Problem 216 - An Introduction to Game Theory

Question

A committee has three members. a. Suppose that there are three alternatives, \(x, y\), and \(z\), and that one member prefers \(x\) to \(y\) to \(z\), another prefers \(y\) to \(z\) to \(x\), and the third prefers \(z\) to \(x\) to \(y\). Find the top cycle set. b. Suppose that there are four alternatives, \(w, x, y\), and \(z\), and that one member prefers \(w\) to \(z\) to \(x\) to \(y\), one member prefers \(y\) to \(w\) to \(z\) to \(x\), and one member prefers \(x\) to \(y\) to \(w\) to \(z\). Find the top cycle set. Show, in particular, that \(z\) is in the top cycle set even though all committee members prefer \(w\). Rephrasing my conclusion for the agenda in Figure 214.1, if an alternative is the outcome of sophisticated voting, then it is in the top cycle set. The argument for this conclusion extends to any binary agenda. In every subgame, the outcome of sophisticated voting must beat the alternative that will be selected if it is rejected. Thus by backward induction, the outcome of sophisticated voting in the whole game must beat every other alternative either directly or indirectly: the outcome of sophisticated voting in any binary agenda is in the top cycle set. Now consider a converse question: for any given alternative \(x\) in the top cycle set, is there a binary agenda for which \(x\) is the outcome of sophisticated voting? The answer is affirmative. The idea behind the construction of an appropriate agenda is illustrated by a simple example. Suppose that there are three alternatives, \(x, y\), and \(z\), and \(x\) beats \(y\) beats \(z\). Then the agenda in Figure \(214.1\) is one for which \(x\) is the outcome of sophisticated voting. Now suppose there are two additional alternatives, \(u\) and \(w\), and \(x\) beats \(u\) beats \(w\). Then we can construct a larger agenda in which \(x\) is the outcome of sophisticated voting by replacing the alternative \(x\) in Figure \(214.1\) with a subgame in which a vote is taken for or against \(x\), and, if \(x\) is rejected, a vote is subsequently taken between \(u\) and \(w .\) If there are other chains through which \(x\) beats other alternatives, we can similarly add further subgames.

Step-by-Step Solution

Answer

For part (a), the top cycle set is \(\{x, y, z\}\). For part (b), the top cycle set is \(\{w, x, y, z\}\).

Step 1: Understanding Preferences for Part (a)

First, take note of each member's preference for the three alternatives. One member prefers \(x\) to \(y\) to \(z\), another prefers \(y\) to \(z\) to \(x\), and the third prefers \(z\) to \(x\) to \(y\).

Step 2: Comparing Each Pair for Part (a)

Compare each pair of alternatives based on majority preference: 1. Comparing \(x\) and \(y\): \(x\) is preferred by members 1 and 3, while \(y\) is preferred by member 2. Thus, \(x\) beats \(y\).2. Comparing \(y\) and \(z\): \(y\) is preferred by members 1 and 2, while \(z\) is preferred by member 3. Thus, \(y\) beats \(z\).3. Comparing \(z\) and \(x\): \(z\) is preferred by members 2 and 3, while \(x\) is preferred by member 1. Thus, \(z\) beats \(x\).

Step 3: Finding the Top Cycle Set for Part (a)

The top cycle set includes alternatives that can beat other alternatives in a cyclic manner. Here, since \(x\) beats \(y\), \(y\) beats \(z\), and \(z\) beats \(x\), all three alternatives \(x, y, z\) are in the top cycle set.

Step 4: Understanding Preferences for Part (b)

Again, take note of each member's preference for the four alternatives. One member prefers \(w\) to \(z\) to \(x\) to \(y\), another prefers \(y\) to \(w\) to \(z\) to \(x\), and the third prefers \(x\) to \(y\) to \(w\) to \(z\).

Step 5: Comparing Each Pair for Part (b)

Compare each pair of alternatives based on majority preference: 1. Comparing \(w\) and \(z\): \(w\) is preferred by members 1 and 2, while \(z\) is preferred by member 3. Thus, \(w\) beats \(z\).2. Comparing \(w\) and \(x\): No majority, \(w\) is preferred by members 1 and 2, \(x\) is preferred by member 3. Thus, \(w\) beats \(x\).3. Comparing \(w\) and \(y\): \(w\) is preferred by members 1 and 2, while \(y\) is preferred by member 3. Thus, \(w\) beats \(y\).4. Comparing \(z\) and \(x\): \(z\) is preferred by members 1 and 2, while \(x\) is preferred by member 3. Thus, \(z\) beats \(x\).5. Comparing \(z\) and \(y\): \(z\) is preferred by members 2 and 3, while \(y\) is preferred by member 1. Thus, \(z\) beats \(y\).6. Comparing \(x\) and \(y\): \(x\) is preferred by members 1 and 3, while \(y\) is preferred by member 2. Thus, \(x\) beats \(y\).

Step 6: Finding the Top Cycle Set for Part (b)

All alternatives that can beat other alternatives in a cyclic manner form the top cycle set. Here, \(w\) beats \(z, x, y\), \(z\) beats \(x, y\), and \(x\) beats \(y\). Since there are interconnections, the top cycle set consists of \(w, x, y, z\), and \(z\) is included even though all members prefer \(w\).

Key Concepts

committee decision-making

Committee decision-making is a process where a group collectively makes choices or judgments. It involves multiple members who have their own preferences and interests. Usually, to make decisions, committees use various voting systems to aggregate these preferences into a single outcome. This aggregation isn't always straightforward. Members' preferences can vary, and when there are multiple alternatives, decision-making often becomes complex. Understanding how committees function and the tools they use can help navigate these complexities.

preferences

Preferences in decision-making refer to the order in which decision-makers value or rank the available choices. Each member in a committee may have different preferences, which can shape the overall decision. For example, one member might prefer alternative A over B over C, another might prefer B over C over A, and a third might prefer C over A over B. These differing preferences highlight the need to find a method to reach a consensus.
In the given exercise, three committee members had distinct preferences over the alternatives. Such scenarios require methods like pairwise comparisons to understand majority preferences for each pair of alternatives. This process underscores how individual preferences influence overall decision outcomes in committees.

sophisticated voting

Sophisticated voting is a strategy where voters consider not just their direct preferences, but also the potential outcomes of their votes. It is an anticipatory approach where voters think ahead about how other members may vote and adjust their votes to maximize their preferred outcomes.
For example, in a binary agenda, a voter might prefer option A over B but might vote for B if they believe it will lead to a more favorable final outcome. This approach makes the voting process strategic rather than straightforward.

• Voters analyze the possible outcomes and vote in a way that improves their chances of achieving the best possible result.
• This involves backward induction, where voters anticipate the sequence of voting rounds and the possible results of each round.

This method ensures that the final decision aligns closely with the voters' overall preferences.

backward induction

Backward induction is a reasoning technique often used in game theory and sophisticated voting processes. It involves analyzing decisions from the end of the decision-making process backward to determine the optimal actions to take at each previous step.
This method is particularly useful in sequential voting scenarios or multi-stage decision processes. By considering the final outcomes first, decision-makers can plan their strategies effectively for earlier stages.
The given exercise demonstrates backward induction in committee decision-making. Members use this technique to anticipate final outcomes and adjust their votes in earlier rounds to influence the final decision. This ensures their highest-ranked alternatives have the best chance of success, effectively guiding the committee to an optimal top cycle set outcome.