Game Theory and Strategy · Quiz 1 · May 2024

Rows are player 1 bids; columns are player 2 bids. Each pair is (player 1 payoff, player 2 payoff).\[\begin{array}{c|ccc} &0&1&2\\ 0&(3/2,5/2)&(0,4)&(0,3)\\ 1&(2,0)&(1,2)&(0,3)\\ 2&(1,0)&(1,0)&(1/2,3/2)\end{array}\] **1. Calculate the payoffs:** A higher bidder earns value minus bid; the loser earns zero. A tie at b gives expected payoff \((v_i-b)/2\), producing the displayed matrix. **2. Check pure-strategy strict dominance:** For player 2, bid 2 gives payoffs (3,3,1.5), strictly above bid 0’s (2.5,0,0). For player 1, no single alternative bid is strictly better than another at every column: bid 1 only weakly dominates 0, while bids 1 and 2 trade off gains. **3. State the convention-sensitive answer:** Under elimination by another pure strategy, player 1 has no strictly dominated strategy in the original matrix. **Answer: A — Player 1.** **Convention note:** The answer uses pure-strategy dominance, as the supplied key intends. If domination by mixed strategies is allowed, player 1’s bid 0 is strictly dominated by a mixture of 0.6 on bid 1 and 0.4 on bid 2: its payoffs (1.6,1,0.2) exceed (1.5,0,0). Under that broader definition, neither player lacks a strictly dominated strategy and option C would apply. The question does not state this restriction explicitly. A: **Correct:** Under pure-strategy comparisons, no row is strictly worse than another row in every column. B: **Incorrect:** Player 2’s bid 0 is strictly dominated by bid 2. C: **Incorrect:** Under the intended pure-strategy convention, player 1 does have the requested property.

Rows are player 1 bids; columns are player 2 bids. Each pair is (player 1 payoff, player 2 payoff).\[\begin{array}{c|ccc} &0&1&2\\ 0&(3/2,5/2)&(0,4)&(0,3)\\ 1&(2,0)&(1,2)&(0,3)\\ 2&(1,0)&(1,0)&(1/2,3/2)\end{array}\] **1. Remove player 2’s bid 0:** Against player 1 bids 0,1,2, player 2’s bid 2 earns (3,3,1.5), strictly exceeding bid 0’s (2.5,0,0). **2. Remove player 1’s bid 0:** In remaining columns 1 and 2, player 1’s bid 0 gives (0,0), while bid 2 gives (1,0.5), both strictly greater. **3. Continue in the reduced game:** Against remaining rows 1 and 2, player 2’s bid 2 gives (3,1.5), strictly exceeding bid 1’s (2,0). With only column 2 left, player 1 prefers bid 2’s 0.5 to bid 1’s 0. **4. Identify survivors:** Each player is left with bid 2. The surviving profile is (2,2), with expected payoffs (0.5,1.5). **Answer: C — (2, 2).** A: **Incorrect:** Both bid-0 strategies are eliminated during the procedure. B: **Incorrect:** Bid 1 is eliminated for both players as the game is reduced. C: **Correct:** Iterated strict elimination leaves only bid 2 for each player. D: **Incorrect:** There is a surviving strategy for each player.

Rows are player 1 actions; columns are player 2 actions. Each pair is (player 1 payoff, player 2 payoff).\[\begin{array}{c|cc}&H&D\\ H&(-1,-1)&(10,0)\\ D&(0,10)&(1,1)\end{array}\] **1. Calculate each outcome:** With v=10, k=6 and d=4, each player gets \(v/2-k=-1\) at HH and \(v/2-d=1\) at DD. A lone hawk gets 10 while the dove gets 0. **2. Find best responses:** Against H, D is better because 0>-1. Against D, H is better because 10>1. Thus HD and DH are the two pure-strategy Nash equilibria. **3. Select the outcomes that are not equilibria:** At HH either player can switch to D and improve to 0. At DD either can switch to H and improve to 10. Therefore select HH and DD. **Answer: A, D — (H, H); (D, D).** A: **Correct:** Select this outcome: either player can deviate from H to D, raising payoff from -1 to 0, so HH is not Nash. B: **Incorrect:** Do not select: the hawk prefers 10 to 1, and the dove prefers 0 to -1, so HD is Nash. C: **Incorrect:** Do not select: the dove prefers 0 to -1, and the hawk prefers 10 to 1, so DH is Nash. D: **Correct:** Select this outcome: either player can deviate from D to H, raising payoff from 1 to 10, so DD is not Nash.

