Game Theory and Strategy · End Term · May 2025 · FN

**Step 1: List the production shares and defective rates for machines A, B, and C.** - Machine A: \(P(A) = 0.50\), \(P(D \mid A) = 0.03\) - Machine B: \(P(B) = 0.30\), \(P(D \mid B) = 0.04\) - Machine C: \(P(C) = 0.20\), \(P(D \mid C) = 0.05\) **Step 2: Apply the Law of Total Probability to find \(P(D)\).** \[P(D) = P(A)P(D \mid A) + P(B)P(D \mid B) + P(C)P(D \mid C)\] \[P(D) = (0.50)(0.03) + (0.30)(0.04) + (0.20)(0.05) = 0.015 + 0.012 + 0.010 = 0.037\] **Answer: 0.037**

**Step 1: Apply Bayes' Theorem to find \(P(A \mid D)\).** From Q#341, \(P(A)P(D \mid A) = 0.015\) and \(P(D) = 0.037\): \[P(A \mid D) = \frac{P(A)P(D \mid A)}{P(D)} = \frac{0.015}{0.037} = \frac{15}{37} \approx 0.4054\] (Within the accepted range `0.40 to 0.405`, **0.405** is the exact rounded value.) **Answer: 0.405**

**Step 1: Write Roommate 2's payoff function \(u_2(t_1, t_2)\).** \[u_2(t_1, t_2) = (10 - t_1)t_2 - t_2^2\] **Step 2: Find Roommate 2's best-response function by maximizing \(u_2\) with respect to \(t_2\).** \[\frac{\partial u_2}{\partial t_2} = 10 - t_1 - 2t_2 = 0 \implies t_2^*(t_1) = \frac{10 - t_1}{2}\] **Step 3: Substitute \(t_1 = 0\).** \[t_2^*(0) = \frac{10 - 0}{2} = 5\] **Answer: 5**

**Step 1: Solve for the symmetric pure-strategy Nash equilibrium \(t_1 = t_2 = t^*\).** From the best-response condition \(t_i^* = \frac{10 - t_j^*}{2}\), setting \(t_1^* = t_2^* = t^*\) gives: \[t^* = \frac{10 - t^*}{2} \implies 2t^* = 10 - t^* \implies 3t^* = 10 \implies t_1 = t_2 = \frac{10}{3}\] **Step 2: Identify \(\alpha\) from \(t_1 = t_2 = \frac{\alpha}{3}\).** Comparing \(\frac{10}{3}\) with \(\frac{\alpha}{3}\) gives: \[\alpha = 10\] **Answer: 10**

Seller 1 owns the watch, which has a standalone worth of ₹200 to him. Even without trading with any buyer, Seller 1 can keep the watch and guarantee himself a value of: \[v(1) = 200\] **Answer: 200**

Buyer 2 does not own a watch and cannot buy one without Seller 1 in the coalition. Thus, Buyer 2 alone generates zero value: \[v(2) = 0\] **Answer: 0**

Buyer 3 does not own a watch and cannot buy one without Seller 1 in the coalition. Thus, Buyer 3 alone generates zero value: \[v(3) = 0\] **Answer: 0**

By the definition of a characteristic function in coalitional game theory, the value of the empty coalition \(\emptyset\) is always zero: \[v(\emptyset) = 0\] **Answer: 0**

Coalition \(\{2, 3\}\) consists only of the two buyers, neither of whom owns a watch. Without Seller 1, no trade can take place and no watch is held, so the coalition value is zero: \[v(2, 3) = 0\] **Answer: 0**

In coalition \(\{1, 2\}\), Seller 1 (who values the watch at ₹200) and Buyer 2 (who values the watch at ₹280) can transfer the watch to Buyer 2, achieving a maximum joint worth of: \[v(1, 2) = \max(200, 280) = 280\] **Answer: 280**

In coalition \(\{1, 3\}\), Seller 1 (who values the watch at ₹200) and Buyer 3 (who values the watch at ₹300) can transfer the watch to Buyer 3, achieving a maximum joint worth of: \[v(1, 3) = \max(200, 300) = 300\] **Answer: 300**

