155_Fall_18_soln (4) - Game Theory Homework#4 All the problems are worth 4 points each and will be graded on a 0\/1\/2\/3\/4 scale Due on Sunday before

# 155_Fall_18_soln (4) - Game Theory Homework#4 All the...

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This preview shows page 1 out of 3 pages. Unformatted text preview: Game Theory Homework #4, 09/22/2018 All the problems are worth 4 points each and will be graded on a 0/1/2/3/4 scale. Due on Sunday 09/30/2018 before 11:59 pm to be uploaded via Gradescope. 1. Two smart students form a study group in some math class where homework is handed in jointly by each group. In the last homework of the semester, each of the two students can choose to either work ("W") or party ("P"). If at least one of them solves the homework that week (chooses "W"), then they will both receive 10 points. But solving the homework incurs a substantial eort, worth −7 points for a student doing it alone, and an eort worth −2 points for each student, if both students work together. Partying involves no eort, and if both students party, they both receive 0 points. Assume that the students do not communicate prior to deciding whether they will work or party. Write this situation as a matrix game and determine all Nash equilibria. Solution. Let's start with writing out the matrix of the game: W P (8, 8) (3, 10) (10, 3) (0, 0) W P We need to nd all Nash equilibria. Consider some Nash equilibrium (x∗ , y∗ ). There are 4 possible cases: • x∗ and y ∗ are both pure. • x∗ and y ∗ are both not pure. • x∗ is pure and y ∗ is not. • y ∗ is pure and x∗ is not. First case: If one student works, the optimal responce of another one is to party, and if one parties, the optimal responce of another is to work. Thus, there are 2 pure Nash equilibria in this game  (W, P) and (P, W). Second case: Since we have a 2 × 2 game, the player's optimal response can be not pure i the other player uses an equalizing strategy. Thus, x∗ and y∗ should be equalizing strategies. Let's nd equalizing strategies: 8x∗1 + 3(1 − x∗1 ) = 10x∗1 =⇒ x∗1 = 0.6. Thus (0.6, 0.4)T is the unique equalizing strategy of the rst player. Since the players are symmetric in this game, it is also the unique equalizing strategy of the second player. So, if the Nash equilibrium where both strategies are not pure exists, then it is (0.6, 0.4)T , (0.6, 0.4)T . However, since any response to the equalizing strategy is optimal, the pair of equalizing strategies is a Nash equilibrium. Finally, (0.6, 0.4)T , (0.6, 0.4)T is the only Nash equilibrium in this game, that has both strategies not pure. Third and fourth case: One of the strategies is not pure, so the other is equalizing. However, that other is also pure. Since 3 6= 0 and 8 6= 10, there are no pure equalizing strategies in this game.  Overall, all the Nash equilibria in this game are (W, P), (P, W), and (0.6, 0.4)T , (0.6, 0.4)T . 2. Consider the two player general sum game with payo matrix  (1, 1) (2, 0)  (2, 0) (−1, 5) Find safety strategies (for both players), and all mixed and pure Nash Equilibriua, and the respective expected payos for both players in each case. Solution. 1-st player safety strategy:  1 argmaxx min x 2 y T  2 y = argmaxx min(x1 + 2x2 , 2x1 − x2 ) = argmaxx min(2 − x1 , 3x1 − 1). −1 Since the function 2 − x1 is decreasing, and 3x1 − 1 is increasing, their minimum is maximal when they are equal. Thus 2 − x1 = 3x1 − 1 =⇒ x = (0.75, 0.25)T . The guaranteed expected payo to the rst player is 2 − 0.75 = 1.25. 2-nd player safety strategy: argmaxy min xT x Analogously to the previous part  1 0  0 y = argmaxy min(y1 , 5y2 ) = argmaxy min(y1 , 5 − 5y1 ). 5 y1 = 5 − 5y1 =⇒ y = (5/6, 1/6)T . The guaranteed expected payo to the second player is 5/6. Next, we search for the Nash equilibria. As in the previous problem there are 4 cases. 1) Pure equilibrium. Discard the elements of the payo matrices, which correspond to suboptimal moves of the players (leave column maxima for the rst player utility, and row maxima for the second):  (1, 1) (2, 0) (2, 0) (−1, 5)  ⇒   (∗, 1) (2, ∗) (2, ∗) (∗, 5) We see, that there is no pair of moves that are optimal responses to each other, so there are no pure equilibria. 