Rows are police choices; columns are dealer choices, with S=street and P=park. Each pair is (police payoff, dealer payoff).\[\begin{array}{c|cc}&S&P\\ S&(80,20)&(0,100)\\ P&(10,90)&(60,40)\end{array}\] **1. Find the officer’s best responses:** If the dealer chooses street, the officer prefers street (80>10). If the dealer chooses park, the officer prefers park (60>0). **2. Find the dealer’s best responses:** Against street patrol, the dealer prefers park (100>20). Against park patrol, the dealer prefers street (90>40). **3. Check mutual best responses:** In every cell at least one player benefits by switching. Therefore no pure-strategy Nash equilibrium exists. **Answer: C — There is no Nash equilibrium in this game in pure strategy.** **Convention note:** Literally zero is a finite number, so B is imprecise rather than logically disjoint from C. C states the exact conclusion and is the intended single best answer; B should say “a positive finite number” to avoid overlap. A: **Incorrect:** No cell is a mutual best response, so there is no unique pure equilibrium. B: **Incorrect:** The intended contrast is a positive finite number of pure equilibria; here the number is zero. C: **Correct:** Every pure profile permits a profitable unilateral deviation. D: **Incorrect:** The explicit no-pure-equilibrium alternative applies.

Rows are police choices; columns are dealer choices, with S=street and P=park. Each pair is (police payoff, dealer payoff).\[\begin{array}{c|cc}&S&P\\ S&(80,20)&(0,100)\\ P&(10,90)&(60,40)\end{array}\] **1. Make the dealer indifferent:** Let alpha be the officer’s probability of street patrol. The dealer’s street payoff is \(20\alpha+90(1-\alpha)=90-70\alpha\); its park payoff is \(100\alpha+40(1-\alpha)=40+60\alpha\). Equality gives \(\alpha=5/13\), hence park patrol probability 8/13. **2. Make the officer indifferent:** Let beta be the dealer’s probability of street. The officer’s street payoff is \(80\beta\); park payoff is \(10\beta+60(1-\beta)=60-50\beta\). Equality gives \(\beta=6/13\), hence dealer park probability 7/13. **3. Establish uniqueness:** There is no pure equilibrium, all best responses to pure actions are strict, and each indifference equation has a unique interior solution. Thus the unique equilibrium has officer mix (5/13,8/13) and dealer mix (6/13,7/13). **Answer: A — There is a unique Nash equilibrium in this game.** **Convention note:** The options overlap literally: a unique equilibrium is also finitely many. A is the more precise single best answer intended by the source. This equilibrium is in mixed, not pure, strategies. A: **Correct:** The two indifference equations identify exactly one mixed-strategy equilibrium. B: **Incorrect:** One is finite, so this is literally true but weaker than the explicit uniqueness statement; it is not the intended alternative. C: **Incorrect:** A mixed equilibrium exists even though no pure equilibrium does. D: **Incorrect:** The uniqueness alternative gives the exact result.

Rows are police choices; columns are dealer choices, with S=street and P=park. Each pair is (police payoff, dealer payoff).\[\begin{array}{c|cc}&S&P\\ S&(80,20)&(0,100)\\ P&(10,90)&(60,40)\end{array}\] **1. Make the dealer indifferent:** Let alpha be the officer’s probability of street patrol. The dealer’s street payoff is \(20\alpha+90(1-\alpha)=90-70\alpha\); its park payoff is \(100\alpha+40(1-\alpha)=40+60\alpha\). Equality gives \(\alpha=5/13\), hence park patrol probability 8/13. **2. Make the officer indifferent:** Let beta be the dealer’s probability of street. The officer’s street payoff is \(80\beta\); park payoff is \(10\beta+60(1-\beta)=60-50\beta\). Equality gives \(\beta=6/13\), hence dealer park probability 7/13. **3. Establish uniqueness:** There is no pure equilibrium, all best responses to pure actions are strict, and each indifference equation has a unique interior solution. Thus the unique equilibrium has officer mix (5/13,8/13) and dealer mix (6/13,7/13). **Answer: C — 5/13, 8/13.** A: **Incorrect:** These are the dealer’s probabilities, not the officer’s. B: **Incorrect:** These are neither the officer’s street/park pair nor a dealer-indifference solution. C: **Correct:** This pair makes the dealer indifferent between its two locations. D: **Incorrect:** This reverses the officer’s correct street and park probabilities.

