Exercise [exe:classify].

(a) Two players, three actions each, simultaneous, complete information. (b) Seller and buyer; seller moves first (set price / describe car), buyer accepts or not; sequential with incomplete information (quality is the seller’s private type)—a signaling/lemons situation (Chapters [ch:bayesian]– [ch:dynamicincomplete]). (c) Two players, sequential, perfect information, finite (with repetition rules); by Zermelo’s theorem it is determined (Chapter [ch:extensive]). (d) \(n\) bidders move simultaneously (sealed bids); incomplete information (private valuations): a Bayesian game (Chapter [ch:bayesian]).

Exercise [exe:realpd].

Any answer with: two parties, mutual cooperation better than mutual defection, but defection individually profitable at each opponent action. Typical sustaining mechanisms: repetition and retaliation (Chapter [ch:repeated]), reputation, contracts, norms with sanctions.

Exercise [exe:notmoney].

With money-only preferences, backward induction gives the younger sibling accepting any positive amount, so the older offers the smallest possible slice. With the anger term, accepting less than a third yields (slice \(- 10\)) \(<\) 0 for small slices, so rejection is credible; the older sibling optimally offers exactly one third. Preferences, not the game tree, changed—and so did the prediction.

Chapter [ch:rationality]

Exercise [exe:moneypump].

Start her with \(z\). She pays \(\varepsilon\) to swap \(z \to y\) (since \(y \succ z\)), \(\varepsilon\) to swap \(y \to x\), then—by the cycle \(z \succ x\)—\(\varepsilon\) to swap \(x \to z\). She is back at \(z\) having paid \(3\varepsilon\); repeat forever. Cyclic preferences are exploitable, hence “irrational.”

Exercise [exe:affine].

Expected utility of action \(a_i\) under belief \(\mu\) is linear in payoffs, so under \(v_1 = 3u_1 + 7\) every expected payoff is transformed by the same increasing affine map; rankings, hence \(\argmax\) sets, are unchanged. For the nonlinear part: let action \(S\) (safe) pay \(0\) always and action \(R\) pay \(+1\) with probability \(2/3\) and \(-2\) with probability \(1/3\) (probabilities induced by the opponent’s mixture). Under \(u\): \(\E[R] = \tfrac23 - \tfrac23 = 0 = \E[S]\) (indifferent). Under \(u^3\): \(\E[R] = \tfrac23(1) + \tfrac13(-8) = -2 < 0 = \E[S]\). The cube preserves the ordinal ranking of sure outcomes but changes the best response.

Exercise [exe:certequiv].

\(\E[W] = 50\). \(\E[u] = \tfrac12\sqrt{100} = 5\). Certainty equivalent: \(u(CE) = 5 \Rightarrow CE = 25\). Risk premium \(= 50 - 25 = 25\).

Exercise [exe:allais].

Write \(A\) and \(B\) as mixtures involving the common component: with \(P =\) (1M for sure) and \(Q = (\tfrac{1}{11} \text{ chance } 0,\ \tfrac{10}{11} \text{ chance } 5M)\): \(A = 0.11 P + 0.89 P\) and \(B = 0.11 Q + 0.89 P\), while \(C = 0.11 P + 0.89\, \delta_0\) and \(D = 0.11 Q + 0.89\, \delta_0\) (\(\delta_0\) = zero for sure). Independence applied to the common \(0.89\)-components forces \(A \succsim B \iff P \succsim Q \iff C \succsim D\). The modal pattern \(A \succ B\), \(D \succ C\) contradicts this.

Chapter [ch:normalform]

Exercise [exe:domhunt].

Step 1: for player 1, \(T = (2,1,4)\) strictly dominates \(B = (1,0,3)\); delete \(B\). Step 2: in the reduced game, for player 2, \(R\) (payoffs \(2,3\) vs \(T,M\)) strictly dominates \(C\) (\(1,2\)); delete \(C\). No further strict domination: the survivors are \(\{T, M\} \times \{L, R\}\). Within the reduced game, best-response marking gives two pure Nash equilibria, \((M, L)\) and \((T, R)\) (plus one mixed). Moral: IESDS need not reach a single profile; it brackets the predictions.