2) Equalizing strategies. As in the previous problem, we are looking for a pair of equalizing strategies, and if we nd one, it is a Nash equilibrium: x1 = 5x2 ⇐⇒ x = (5/6, 1/6), Thus, y1 + 2y2 = 2y1 − y2 ⇐⇒ 3y2 = y1 ⇐⇒ y = (0.75, 0.25)T .  (5/6, 1/6)T , (0.75, 0.25)T        1 2 0.75 1 0 0.75 = (1.25, 5/6). (5/6, 1/6) , (5/6, 1/6) 2 −1 0.25 0 5 0.25 is a Nash equilibrium. The corresponding payos are 3,4) As in the previous case, there are no pure equalizing strategies.  Thus, the only Nash equilibrium is (5/6, 1/6)T , (0.75, 0.25)T with payos (1.25, 5/6). 3. Two cheetahs and three antelopes: Two cheetahs each chase one of three antelopes. If they catch the same one, they have to share. The antelopes are Large, Small, and Tiny, and their values to the cheetahs are l, s and t. Write the 3 × 3 matrix for this game. Assume that t < s < l < 2s: Find a condition for a symmetric fully mixed equilibrium to exist and nd the latter as well as all the pure equilibria in that case. Solution. First, let's nd all the pure equilibria. Underscore the maximal payos to the PI in columns and to the PII in rows (since l < 2s, it holds s > 2l ): ll ( 2 2 ) ( l, s) ( l, t) ( s, l) ( s , s ) (s, t) 2 2 (t, l) (t, s) ( 2t , 2t ) We see that there are 2 pure NE: (s, l) and (l, s). Next, we look for a symmetric fully mixed equilibrium. For a strategy x the pair (x, x) is a fuppy mixed symmetric equilibrium i x is a fully equalizing strategy. Write out the equations: x1 + x2 + x3 = 1, x1 , x2 , x3 > 0, l s t x1 + lx2 + lx3 = sx1 + x2 + sx3 = tx1 + tx2 + x3 2 2 2 Using the rst equation we can rewrite the last line as l(1 − x1 /2) = s(1 − x2 /2) = t(1 − x3 /2) = A, where we introduced a new variable A. Now we can express x1 = 2(1 − A/l), x2 = 2(1 − A/s), x3 = 2(1 − A/t). Plugging that into the rst equation, we can nd A: x1 + x2 + x3 = 1 = 2(3 − A(l−1 + s−1 + t−1 )) =⇒ A = 2.5 . l−1 + s−1 + t−1 Thus, 2t−1 + 2s−1 − 3l−1 , l−1 + s−1 + t−1 2t−1 + 2l−1 − 3s−1 x2 =2(1 − As−1 ) = −1 , l + s−1 + t−1 2l−1 + 2s−1 − 3t−1 x3 =2(1 − At−1 ) = −1 l + s−1 + t−1 x1 =2(1 − Al−1 ) = The only condition that should be met for this to be a valid strategy is x1 , x2 , x3 > 0. Note that since t < s < l, x3 is the smallest component. Thus the only condition we should have is x3 > 0, which is 2l−1 + 2s−1 > 3t−1 . 4. Volunteering dilemma: There are n players in a game show. Each player is put in a separate room. If some of the players volunteer to help the others, then each volunteer will receive 1000 and each of the remaining players will receive 1500. If no player volunteers, then they all get zero. Show that for this game the set of symmetric (mixed) Nash equilibria contains exactly one element. Let pn denote the probability that player 1 volunteers in this equilibrium. Find p2 and show that lim npn = log(3). n→∞ Solution. First, let's show that there are no pure symmetric equilibria. Indeed, if all players volunteer, then they all get 1000, any player can deviate and get 1500 instead. On the other hand, if all players don't volunteer, they all get 0 and any player can deviate and get 1000. Thus, the set of the symmetric equilibria consists of fully mixed equilibria (since a player has only 2 options, any startegy is either pure or fully mixed), and pn 6= 0, pn 6= 1. When every player volunteers with probability pn ∈ (0, 1), it is a NE i both options of every player are equalized. Let's write that out with a formula: • If a player volunteers, then she gets 1000, no matter what others do. • If a player does not volunteer, she gets 1500 unless all the others don't volunteer, which happens with probability (1 − pn )n−1 . Thus, shwe gets 1500 − 1500(1 − pn )n−1 . Overall, pn gives a symmetric mixed NE i 1000 = 1500 − 1500(1 − pn )n−1 , (1 − pn )n−1 = 3−1 , 1 pn = 1 − 3− n−1 . Thus, p2 = 2/3 and      log(3) log(3) n npn = n 1 − e− n−1 = n 1 − 1 − + o(n−1 ) = log 3 + o(1), n−1 n−1 where we used the Taylor formula ex = 1 + x + o(x) when x → 0. Thus, lim npn = log(3). n→∞ ...
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