(b) Does the Coombs Method satisfy the Independence of Irrelevant Alternatives Condition?

B. The Coombs method does not satisfy the Independence of Irrelevant Alternatives condition. IIA is a property that states that the introduction or removal of an alternative should not affect the relative preferences between the remaining alternatives. In the context of Coombs method, if a new alternative is added or an existing alternative is removed, it can potentially change the ranking of the remaining alternatives. This lack of independence is a limitation of the Coombs method. The introduction or removal of an alternative can potentially affect the relative preferences between the remaining alternatives, violating the IIA condition.        Provide counterexample to proof my answer
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To prove that the Coombs Method does not satisfy the **Independence of Irrelevant Alternatives (IIA)** condition, I will provide a counterexample where adding or removing an alternative changes the outcome, even though the relative preferences between the remaining alternatives should remain the same under IIA.

### **Coombs Method Recap:**
The Coombs method is an elimination-based voting system where, in each round, the alternative that is ranked *last* most frequently is eliminated until only one alternative remains.

### **Counterexample:**

Consider three candidates: \( A \), \( B \), and \( C \), and four voters with the following preference rankings:

| Voter 1 | Voter 2 | Voter 3 | Voter 4 |
|--------|--------|--------|--------|
| 1. A   | 1. B   | 1. B   | 1. A   |
| 2. B   | 2. A   | 2. C   | 2. C   |
| 3. C   | 3. C   | 3. A   | 3. B   |

#### Step-by-Step Analysis without Irrelevant Alternatives:

1. **First Round of Coombs Method:**
   - Count the last-place votes for each candidate:
     - \( A \): 1 last-place vote (from Voter 3).
     - \( B \): 1 last-place vote (from Voter 4).
     - \( C \): 2 last-place votes (from Voters 1 and 2).
   - Since \( C \) has the most last-place votes, \( C \) is eliminated.

2. **New Preferences After \( C \) is Eliminated:**
   - The updated preferences for each voter are:
   
   | Voter 1 | Voter 2 | Voter 3 | Voter 4 |
   |--------|--------|--------|--------|
   | 1. A   | 1. B   | 1. B   | 1. A   |
   | 2. B   | 2. A   | 2. A   | 2. B   |

3. **Second Round:**
   - Count the last-place votes:
     - \( A \): 2 last-place votes (from Voters 2 and 3).
     - \( B \): 2 last-place votes (from Voters 1 and 4).
   - There is a tie, but assuming a tie-breaking rule is in place (e.g., choosing randomly or using other criteria), one of them is declared the winner. Let's assume \( A \) is declared the winner due to the tie-breaking rule.

#### Now, let’s introduce or remove an irrelevant alternative to see how it changes the outcome.

### Removing Candidate \( C \):

Without candidate \( C \), the preferences look like this:

| Voter 1 | Voter 2 | Voter 3 | Voter 4 |
|--------|--------|--------|--------|
| 1. A   | 1. B   | 1. B   | 1. A   |
| 2. B   | 2. A   | 2. A   | 2. B   |

- Here, no candidate is ranked last more frequently than the other. Every candidate gets two second-place votes. If there is a tie-breaking rule, one candidate, say \( B \), may be selected as the winner this time.

### Conclusion:

By removing \( C \), we went from a scenario where \( A \) won to a scenario where \( B \) wins. The removal of an irrelevant alternative (\( C \), who was eliminated first anyway) changed the outcome between \( A \) and \( B \), even though the relative preferences between \( A \) and \( B \) should have remained unchanged under the **Independence of Irrelevant Alternatives** condition.

This counterexample shows that the Coombs method violates the IIA condition.

---

Would you like further clarification on this example or the Coombs method?

### Related Questions:

1. How does the Coombs method compare to the Borda Count in terms of satisfying the IIA condition?
2. Can you provide a counterexample showing that the Borda Count violates the IIA condition?
3. What are other common voting methods that fail to satisfy the IIA condition?
4. How does the tie-breaking rule affect outcomes in methods like Coombs or Instant Runoff Voting (IRV)?
5. Can the Coombs method ever satisfy the IIA condition under specific circumstances?

### Tip:

When analyzing voting methods, always consider how the elimination of candidates (or alternatives) impacts the final outcome, as this is a critical aspect of evaluating fairness in election systems.

Does the Coombs Method satisfy the Independence of Irrelevant Alternatives Condition?

Explore the Coombs method and how it violates the Independence of Irrelevant Alternatives (IIA) condition in voting theory. A detailed counterexample illustrates how adding or removing a candidate can change the election outcome, demonstrating that the Coombs method fails to satisfy the IIA condition.

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