Rows are police choices; columns are dealer choices, with S=street and P=park. Each pair is (police payoff, dealer payoff).\[\begin{array}{c|cc}&S&P\\ S&(80,20)&(0,100)\\ P&(10,90)&(60,40)\end{array}\] **1. Make the dealer indifferent:** Let alpha be the officer’s probability of street patrol. The dealer’s street payoff is \(20\alpha+90(1-\alpha)=90-70\alpha\); its park payoff is \(100\alpha+40(1-\alpha)=40+60\alpha\). Equality gives \(\alpha=5/13\), hence park patrol probability 8/13. **2. Make the officer indifferent:** Let beta be the dealer’s probability of street. The officer’s street payoff is \(80\beta\); park payoff is \(10\beta+60(1-\beta)=60-50\beta\). Equality gives \(\beta=6/13\), hence dealer park probability 7/13. **3. Establish uniqueness:** There is no pure equilibrium, all best responses to pure actions are strict, and each indifference equation has a unique interior solution. Thus the unique equilibrium has officer mix (5/13,8/13) and dealer mix (6/13,7/13). **Answer: B — 6/13, 7/13.** A: **Incorrect:** This reverses the dealer’s street and park probabilities. B: **Correct:** This makes the officer indifferent between the two patrol choices. C: **Incorrect:** At beta=4/13, officer street and park payoffs are unequal. D: **Incorrect:** At beta=9/13, officer street and park payoffs are unequal.

Payoffs are (player 1, player 2). Each downward arrow continues play; each rightward arrow ends it. All three decision nodes are singleton information sets.\[\begin{array}{rcl}P_1&\xrightarrow{A}&(2,0)\\ {\scriptstyle B}\downarrow&&\\ P_2&\xrightarrow{C}&(3,1)\\ {\scriptstyle D}\downarrow&&\\ P_1&\xrightarrow{E}&(0,0)\\ {\scriptstyle F}\downarrow&&\\ (1,2)&&\end{array}\] **1. Identify what each player knows:** Player 2 moves only after B, and player 1’s final decision occurs only after the history BD. Each moving player can identify the exact decision node. **2. Classify information:** Each information set contains a single decision node. The game therefore has perfect information. **Answer: A — Perfect information.** A: **Correct:** Every decision maker knows the full history when acting. B: **Incorrect:** No indistinguishable decision nodes or hidden past moves are specified. C: **Incorrect:** The game fits the perfect-information category. D: **Incorrect:** The sequence and observations determine the classification.

Payoffs are (player 1, player 2). Each downward arrow continues play; each rightward arrow ends it. All three decision nodes are singleton information sets.\[\begin{array}{rcl}P_1&\xrightarrow{A}&(2,0)\\ {\scriptstyle B}\downarrow&&\\ P_2&\xrightarrow{C}&(3,1)\\ {\scriptstyle D}\downarrow&&\\ P_1&\xrightarrow{E}&(0,0)\\ {\scriptstyle F}\downarrow&&\\ (1,2)&&\end{array}\] **1. List the complete terminal histories:** The game can end after A, after BC, after BDE, or after BDF. **2. Count leaves:** These four histories end at four separate terminal nodes, with payoffs (2,0),(3,1),(0,0),(1,2). **Answer: C — 4.** A: **Incorrect:** This counts the three decision nodes, not the terminal leaves. B: **Incorrect:** The final E/F choices are not the only ways the game can end. C: **Correct:** The four terminal histories are A, BC, BDE and BDF. D: **Incorrect:** There is no fifth terminal history in the tree.