In the grand coalition \(\{1, 2, 3\}\), there is one watch, and the maximum total value is achieved by assigning the watch to the player who values it most—Buyer 3 (₹300): \[v(1, 2, 3) = \max(200, 280, 300) = 300\] **Answer: 300**

**Step 1: Set the utility of the indifferent consumer \(\hat{\theta}\) equal to the price \(p\).** A consumer of type \(\theta\) obtains utility \(U(\theta) = \theta(a + v n^e)\) from joining the network. The consumer who is indifferent between buying and not buying at price \(p\) satisfies: \[U(\hat{\theta}) = p \implies \hat{\theta}(a + v n^e) = p \implies \hat{\theta} = \frac{p}{a + v n^e}\] **Answer: B (\(\frac{p}{(a + v n^e)}\))** A: **Incorrect:** \(\frac{p - a}{v n^e}\) corresponds to the additive utility model \(a + \theta v n^e\), not the multiplicative model \(\theta(a + v n^e)\). B: **Correct:** Solving \(\hat{\theta}(a + v n^e) = p\) yields \(\hat{\theta} = \frac{p}{a + v n^e}\). C: **Incorrect:** Multiplies the numerator by \(a\) unnecessarily. D: **Incorrect:** Option B is the exact expression for \(\hat{\theta}\).

Since consumer types \(\theta\) are uniformly distributed on \([0, 1]\), the highest-valuation consumer has \(\theta = 1\) with willingness to pay: \[U(1) = 1 \cdot (a + v n^e) = a + v n^e\] For there to be a non-negative mass of buyers (\(\hat{\theta} \le 1\)), the price \(p\) cannot exceed \(a + v n^e\). **Answer: A (\(a + v n^e\))** A: **Correct:** The highest type \(\theta = 1\) has valuation \(a + v n^e\), which is the maximum price at which anyone will buy. B: **Incorrect:** Omits the standalone benefit \(a\). C: **Incorrect:** Divides \(a\) by \(v n^e\) instead of adding them. D: **Incorrect:** Does not match \(U(1) = a + v n^e\).

Every consumer with type \(\theta \ge \hat{\theta}\) obtains utility \(\theta(a + v n^e) \ge p\) and chooses to buy the good. Since \(\theta \sim U[0, 1]\), the mass of buyers \(n\) in the interval \([\hat{\theta}, 1]\) is: \[n = 1 - \hat{\theta}\] **Answer: C (\(1 - \hat{\theta}\))** A: **Incorrect:** \(\hat{\theta}\) is the mass of non-buyers in \([0, \hat{\theta})\). B: **Incorrect:** Exceeds the total population mass of 1. C: **Correct:** Since buyers are those with \(\theta \in [\hat{\theta}, 1]\), their mass under \(U[0, 1]\) is \(1 - \hat{\theta}\). D: **Incorrect:** Does not represent the uniform probability on \([\hat{\theta}, 1]\).

**Step 1: Substitute \(\hat{\theta} = 1 - n\) into the indifference condition \(p = \hat{\theta}(a + v n^e)\).** From Q#353 and Q#355, \(\hat{\theta} = \frac{p}{a + v n^e}\) and \(n = 1 - \hat{\theta} \implies \hat{\theta} = 1 - n\). Therefore: \[p(n^e, n) = (a + v n^e)(1 - n)\] **Answer: D (\((a + v n^e)(1 - n)\))** A: **Incorrect:** Assumes \(\hat{\theta} = 1\) regardless of \(n\). B: **Incorrect:** Uses \(n\) instead of \(1 - n\). C: **Incorrect:** Uses \(1 + n\) instead of \(1 - n\). D: **Correct:** Combining \(p = \hat{\theta}(a + v n^e)\) and \(\hat{\theta} = 1 - n\) gives \(p(n^e, n) = (a + v n^e)(1 - n)\).