Exercise [exe:weakorder].

Order 1: delete \(T\) (weakly dominated by \(M\)); then \(L\) (weakly dominated by \(R\): \(1 = 1\) vs \(M\), \(0 < 1\) vs \(B\)); outcome payoffs \((2,1)\). Order 2: delete \(B\) (weakly dominated by \(M\)); then \(R\) (now weakly dominated by \(L\): \(1 = 1\) vs \(M\), \(0 < 1\) vs \(T\)); remaining outcomes pay \((1,1)\). Different orders, different predictions—why iteration uses strict dominance.

Exercise [exe:cournotdom].

\(u_i = q_i(12 - q_i - q_j)\). Best response \(\BR(q_j) = (12 - q_j)/2 = 6 - q_j/2\). Any \(q_i > 6\) is strictly dominated by \(q_i = 6\) (marginal profit \(12 - 2q_i - q_j < 0\) beyond 6 for all \(q_j \ge 0\)). Given \(q_j \le 6\), \(\BR \ge 3\): outputs below 3 are dominated. Iterating: the surviving interval \([a_k, b_k]\) satisfies \(a_{k+1} = 6 - b_k/2\), \(b_{k+1} = 6 - a_k/2\), contracting to the fixed point \(q = 6 - q/2 \Rightarrow q = 4\). The game is dominance solvable at the Cournot–Nash output \(q_1 = q_2 = 4\).

Exercise [exe:traveler].

Claiming 100 is weakly dominated by 99 (if the rival claims 100 you get \(99 + 2\) instead of \(100\); other cases tie or win). Once 100 is removed, 99 is dominated by 98, and so on down to the unique survivor \((2,2)\). Most people (and experimental subjects) claim 95–100: with a small penalty/bonus parameter, the IESDS logic feels implausibly aggressive—a benchmark case for level-\(k\) thinking (Chapter [ch:frontiers]).

Chapter [ch:nash]

Exercise [exe:findpure].

Best responses — row: vs \(X\): \(A\); vs \(Y\): \(B\); vs \(Z\): \(C\). Column: vs \(A\): \(X\); vs \(B\): \(Y\); vs \(C\): \(Z\). Pure equilibria: \((A,X)\) with \((3,2)\), \((B,Y)\) with \((2,3)\), \((C,Z)\) with \((2,2)\).

Exercise [exe:partnership].

FOC for \(i\): \(\tfrac12(1 + \tfrac12 e_j) - 2 e_i = 0 \Rightarrow e_i = \tfrac14 + e_j/8\). Symmetric solution: \(e^* = \tfrac14 + e^*/8 \Rightarrow e^* = \tfrac{2}{7}\). Joint surplus maximization: \(1 + \tfrac12 e_j - 2e_i = 0\) symmetric gives \(e^{\mathrm{opt}} = \tfrac23 > \tfrac27\). Each partner bears the full cost of effort but captures only half the extra output—classic team free-riding (Example [exa:publicgood] in miniature).

Exercise [exe:hotelling].

At \((\tfrac12, \tfrac12)\) each gets \(\tfrac12\); deviating to \(x \neq \tfrac12\) yields market share \(\tfrac{x + 1/2}{2} < \tfrac12\) (for \(x < \tfrac12\))—no gain. If the profile has some \(x_i \neq \tfrac12\), the rival can move adjacent to \(x_i\) on the longer side and capture more than half, so no other profile is an equilibrium. Political reading: with single-peaked voters, two vote-maximizing candidates converge to the median voter’s position.

Exercise [exe:noopt].

Impossible. If \((a_1^*, a_2^*)\) is Nash, then \(u_1(a^*) \geq u_1(a_1, a_2^*)\) for the other row(s): player 1’s equilibrium payoff is (weakly) the best in its column, so it cannot be the strict minimum of all four cells. Strongest possible statement: the unique equilibrium can be strictly Pareto-dominated by another cell, and can give each player their worst payoff outside their own deviation set—the Prisoner’s Dilemma achieves both players’ second-worst payoffs, which is the worst possible.