Payoffs are (player 1, player 2). Each downward arrow continues play; each rightward arrow ends it. All three decision nodes are singleton information sets.\[\begin{array}{rcl}P_1&\xrightarrow{A}&(2,0)\\ {\scriptstyle B}\downarrow&&\\ P_2&\xrightarrow{C}&(3,1)\\ {\scriptstyle D}\downarrow&&\\ P_1&\xrightarrow{E}&(0,0)\\ {\scriptstyle F}\downarrow&&\\ (1,2)&&\end{array}\] **1. Locate decision nodes:** There is an initial player 1 node, a player 2 node after B, and a later player 1 node after BD. **2. Count information sets:** Under perfect information each decision node is its own singleton information set, giving 3. Terminal nodes are not decision information sets. **Answer: B — 3.** A: **Incorrect:** This counts players rather than information sets; player 1 has two distinct decision situations. B: **Correct:** Each of the three decision nodes forms a separate information set. C: **Incorrect:** Four is the number of terminal nodes, not information sets. D: **Incorrect:** All decision nodes cannot share one information set across different players and histories.

Payoffs are (player 1, player 2). Each downward arrow continues play; each rightward arrow ends it. All three decision nodes are singleton information sets.\[\begin{array}{rcl}P_1&\xrightarrow{A}&(2,0)\\ {\scriptstyle B}\downarrow&&\\ P_2&\xrightarrow{C}&(3,1)\\ {\scriptstyle D}\downarrow&&\\ P_1&\xrightarrow{E}&(0,0)\\ {\scriptstyle F}\downarrow&&\\ (1,2)&&\end{array}\] **1. Use a complete contingent plan:** A pure strategy specifies what player 1 would do initially and what they would do later after BD, even if the later node is never reached. **2. Multiply independent choices:** There are 2 initial actions and 2 later actions, giving \(2\times2=4\) strategies: AE, AF, BE and BF. **Answer: D — 4.** A: **Incorrect:** Player 1 has choices at two separate decision nodes. B: **Incorrect:** This counts actions at a single node rather than complete plans. C: **Incorrect:** AE and AF remain distinct full strategies even though both stop play immediately. D: **Correct:** The four complete plans are AE, AF, BE and BF.

Payoffs are (player 1, player 2). Each downward arrow continues play; each rightward arrow ends it. All three decision nodes are singleton information sets.\[\begin{array}{rcl}P_1&\xrightarrow{A}&(2,0)\\ {\scriptstyle B}\downarrow&&\\ P_2&\xrightarrow{C}&(3,1)\\ {\scriptstyle D}\downarrow&&\\ P_1&\xrightarrow{E}&(0,0)\\ {\scriptstyle F}\downarrow&&\\ (1,2)&&\end{array}\] **1. Locate player 2’s choice:** Player 2 acts at only one information set, after player 1 chooses B. **2. List complete plans:** Choosing C or choosing D are the two possible pure strategies. No additional contingency for player 2 is needed. **Answer: B — 2.** A: **Incorrect:** Player 2 has two available actions, so there is more than one strategy. B: **Correct:** The complete strategy set is {C,D}. C: **Incorrect:** Player 2 has no third action or extra information set. D: **Incorrect:** Four is player 1’s strategy count, not player 2’s.

Payoffs are (player 1, player 2). Each downward arrow continues play; each rightward arrow ends it. All three decision nodes are singleton information sets.\[\begin{array}{rcl}P_1&\xrightarrow{A}&(2,0)\\ {\scriptstyle B}\downarrow&&\\ P_2&\xrightarrow{C}&(3,1)\\ {\scriptstyle D}\downarrow&&\\ P_1&\xrightarrow{E}&(0,0)\\ {\scriptstyle F}\downarrow&&\\ (1,2)&&\end{array}\] Rows are complete player 1 plans (initial action, later action); columns are player 2 plans. Each pair is (player 1 payoff, player 2 payoff).\[\begin{array}{c|cc}&C&D\\ AE&(2,0)&(2,0)\\ AF&(2,0)&(2,0)\\ BE&(3,1)&(0,0)\\ BF&(3,1)&(1,2)\end{array}\] **1. Find player 1’s best responses:** Against C, BE and BF earn 3, exceeding AE/AF’s 2. Against D, AE and AF earn 2, exceeding BE’s 0 and BF’s 1. **2. Find player 2’s best responses:** Against AE or AF, C and D both yield 0. Against BE, C is better (1>0). Against BF, D is better (2>1). **3. Intersect the best responses:** The pure equilibria are (AE,D), (AF,D), and (BE,C), giving exactly three. **Answer: C — 3.** A: **Incorrect:** This counts only one equilibrium and misses two other mutual best responses. B: **Incorrect:** There are two equilibria with initial A plus one with initial B. C: **Correct:** Exactly (AE,D), (AF,D) and (BE,C) are mutual best responses. D: **Incorrect:** (BF,C) is not an equilibrium because player 2 gains by switching to D.