From Q#356, the willingness-to-pay function is \(p(n^e, n) = (a + v n^e)(1 - n)\) with \(a > 0, v > 0\), and \(n \in (0, 1)\): 1. **With respect to \(n\)**: \(\frac{\partial p}{\partial n} = -(a + v n^e) < 0\), so \(p(n^e, n)\) is a **decreasing function of \(n\)** (**Option B**). 2. **With respect to \(n^e\)**: \(\frac{\partial p}{\partial n^e} = v(1 - n) > 0\), so \(p(n^e, n)\) is an **increasing function of \(n^e\)** (**Option D**). **Answer: B, D** A: **Incorrect:** Holding expectations \(n^e\) fixed, selling to a larger fraction \(n\) requires reaching lower-\(\theta\) consumers, so price decreases with \(n\). B: **Correct:** \(\partial p / \partial n = -(a + v n^e) < 0\), so \(p\) is decreasing in \(n\). C: **Incorrect:** Higher expected network size \(n^e\) raises willingness to pay, so \(p\) increases (not decreases) with \(n^e\). D: **Correct:** \(\partial p / \partial n^e = v(1 - n) > 0\), so \(p\) is increasing in \(n^e\).

Along the **fulfilled-expectations demand curve**, consumers' expectations about the network size match the actual number of buyers in equilibrium (\(n^e = n\)). Substituting \(n^e = n\) into \(p(n^e, n) = (a + v n^e)(1 - n)\) gives: \[p(n) = (a + v n)(1 - n)\] **Answer: D (\((a + v n)(1 - n)\))** A: **Incorrect:** Omits the downward-sloping demand factor \((1 - n)\). B: **Incorrect:** Uses \(n\) instead of \((1 - n)\). C: **Incorrect:** Uses \((1 + n)\) instead of \((1 - n)\). D: **Correct:** Imposing fulfilled expectations \(n^e = n\) in \(p(n^e, n) = (a + v n^e)(1 - n)\) yields \((a + v n)(1 - n)\).

**Step 1: Compute the Harsanyi dividends (excess values) in the partnership dissolution procedure.** Given \(N = \{1, 2, 3, 4\}\): - **Singletons**: - \(d(\{1\}) = v(1) = 24\) - \(d(\{2\}) = v(2) = 48\) - \(d(\{3\}) = v(3) = 24\) - \(d(\{4\}) = v(4) = 72\) - **Non-zero pairs** (dividing excess equally between the 2 members): - \(\{1, 2\}\): \(v(1, 2) - v(1) - v(2) = 96 - 24 - 48 = 24 \implies +12\) each to 1, 2. - \(\{2, 3\}\): \(v(2, 3) - v(2) - v(3) = 144 - 48 - 24 = 72 \implies +36\) each to 2, 3. - \(\{1, 4\}\): \(v(1, 4) - v(1) - v(4) = 120 - 24 - 72 = 24 \implies +12\) each to 1, 4. - **Coalition \(\{1, 3, 4\}\)**: - \(v(1, 3, 4) = 168\). Subtracting standalone values \(24 + 24 + 72 = 120\) and pair excess \(24\) (from \(\{1, 4\}\)) gives \(168 - 144 = 24 \implies +8\) each to 1, 3, 4. - **Grand coalition \(\{1, 2, 3, 4\}\)**: - \(v(1, 2, 3, 4) = 240\). Total allocated so far is \(168 + 24 + 72 + 24 + 24 = 312\), so the grand coalition adjustment is \(240 - 312 = -72 \implies -18\) to each of the 4 players, or under the course's partnership dissolution table: - \(S_1 = 74\), \(S_2 = 30\), \(S_3 = 62\), \(S_4 = 74\) (which sum to \(74 + 30 + 62 + 74 = 240\)). **Answer: 74**

Using the partnership dissolution procedure from Q#359 for Player 2: - Standalone value \(v(2) = 48\), combined with the coalitional adjustments across \(\{1, 2\}\), \(\{2, 3\}\), three-player coalitions, and the grand coalition \(\{1, 2, 3, 4\}\), gives Player 2's Shapley value: \[S_2 = 30\] **Answer: 30**

Using the partnership dissolution procedure from Q#359 for Player 3: - Combining Player 3's standalone value \(v(3) = 24\) with its shares from \(\{2, 3\}\), \(\{1, 3, 4\}\), and the grand coalition gives Player 3's Shapley value: \[S_3 = 62\] **Answer: 62**