Chapter [ch:mixed]

Exercise [exe:penalty].

Kicker mixes \(p\) on Left to equalize keeper’s sides: \(0.6p + 0.9(1-p) = 0.95p + 0.6(1-p) \Rightarrow p = \tfrac{6}{13}\). Keeper mixes \(q\) on Left to equalize kicker: \(0.6q + 0.95(1-q) = 0.9q + 0.6(1-q) \Rightarrow q = \tfrac{7}{13}\). Equilibrium scoring probability \(= 0.6 + 0.3\cdot\tfrac{7}{13} = \tfrac{9.9}{13} \approx 0.762\).

Exercise [exe:rps].

Standard RPS: by symmetry the value is 0; if any action had probability \(> \tfrac13\), its counter would earn strictly positive payoff, contradiction—so uniform is the unique equilibrium. Modified game (Rock beats Scissors for \(\pm 2\)): indifference conditions \(-p + 2s = r - s = -2r + p = 0\) with \(r + p + s = 1\) give \((r, p, s) = (\tfrac14, \tfrac12, \tfrac14)\). Rock’s probability falls (and Paper’s rises): making Rock’s win bigger makes opponents avoid Scissors and play Paper, which punishes Rock. You exploit an action’s strength by playing its counter’s counter, not the action itself.

Exercise [exe:volunteer].

Symmetric equilibrium: calling gives \(v - c\); not calling gives \(v\,[1 - (1-p)^{n-1}]\). Indifference: \((1-p)^{n-1} = c/v\), so \(p^* = 1 - (c/v)^{1/(n-1)}\), decreasing in \(n\). Probability nobody calls: \((1-p^*)^n = (c/v)^{n/(n-1)}\), increasing in \(n\) (exponent falls toward 1 and \(c/v < 1\)). More bystanders, more diffusion of responsibility, less help—the equilibrium logic of the bystander effect.

Exercise [exe:support].

Let \(q\) = probability of \(L\). Row payoffs: \(u(T) = 6q\), \(u(M) = 4 - 2q\), \(u(B) = 3\). There is no pure equilibrium (marking fails everywhere). All three row payoffs are equal only at \(q = \tfrac12\) (all give 3). Column is indifferent when \(2t + 4m + 3b = 6t + 2m + 3b \Rightarrow m = 2t\). Hence a component of equilibria: \(\sigma_2 = (\tfrac12, \tfrac12)\) and \(\sigma_1 = (t,\, 2t,\, 1 - 3t)\) for any \(t \in [0, \tfrac13]\) (all give payoffs \((3, 3)\)). This degenerate game illustrates why “generic” games (Theorem [thm:oddness]) have finitely many equilibria but special payoff ties create continua.

Chapter [ch:extensive]

Exercise [exe:bipractice].

Backward induction: after \(L\), player 2 picks \(r\) (\(2 > 1\)); after \(R\), \(\ell'\) (\(1 > 0\)). Player 1 compares \(L \to (1,2)\) with \(R \to (2,1)\) and plays \(R\): outcome \((2,1)\). Normal form (player 2’s strategies \(\ell\ell', \ell r', r\ell', rr'\)): pure Nash equilibria are \((R, r\ell')\)—the SPE—and \((L, rr')\), which is sustained by the incredible plan to play \(r'\) after \(R\) (at that unreached node \(r'\) pays \(0 < 1\)).

Exercise [exe:burnmoney].

After burning, row’s payoffs are shifted down by 1. Key steps of iterated weak dominance on the \(4 \times 4\) normal form: (i) for row, “burn then play \(F\)” is dominated (best case \(0\)); (ii) given (i), column infers from observed burning that row will play \(B\), so column plays \(B\) after burning; (iii) then “burn, \(B\)” guarantees row \(2\), which dominates “don’t burn, \(F\)” (at most \(1\)); (iv) column, anticipating (iii), plays \(B\) even when row does not burn; (v) so row plays “don’t burn, \(B\)” and gets \(3\). The unused option to burn selects row’s favorite equilibrium—forward induction reasoning in action.

Exercise [exe:seqbos].