Payoffs are (player 1, player 2). Each downward arrow continues play; each rightward arrow ends it. All three decision nodes are singleton information sets.\[\begin{array}{rcl}P_1&\xrightarrow{A}&(2,0)\\ {\scriptstyle B}\downarrow&&\\ P_2&\xrightarrow{C}&(3,1)\\ {\scriptstyle D}\downarrow&&\\ P_1&\xrightarrow{E}&(0,0)\\ {\scriptstyle F}\downarrow&&\\ (1,2)&&\end{array}\] **1. Solve the last decision:** After BD, player 1 chooses F because 1>0. Any subgame-perfect strategy must prescribe F there, including off the equilibrium path. **2. Solve player 2’s decision:** Anticipating F, choosing D yields 2 while C yields 1, so player 2 chooses D. **3. Solve the initial decision:** Player 1 compares A’s payoff 2 with the continuation BF payoff 1 and chooses A. The full subgame-perfect strategy profile is (AF,D); the realized outcome is A with payoffs (2,0). **Answer: B — (AF, D).** A: **Incorrect:** Prescribing E after BD is not optimal in that subgame, since F gives player 1 a larger payoff. B: **Correct:** Every prescribed decision is optimal in its continuation subgame. C: **Incorrect:** After D, E is not credible; backward induction selects F and then D instead of C. D: **Incorrect:** Given the continuation F, player 2 prefers D to C, so this profile is not even Nash.

Payoffs are (player 1, player 2). Each downward arrow continues play; each rightward arrow ends it. All three decision nodes are singleton information sets.\[\begin{array}{rcl}P_1&\xrightarrow{A}&(2,0)\\ {\scriptstyle B}\downarrow&&\\ P_2&\xrightarrow{C}&(3,1)\\ {\scriptstyle D}\downarrow&&\\ P_1&\xrightarrow{E}&(0,0)\\ {\scriptstyle F}\downarrow&&\\ (1,2)&&\end{array}\] **1. Apply the subgame definition:** In this perfect-information tree, each decision node can start a subgame consisting of that node and all its successors. **2. List all roots:** They are the initial player 1 node, the player 2 node after B, and the player 1 node after BD. Counting the whole game gives 3 subgames; excluding it would leave 2 proper subgames. **Answer: C — 3.** **Convention note:** The standard count here includes the entire game, as the supplied answer does. The question does not ask specifically for proper subgames. A: **Incorrect:** This counts only the entire game and misses two continuation subgames. B: **Incorrect:** This is the number of proper subgames, excluding the whole game. C: **Correct:** There are three subgames when the whole game is included. D: **Incorrect:** Terminal nodes alone are not counted as strategic subgames in this convention.

One ordinal payoff assignment is: no entry (1,2), accommodated entry (2,1), and fought entry (0,0). Pairs are (entrant payoff, incumbent payoff); only their order matters.\[\begin{array}{rcl}P_1&\xrightarrow{N}&(1,2)\\ {\scriptstyle E}\downarrow&&\\ P_2&\xrightarrow{A}&(2,1)\\ {\scriptstyle F}\downarrow&&\\ (0,0)&&\end{array}\] **1. Represent the preferences:** The entrant ranks accommodated entry above no entry above fought entry. The incumbent ranks no entry above accommodation above fighting. The displayed ordinal payoffs respect both rankings. **2. Solve the incumbent’s subgame:** Once entry occurs, player 2 prefers accommodation to fighting, since accommodation is not its least-preferred outcome whereas fighting is. It therefore chooses A. **3. Solve the entrant’s decision:** Anticipating accommodation, player 1 prefers E to N. Thus the unique subgame-perfect equilibrium is (E,A), producing accommodated entry. **Answer: B — (E, A).** A: **Incorrect:** Fighting after entry is not credible: the incumbent would prefer accommodation in that subgame. B: **Correct:** Entry followed by accommodation is optimal at both decision nodes. C: **Incorrect:** The no-entry/fight profile can be Nash in the strategic form, but it is not subgame perfect. D: **Incorrect:** Option B is the backward-induction solution.