willingness to pay exceeds the cost of the project. (15 October 2007, The Royal Swedish Academy of Sciences, Press Release, Scientific Background) Strategic Games: a Mini-Course – p. 76/11

Public Project Problem, formally c : cost of the public project (e.g., building a bridge), D = { 0 , 1 } , each Θ i is R + , v i ( d, θ i ) := d ( θ i − c n ) , � if � n 1 i =1 θ i ≥ c f ( θ ) := 0 otherwise Note f is efficient. Strategic Games: a Mini-Course – p. 77/11

Ex. 3: Reversed Sealed-bid Auction Set up Each player offers the same service. The decision is taken by means of a sealed-bid auction. The service is purchased from the lowest bidder. D = { 1 , . . . , n } , each Θ i is R − ; − θ i is the price player i offers, � θ i if d = i v i ( d, θ i ) := 0 otherwise f ( θ ) := argsmax θ. Example f ( − 8 , − 5 , − 4 , − 6) = 3 . That is, given the offers 8 , 5 , 4 , 6 , the service is bought from player 3. Strategic Games: a Mini-Course – p. 78/11

Ex. 4: Buying a Path in a Network Set up Given a graph G := ( V, E ) . • Each edge e ∈ E is owned by player e . • Two distinguished vertices s, t ∈ V . • Each player e submits the cost θ e of using the edge e . • The central authority selects the shortest s − t path in G . D = { p | p is a s − t path in G } , each Θ i is R + , � − θ i if i ∈ p v i ( p, θ i ) := 0 otherwise f ( θ ) := p , where p is the shortest s − t path in G . Strategic Games: a Mini-Course – p. 79/11

Manipulations Example An optimal strategy for player i in public project problem: if θ i ≥ c n submit θ ′ i = c . if θ i < c n submit θ ′ i = 0 . For example, assume c = 30 . player type A 6 B 7 C 25 Players A and B should submit 0. Player c should submit 30. Strategic Games: a Mini-Course – p. 80/11

Revised Set-up: Direct Mechanisms Each player i receives/has a type θ i , each player i submits to the central authority a type θ ′ i , the central authority computes decision d := f ( θ ′ 1 , . . ., θ ′ n ) , and taxes ( t 1 , . . ., t n ) := g ( θ ′ 1 , . . ., θ ′ n ) ∈ R n , and communicates to each player i both d and t i . final utility function for player i : u i ( d, θ i ) := v i ( d, θ i ) + t i . Strategic Games: a Mini-Course – p. 81/11

Direct Mechanisms, ctd Direct mechanism ( f, t ) is incentive compatible if for all θ ∈ Θ , i ∈ { 1 , . . ., n } and θ ′ i ∈ Θ i u i (( f, t )( θ i , θ − i ) , θ i ) ≥ u i (( f, t )( θ ′ i , θ − i ) , θ i ) . Intuition Submitting false type (so θ ′ i � = θ i ) does not pay off. Direct mechanism ( f, t ) is feasible if � n i =1 t i ( θ ) ≤ 0 for all θ . Intuition External financing is never needed. Strategic Games: a Mini-Course – p. 82/11

Groves Mechanisms t i ( θ ) := � j � = i v j ( f ( θ ) , θ j ) + h i ( θ − i ) , where h i : Θ − i → R is an arbitrary function. Note u i (( f, t )( θ ) , θ i ) = � n j =1 v j ( f ( θ ) , θ j ) + h i ( θ − i ) . Intuitions Player i cannot manipulate the value of h i ( θ − i ) . Suppose h i ( θ − i ) = 0 . When the individual types are θ 1 , . . ., θ n u i (( f, t )( θ ) , θ i ) is the social welfare from f ( θ ) . Strategic Games: a Mini-Course – p. 83/11

Groves Theorem Theorem (Groves ’73) Suppose f is efficient. Then each Groves mechanism is incentive compatible. Proof. For all θ ∈ Θ , i ∈ { 1 , . . . , n } and θ ′ i ∈ Θ i n � u i (( f, t )( θ i , θ − i ) , θ i ) = v j ( f ( θ i , θ − i ) , θ j ) + h i ( θ − i ) j =1 n � v j ( f ( θ ′ ( f is efficient ) ≥ i , θ − i ) , θ j ) + h i ( θ − i ) j =1 = u i (( f, t )( θ ′ i , θ − i ) , θ i ) . Strategic Games: a Mini-Course – p. 84/11

Special Case: Pivotal Mechanism � j � = i v j ( d, θ j ) . h i ( θ − i ) := − max d ∈ D Then � � t i ( θ ) := v j ( f ( θ ) , θ j ) − max v j ( d, θ j ) ≤ 0 . d ∈ D j � = i j � = i Note Pivotal mechanism is feasible. Strategic Games: a Mini-Course – p. 85/11

Re: Sealed-Bid Auction Note In the pivotal mechanism � − max j � = i θ j if i = argsmax θ. t i ( θ ) = 0 otherwise So the pivotal mechanism is Vickrey auction (Vickrey ’61): the winner pays the 2nd highest bid. Strategic Games: a Mini-Course – p. 86/11

Example player bid tax to authority util. A 18 0 0 B 24 − 21 3 C 21 0 0 Social welfare: 0 + 0 + 3 = 3. Strategic Games: a Mini-Course – p. 87/11