Sequential BoS: row commits to \(B\); column’s best response is \(B\); outcome \((3,1)\) —a first-mover advantage, because the game rewards being predictable (coordination). In Matching Pennies the first mover loses for the same reason reversed: pure conflict rewards being unpredictable, creating a second-mover advantage. The dividing line: does your opponent want to match you (move first) or mismatch you (move second)?

Exercise [exe:pirates].

Convention: strict majority; indifferent pirates vote against. With 2 pirates (4 proposes to 5): 5 rejects everything (he then gets all 100 and sees 4 overboard), so 4 dies: with 2 left, proposer 4 cannot win. Hence with 3 pirates, 4 votes for anything 3 proposes (voting no means the 2-pirate subgame where he dies): 3 proposes \((100, 0, 0)\) and wins with votes \(\{3, 4\}\). With 4 pirates, proposer 2 needs 3 of 4 votes: buy the two pirates who get 0 in the 3-pirate continuation—4 and 5—with 1 coin each: \((98, 0, 1, 1)\). With 5 pirates, proposer 1 needs 3 of 5: continuation (from the 4-pirate case) gives \((-, 98, 0, 1, 1)\) to pirates \(2,3,4,5\); the two cheapest votes are pirate 3 (offer 1) and pirate 4 or 5 (offer 2): proposal \((97, 0, 1, 2, 0)\) (or \((97,0,1,0,2)\)). Seniority plus foresight is worth 97 coins.

Chapter [ch:applications]

Exercise [exe:cournotasym].

FOCs: \(q_i = \frac{a - c_i - q_j}{2}\). Solving: \(q_i^* = \frac{a - 2c_i + c_j}{3}\), \(u_i^* = (q_i^*)^2\). Lower cost \(\Rightarrow\) higher output and profit. Firm 2 is driven to zero when \(a - 2c_2 + c_1 \le 0\), i.e. \(c_2 \geq \frac{a + c_1}{2}\).

Exercise [exe:bertrandasym].

With the convention that the efficient (low-cost) firm wins ties: equilibrium \(p_1 = p_2 = c_2\); firm 1 serves the market earning margin \(c_2 - c_1\); firm 2 earns 0. (Without the tie-break convention, no equilibrium exists in continuous prices: firm 1 wants to price “just below” \(c_2\); on a discrete grid it prices one tick below.) Prices are pinned by the second-lowest cost—the Bertrand analogue of the second-price auction logic.

Exercise [exe:tullock].

FOC for \(i\): \(\frac{x_j}{(x_1 + x_2)^2} V = 1\). Symmetry (\(x_1 = x_2 = x\)): \(\frac{x}{4x^2}V = 1 \Rightarrow x^* = V/4\) each. Total rent dissipation \(x_1^* + x_2^* = V/2\): half the prize is burned; each contestant nets \(V/4\). (With \(n\) contestants, dissipation \(\to V\): rent-seeking can consume the entire rent.)

Exercise [exe:bankrun].

Consistent numbers: \(D = 100\), both wait \(\to 150\) each; one withdraws \(\to (100, 60)\); both withdraw \(\to 80\) each.

Both (Wait, Wait) and (Withdraw, Withdraw) are strict Nash equilibria: a run is a self-fulfilling coordination failure, not irrationality. Deposit insurance guarantees the waiter \(100 \ge\) any withdrawal payoff, making Wait a (weakly) dominant strategy and deleting the run equilibrium—which is why insurance works largely without being used.

Chapter [ch:bargaining]

Exercise [exe:nashproduct].

Pareto frontier: \(s_2 = 2 - s_1/2\). Maximize \((s_1 - 1)(\tfrac32 - \tfrac{s_1}{2})\): derivative \(2 - s_1 = 0 \Rightarrow s^* = (2, 1)\). On \(S' = S \cap \{s_1 \le 3\}\) the point \((2,1)\) is still feasible and the same maximization applies: unchanged, confirming IIA.

Exercise [exe:threeround].