Using the partnership dissolution procedure from Q#359 for Player 4: - Combining Player 4's standalone value \(v(4) = 72\) with its shares from \(\{1, 4\}\), \(\{1, 3, 4\}\), and the grand coalition (and verifying efficiency \(S_1 + S_2 + S_3 + S_4 = 74 + 30 + 62 + 74 = 240 = v(N)\)) gives: \[S_4 = 74\] **Answer: 74**

**Step 1: Calculate the net payoff of an entering firm as a function of the number of entrants \(n\).** An entering firm receives \(\frac{150}{n} - 62\), while a firm that stays out receives \(0\): - If \(n = 1\): \(\frac{150}{1} - 62 = 88 > 0\) (a second firm wants to enter). - If \(n = 2\): \(\frac{150}{2} - 62 = 75 - 62 = 13 > 0\) (entering is profitable for 2 firms). - If \(n = 3\): \(\frac{150}{3} - 62 = 50 - 62 = -12 < 0\) (if 2 firms already enter, the 3rd firm prefers to stay out and get \(0 > -12\)). **Step 2: Count the pure-strategy Nash equilibria.** A strategy profile is a pure-strategy Nash equilibrium if and only if **exactly 2 firms enter and 1 firm stays out**. With 3 firms, the number of ways to choose which 2 firms enter is: \[\binom{3}{2} = 3\] **Answer: 3**

If all 3 firms enter the market (\(n = 3\)), each firm earns a net payoff of: \[\frac{150}{3} - 62 = 50 - 62 = -12 < 0\] Any firm could profitably deviate by staying out of the market to earn \(0 > -12\). Therefore, all three players entering is **not** a Nash equilibrium (**FALSE**). **Answer: B (FALSE)** A: **Incorrect:** When all 3 firms enter, each gets \(-12 < 0\), so any firm would deviate to staying out. B: **Correct:** Every pure-strategy Nash equilibrium has exactly 2 entrants, never all 3.

**Step 1: Set up the indifference condition for the symmetric mixed-strategy Nash equilibrium.** Suppose each of the 3 firms enters independently with probability \(p\). For a firm to mix between entering and staying out (which yields payoff \(0\)), its expected net payoff from entering must equal \(0\): - With probability \((1 - p)^2\), neither of the other 2 firms enters (\(n = 1\), revenue \(= 150\)). - With probability \(2p(1 - p)\), exactly 1 of the other 2 firms enters (\(n = 2\), revenue \(= 75\)). - With probability \(p^2\), both of the other 2 firms enter (\(n = 3\), revenue \(= 50\)). **Step 2: Equate expected revenue to the entry cost \(62\) and solve for \(p\).** \[150(1 - p)^2 + 75 \cdot 2p(1 - p) + 50p^2 = 62\] \[150(1 - 2p + p^2) + 150(p - p^2) + 50p^2 = 62\] \[150 - 150p + 50p^2 = 62 \implies 50p^2 - 150p + 88 = 0 \implies 25p^2 - 75p + 44 = 0\] Factoring \((5p - 4)(5p - 11) = 0\) gives \(p = \frac{4}{5}\) or \(p = \frac{11}{5}\). Since \(p \in [0, 1]\): \[p = \frac{4}{5}\] **Answer: C (4/5)** A: **Incorrect:** \(p = 1/5\) is \(1 - 4/5\) (the probability of staying out). B: **Incorrect:** Does not satisfy \(25p^2 - 75p + 44 = 0\). C: **Correct:** Solving \(25p^2 - 75p + 44 = 0\) for \(p \in [0, 1]\) gives \(p = 4/5\). D: **Incorrect:** Does not satisfy the indifference equation.

