Question 1198578: The New York City Council is ready to vote on two bills that authorize the construction of
new roads in Manhattan and Brooklyn. If the two boroughs join forces, they can pass both
bills, but neither borough by itself has enough power to pass a bill. If a bill is passed, then it
will cost the taxpayers of each borough $1 million, but if roads are built in a borough, the
benefits to the borough are estimated to be $10 million. The council votes on both bills simultaneously, and each councilperson must vote on the bills without knowing how
anybody else will vote. Assuming that each borough supports its own bill, determine whether
this game has any equilibrium points. Is this game analogous to the Prisoner’s Dilemma?
Explain why or why not.
Answer by CPhill(1987)   (Show Source):
You can put this solution on YOUR website! This scenario is indeed analogous to a **game-theoretic problem** and can be analyzed using concepts such as **Nash equilibrium** and the **Prisoner’s Dilemma**. Let's explore this systematically:
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### **Step 1: Players, Strategies, and Payoffs**
- **Players**: The two boroughs, Manhattan (M) and Brooklyn (B).
- **Strategies**: Each borough has two possible strategies for voting on the bills:
1. **Vote "Yes"** for both bills.
2. **Vote "No"** for both bills.
- **Payoffs**:
- If both boroughs vote "Yes," both bills are passed, and each borough benefits from its own road (\(+10\)) but incurs the cost of both bills (\(-2\)).
- If a borough votes "No" and the other votes "Yes," no bills pass, so there are no costs or benefits (\(0\)).
- If both vote "No," neither bill passes, so no costs or benefits (\(0\)).
| Manhattan/Brooklyn | Yes | No |
|---------------------|-----|-----|
| **Yes** | \(+8, +8\) | \(0, -1\) |
| **No** | \(-1, 0\) | \(0, 0\) |
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### **Step 2: Nash Equilibrium**
A **Nash equilibrium** occurs when neither player can unilaterally change their strategy to improve their payoff, given the other player’s strategy. Let’s analyze:
1. If **both vote "Yes"**:
- Payoff: \(+8\) for both boroughs.
- If either switches to "No," the payoff becomes \(0\), which is worse. So, neither has an incentive to deviate.
2. If **Manhattan votes "Yes"** and **Brooklyn votes "No"**:
- Manhattan’s payoff: \(0\), Brooklyn’s payoff: \(-1\).
- Manhattan would prefer "No" for a payoff of \(0\), so this is not an equilibrium.
3. If **both vote "No"**:
- Payoff: \(0\) for both.
- If either switches to "Yes," their payoff becomes \(-1\), which is worse. So, this is a stable outcome.
Thus, the game has **two Nash equilibria**:
- Both vote "Yes" (\(+8, +8\)).
- Both vote "No" (\(0, 0\)).
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### **Step 3: Comparison to the Prisoner’s Dilemma**
This game **is analogous to the Prisoner’s Dilemma**, but there are differences:
1. **Similarities**:
- The individually rational strategy (voting "No") leads to a worse collective outcome (\(0, 0\)).
- The collectively optimal strategy (voting "Yes") requires cooperation and trust.
2. **Differences**:
- Unlike the Prisoner’s Dilemma, where the payoff for defecting (betrayal) is higher than cooperating, here the penalty for voting "No" is purely the failure to benefit.
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### **Step 4: Conclusion**
- The game has two Nash equilibria: \( (Yes, Yes) \) and \( (No, No) \).
- However, the "No, No" equilibrium is **socially suboptimal**, similar to the Prisoner’s Dilemma.
- To achieve the optimal outcome (\(+8, +8\)), the boroughs must trust each other to vote "Yes" simultaneously, which may require pre-vote negotiation or external enforcement mechanisms.
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