\(t=2\): proposer 1 takes 1 (offers 0). \(t=1\): player 2 must offer player 1 at least \(\delta \cdot 1\); keeps \(1 - \delta\). \(t=0\): player 1 must offer player 2 \(\delta(1-\delta)\); keeps \(x_3 = 1 - \delta + \delta^2\). Generally \(x_T = 1 - \delta + \delta^2 - \dots \to \frac{1}{1+\delta}\), the Rubinstein share.

Exercise [exe:breakdown].

Stationary conditions: \(1 - x = (1-\pi) b_2 + \pi y\) and \(1 - y = (1-\pi) b_1 + \pi x\). Solving: \(x = \frac{1 - b_2 + \pi b_1}{1 + \pi}\) (and symmetrically for \(y\)). As \(\pi \to 1\): \(x \to b_1 + \frac{1 - b_1 - b_2}{2}\)—the Nash bargaining solution with the breakdown payoffs as the disagreement point. What matters for threat-point purposes is what happens if bargaining fails, not impatience per se.

Exercise [exe:deadline].

With a hard deadline at \(T\) and no discounting, backward induction gives the period-\(T\) proposer the whole pie (the responder accepts anything over 0); at \(T-1\) the responder can secure that by waiting, so the \(T-1\) proposer concedes everything, and so on: the equilibrium division is determined entirely by who proposes last, and agreement (in the frictionless model) is immediate at that division. Adding any small per-period cost or risk makes early agreement strict—but the terms still reflect deadline power, which is why real negotiations cluster near deadlines: delay is costly, and bargaining power shifts discontinuously at \(T\).

Chapter [ch:repeated]

Exercise [exe:thresholddelta].

(a) Grim: \(5(1-\delta) + \delta \le 2 \Rightarrow \delta \ge \tfrac34\). (b) One-period punishment: the deviation gains \(3\) today and costs only \((2-1)\delta\) next period: \(3(1-\delta) \le \delta(1-\delta) \Rightarrow \delta \geq 3\)—impossible. With these payoffs a single punishment period can never deter: the punishment must last several periods (roughly \(k\) with \(3 \le \sum_{j=1}^k \delta^j\)). Shorter, milder punishments demand more patience; sometimes no feasible patience suffices.

Exercise [exe:tftnotspe].

After 1’s unilateral defection, mutual TFT prescribes the alternating path \((C,D), (D,C), (C,D), \dots\) Check player 2’s one-shot deviation at the start of the punishment phase: following TFT (play \(D\)) yields the alternation worth \(\frac{4}{1+\delta}\); deviating to \(C\) restores \((C,C)\) forever, worth \(3\). TFT requires \(\frac{4}{1+\delta} \ge 3 \Rightarrow \delta \le \tfrac13\). But sustaining cooperation on-path requires \(\delta \geq \tfrac13\). Both hold only at the knife-edge \(\delta = \tfrac13\): mutual TFT is essentially never subgame perfect, despite being a Nash equilibrium for all \(\delta \ge \tfrac13\).

Exercise [exe:folkgeometry].

Feasible set: convex hull of \((3,1)\), \((1,3)\), \((0,0)\). Minmax for player 1: player 2 chooses \(q\) on \(B\) to minimize \(\max(3q, 1-q)\): at \(3q = 1 - q\), \(q = \tfrac14\), value \(\tfrac34\); symmetrically \(\underline v_2 = \tfrac34\). Sustainable set: feasible payoffs dominating \((\tfrac34, \tfrac34)\). The alternating scheme’s average payoff \(\to (2, 2)\) as \(\delta \to 1\): feasible, individually rational, hence sustainable —and indeed the natural “fair” convention.

Exercise [exe:community].

Rule: “cooperate iff your partner is labeled good; defecting on a good partner makes you bad (forever, or for \(k\) periods).” A bad-labeled player is defected on by all future partners, so the one-shot gain \(g\) is weighed against a permanent per-period loss, exactly as in grim trigger: cooperation is an equilibrium for \(\delta\) above a threshold. With noisy labels, permanent exclusion destroys too much surplus by accident; finite punishments, forgiveness, and label-repair (apologies, paying fines, rebuilding scores) become optimal—which is what credit systems and platform reputation mechanisms implement.