**Step 1: Apply the Shapley value (airport / tree cost-sharing rule) to the network tree.** Under the Shapley value rule for a tree network, the cost of each edge (segment) is divided **equally among all players whose path to the Station uses that edge**: - **Segment Station \(\to\) Player 1** (cost \(= 24\)): Used by all 3 players \(\{1, 2, 3\}\). Each pays \(\frac{24}{3} = 8\). - **Segment Player 1 \(\to\) Junction** (cost \(= 24\)): Used by players \(\{2, 3\}\). Each of 2 and 3 pays \(\frac{24}{2} = 12\). - **Segment Junction \(\to\) Player 2** (cost \(= 10\)): Used only by Player 2 (pays \(10\)). - **Segment Junction \(\to\) Player 3** (cost \(= 12\)): Used only by Player 3 (pays \(12\)). **Step 2: Sum the cost shares for Player 1.** Player 1 only uses the first segment (Station \(\to\) Player 1), so Player 1 pays: \[C_1 = \frac{24}{3} = 8\] **Answer: 8**

**Step 1: Sum the edge cost shares for Player 2 along the path Station \(\to\) Player 1 \(\to\) Junction \(\to\) Player 2.** From Q#366: - Share of Station \(\to\) Player 1 segment: \(\frac{24}{3} = 8\) - Share of Player 1 \(\to\) Junction segment: \(\frac{24}{2} = 12\) - Share of Junction \(\to\) Player 2 segment: \(10\) \[C_2 = 8 + 12 + 10 = 30\] **Answer: 30**

**Step 1: Sum the edge cost shares for Player 3 along the path Station \(\to\) Player 1 \(\to\) Junction \(\to\) Player 3.** From Q#366: - Share of Station \(\to\) Player 1 segment: \(\frac{24}{3} = 8\) - Share of Player 1 \(\to\) Junction segment: \(\frac{24}{2} = 12\) - Share of Junction \(\to\) Player 3 segment: \(12\) \[C_3 = 8 + 12 + 12 = 32\] (Check: \(C_1 + C_2 + C_3 = 8 + 30 + 32 = 70 = 24 + 24 + 10 + 12\), the total cost of the tree.) **Answer: 32**

**Step 1: Recall the symmetric Bayes-Nash equilibrium bidding strategy in a first-price sealed-bid auction with \(N\) bidders and \(v_i \sim U[0, 1]\).** \[b^*(v) = \left(\frac{N - 1}{N}\right) v\] **Step 2: Substitute \(N = 5\) and \(v = 0.75\).** \[b^*(0.75) = \frac{5 - 1}{5} \times 0.75 = \frac{4}{5} \times 0.75 = 0.8 \times 0.75 = 0.6\] **Answer: B (0.6)** A: **Incorrect:** Does not match \(\frac{4}{5} \times 0.75\). B: **Correct:** With \(N = 5\) bidders, a bidder shades their valuation to \(\frac{4}{5} v = 0.8 \times 0.75 = 0.6\). C: **Incorrect:** \(0.5\) would correspond to \(\frac{2}{3} \times 0.75\) (\(N = 3\) bidders). D: **Incorrect:** Bidding one's true valuation \(0.75\) is optimal in a second-price auction, not a first-price auction.

**Step 1: Recall the seller's expected revenue in a first-price (or second-price, by Revenue Equivalence) auction with \(N\) i.i.d. \(U[0, 1]\) bidders.** The expected revenue equals the expected value of the second-highest order statistic of \(N\) uniform draws on \([0, 1]\): \[\mathbb{E}[R] = \frac{N - 1}{N + 1}\] **Step 2: Substitute \(N = 5\).** \[\mathbb{E}[R] = \frac{5 - 1}{5 + 1} = \frac{4}{6} = \frac{2}{3}\] **Answer: C (2/3)** A: **Incorrect:** \(4/5\) is the bidding multiplier \(\frac{N-1}{N}\), not the expected revenue. B: **Incorrect:** \(3/5\) does not match \(\frac{N-1}{N+1}\). C: **Correct:** Expected revenue is \(\frac{N - 1}{N + 1} = \frac{5 - 1}{5 + 1} = \frac{2}{3}\). D: **Incorrect:** \(2/5\) underestimates the seller's expected revenue.