Maximizing Social Welfare Question: Does Vickrey auction maximize social welfare? Notation θ ∗ : the reordering of θ is descending order. Example For θ = (1 , 4 , 2 , 3 , 1) we have θ − 2 = (1 , 2 , 3 , 0) , ( θ − 2 ) ∗ = (3 , 2 , 1 , 0) , so ( θ − 2 ) ∗ 2 = 2 . Intuition ( θ − 2 ) ∗ 2 is the second highest bid among other bids. Strategic Games: a Mini-Course – p. 88/11

Bailey-Cavallo Mechanism (Bailey ’97, Cavallo ’06) Assume n ≥ 3 . i ( θ ) + ( θ − i ) ∗ t i ( θ ) := t p 2 n Note Bailey-Cavallo mechanism is a Groves mechanism. Example player bid tax to authority util. why? A (= 1 / 3 of 21 ) 18 0 7 B (= 24 − 2 − 7 − 6 ) 24 − 2 9 C (= 1 / 3 of 18 ) 21 0 6 Strategic Games: a Mini-Course – p. 89/11

Bailey-Cavallo Mechanism, ctd Note Bailey-Cavallo mechanism is feasible. Proof. For all i and θ , ( θ − i ) ∗ 2 ≤ θ ∗ 2 , so n n n ( θ − i ) ∗ − θ ∗ 2 + ( θ − i ) ∗ � t i ( θ ) = − θ ∗ � 2 � 2 2 + = ≤ 0 . n n i =1 i =1 i =1 Strategic Games: a Mini-Course – p. 90/11

Re: Public Project Problem Assume the pivotal mechanism. Examples Suppose c = 30 and n = 3 . player type tax u i A 6 0 − 4 B 7 0 − 3 C 25 − 7 8 Social welfare can be negative. player type tax u i A 4 − 5 − 5 B 3 − 6 − 6 C 22 0 0 Strategic Games: a Mini-Course – p. 91/11

Formally Note In the pivotal mechanism n c and � n j � = i θ j ≥ n − 1  if � 0 j =1 θ j ≥ c   n c and � n j � = i θ j − n − 1 j � = i θ j < n − 1  � n c if � j =1 θ j ≥ c  t i ( θ ) = n c and � n j � = i θ j ≤ n − 1 if � 0 j =1 θ j < c   n c and � n  n − 1 j � = i θ j > n − 1 n c − � j � = i θ j if � j =1 θ j < c  This is the mechanism essentially proposed in Clarke ’71). Strategic Games: a Mini-Course – p. 92/11

Optimality Result (1) Theorem (Apt, Conitzer, Guo and Markakis ’08) Consider the sealed bid auction. No tax-based mechanism exists that is feasible, incentive compatible, ‘better’ than Bailey-Cavallo mechanism. Strategic Games: a Mini-Course – p. 93/11

Optimality Result (2) Theorem (Apt, Conitzer, Guo and Markakis ’08) Consider the public project problem. No tax-based mechanism exists that is feasible, incentive compatible, ‘better’ than Clarke’s tax. Strategic Games: a Mini-Course – p. 94/11

Proof Steps 1. Limit attention to Groves mechanisms. (B. Holmström, ’79) 2. Introduce anonymous Groves mechanisms. 3. Pivotal mechanism t is here anonymous. 4. Each Groves mechanism t entails an anonymous Groves mechanism t ′ . 5. Lemma If t is feasible, then so is t ′ . If t is ‘better’ than t 0 , then so is t ′ . 6. Lemma No feasible anonymous Groves mechanism is ‘better’ than the pivotal mechanism t . Strategic Games: a Mini-Course – p. 95/11

However . . . Pivotal mechanism is not optimal in the public project problem when the payments per player can differ. Note: Pivotal mechanism then ceases to be anonymous. Strategic Games: a Mini-Course – p. 96/11

Re: Reversed Sealed-Bid Auction Take � � t i ( θ ) := v j ( f ( θ ) , θ j ) − max v j ( d, θ j ) . d ∈ D \{ i } j � = i j � = i Note � − max j � = i θ j if i = argsmax θ. t i ( θ ) = 0 otherwise So in this mechanism the winner receives the amount equal to the 2nd lowest offer. Example Consider Θ = ( − 8 , − 5 , − 4 , − 6) . The service is bought from player 3 who receives for it 5. Strategic Games: a Mini-Course – p. 97/11

Re: Buying a Path in a Network (Nisan, Ronen ’99) Take � � t i ( θ ) := v j ( f ( θ ) , θ j ) − max v j ( p, θ j ) . p ∈ D ( G \{ i } ) j � = i j � = i Note � cost ( p 2 ) − cost ( p 1 − { i } ) if i ∈ p 1 t i ( θ ) = 0 otherwise where p 1 is the shortest s − t path in G ( θ ) , p 2 is the shortest s − t path in ( G \ { i } )( θ − i ) . Strategic Games: a Mini-Course – p. 98/11

Example Consider the player owning the edge e . To compute the payment he receives determine the shortest s − t path. Its length is 7. It contains e . determine the shortest s − t path that does not include e . Its length is 12. So player e receives 12 − (7 − 4) = 9 . His final utility is 9 − 4 = 5 . Strategic Games: a Mini-Course – p. 99/11

Pre-Bayesian Games Strategic Games: a Mini-Course – p. 100/11