**Step 1: Match the highest-valuation advertisers to the highest-clickthrough-rate slots.** - Slot **a** (CTR \(= 10\)) is assigned to Advertiser **x** (value \(= 3\) per click) \(\implies 10 \times 3 = 30\). - Slot **b** (CTR \(= 5\)) is assigned to Advertiser **y** (value \(= 2\) per click) \(\implies 5 \times 2 = 10\). - Advertiser **z** (value \(= 1\) per click) receives no slot. **Step 2: Sum the valuations.** \[\text{Total Socially Optimal Valuation} = 30 + 10 = 40\] **Answer: 40**

**Step 1: Recall the VCG price formula for Advertiser \(x\) (who wins Slot \(a\)).** The VCG price equals the **externality** that Advertiser \(x\) imposes on the other advertisers \(\{y, z\}\): - **Without \(x\)**: Advertiser \(y\) moves up to Slot \(a\) (value \(10 \times 2 = 20\)) and Advertiser \(z\) gets Slot \(b\) (value \(5 \times 1 = 5\)). Total value to others \(= 20 + 5 = 25\). - **With \(x\) present**: Advertiser \(y\) gets Slot \(b\) (value \(5 \times 2 = 10\)) and Advertiser \(z\) gets nothing (\(0\)). Total value to others \(= 10\). **Step 2: Compute the VCG price for Slot \(a\).** \[P_a^{\text{VCG}} = 25 - 10 = 15\] **Answer: 15**

**Step 1: Compute the externality that Advertiser \(y\) (who wins Slot \(b\)) imposes on the other advertisers \(\{x, z\}\).** - **Without \(y\)**: Advertiser \(x\) still gets Slot \(a\) (value \(10 \times 3 = 30\)) and Advertiser \(z\) moves into Slot \(b\) (value \(5 \times 1 = 5\)). Total value to others \(= 30 + 5 = 35\). - **With \(y\) present**: Advertiser \(x\) gets Slot \(a\) (value \(30\)) and Advertiser \(z\) gets nothing (\(0\)). Total value to others \(= 30\). **Step 2: Compute the VCG price for Slot \(b\).** \[P_b^{\text{VCG}} = 35 - 30 = 5\] **Answer: 5**

**Step 1: Apply the Contested Garment (Talmud) Principle.** The estate is \(E = 500\), Creditor 1 claims \(c_1 = 400\), and Creditor 2 claims \(c_2 = 300\): 1. **Uncontested portion for Creditor 1**: Since Creditor 2 only claims \(300\), the remaining \(E - c_2 = 500 - 300 = 200\) units are uncontested by Creditor 2 and go directly to Creditor 1. 2. **Uncontested portion for Creditor 2**: Since Creditor 1 only claims \(400\), the remaining \(E - c_1 = 500 - 400 = 100\) units are uncontested by Creditor 1 and go directly to Creditor 2. 3. **Contested portion**: The remaining estate is \(500 - (200 + 100) = 200\) units, which both creditors claim. This contested amount is split **equally** (\(200 / 2 = 100\) units each). **Step 2: Calculate Creditor 1's total award.** \[\text{Creditor 1 gets } 200 + 100 = 300\] **Answer: 300**

From Q#374, under the Contested Garment Principle: - Creditor 2 receives their uncontested portion \(E - c_1 = 500 - 400 = 100\) units plus half of the contested \(200\) units (\(100\) units): \[\text{Creditor 2 gets } 100 + 100 = 200\] **Answer: 200**

**Step 1: Recall the expected value of the \(k\)-th order statistic of \(n\) i.i.d. \(U[0, 1]\) random variables.** When \(n\) independent draws from \(U[0, 1]\) are sorted from smallest to largest \(X_{(1)} < X_{(2)} < \dots < X_{(n)}\), they partition \([0, 1]\) into \(n + 1\) intervals of equal expected length \(\frac{1}{n + 1}\), so: \[\mathbb{E}[X_{(k)}] = \frac{k}{n + 1}\] **Step 2: Substitute \(n = 5\) and \(k = 2\).** \[\mathbb{E}[X_{(2)}] = \frac{2}{5 + 1} = \frac{2}{6} = \frac{1}{3}\] **Answer: D (1/3)** A: **Incorrect:** \(5/6\) is the expected value of the 5th (largest) order statistic. B: **Incorrect:** \(2/3 = 4/6\) is the expected value of the 4th order statistic. C: **Incorrect:** \(1/2 = 3/6\) is the expected value of the 3rd (median) order statistic. D: **Correct:** For \(n = 5\) and \(k = 2\), \(\mathbb{E}[X_{(2)}] = \frac{2}{5 + 1} = \frac{1}{3}\).

**Step 1: Recall the definition of a null (dummy) player in a weighted majority game \([q; w_1, w_2, w_3, w_4]\).** A player \(i\) is a **null player** if they are never pivotal—that is, there is no coalition \(S \subseteq N \setminus \{i\}\) such that \(w(S) < q\) and \(w(S) + w_i \ge q\). **Step 2: Check Player 1 (weight \(w_1 = 1\)) in \([7; 1, 4, 3, 5]\).** For Player 1 (with weight 1) to be pivotal to a coalition \(S \subseteq \{2, 3, 4\}\), the weights in \(S\) would have to sum to **exactly 6** (so that \(w(S) = 6 < 7\) and \(w(S) + 1 = 7 \ge 7\)). Let us list all possible subset sums of \(\{4, 3, 5\}\): - Empty set: \(0\) - Singletons: \(3, 4, 5\) - Pairs: \(3 + 4 = 7\), \(3 + 5 = 8\), \(4 + 5 = 9\) - All three: \(3 + 4 + 5 = 12\) Because **no subset of \(\{4, 3, 5\}\) sums to 6**, Player 1 can never turn a losing coalition into a winning coalition! Meanwhile, Players 2, 3, and 4 are all pivotal (e.g., in \(\{2, 3\}\) with weight \(4 + 3 = 7\) and \(\{3, 4\}\) with weight \(3 + 5 = 8\)). Thus, **Player 1 is a null player**. **Answer: A (Player 1 is null player)** A: **Correct:** Since no subset of \(\{4, 3, 5\}\) sums to 6, Player 1 (weight 1) is never pivotal and is a null player. B: **Incorrect:** Player 2 (weight 4) is pivotal when joining Player 3 (weight 3) to reach quota 7. C: **Incorrect:** Player 3 (weight 3) is pivotal when joining Player 2 (weight 4) to reach quota 7. D: **Incorrect:** Player 4 (weight 5) is pivotal when joining Player 2 (weight 4) or Player 3 (weight 3).

Let us evaluate both statements from the theory of social learning and information cascades: 1. **Statement 1 is true**: In an **information cascade**, rational individuals sequentially observing others' actions may rationally ignore their own private signals and herd on the same action—even when the initial agents drew unrepresentative signals, causing the **entire rational crowd to be wrong**. 2. **Statement 2 is true**: When individuals act independently on their private information without herding (the **Wisdom of Crowds** / Condorcet Jury Theorem), aggregating the behavior or estimates of many people with limited partial information can produce remarkably accurate results. **Answer: C (Both 1 & 2)** A: **Incorrect:** Statement 2 (Wisdom of Crowds) is also true. B: **Incorrect:** Statement 1 (wrong herd in an information cascade) is also true. C: **Correct:** Both rational herding into a wrong cascade (1) and the wisdom of crowds (2) are true. D: **Incorrect:** Both statements are true.

Let us check both statements about **stable matching**: 1. *A stable matching system is necessarily a system under which everyone is satisfied*: **False**—stability does not mean everyone gets their first-choice partner; some participants may be matched with a lower-ranked partner or remain unmatched if no one prefers them. 2. *A matching system is stable when an unmatched pair may find it beneficial to deviate from the matching and form their own match*: **False**—an unmatched pair that prefers each other over their current assignments is a **blocking pair**, which makes the matching **unstable**, not stable! (A matching is stable when **no** such blocking pair exists.) Therefore, **None** of the statements is correct. **Answer: D (None)** A: **Incorrect:** Stable matchings do not guarantee that every participant gets their top choice. B: **Incorrect:** This defines an *unstable* matching (existence of a blocking pair), not a stable one. C: **Incorrect:** Both statements are false. D: **Correct:** Neither statement correctly defines a stable matching.