Violation of the irrelevant alternative criterion of fairness in a pairwise comparison

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I am teaching my students about the fairness criteria for voting system, working up towards arrow's impossibility theorem.

One of the voting methods is called the pairwise comparison method: voters rank each of the candidates from most to least favourite. To tally the votes, talliers compare each pair of candidates. If candidate X is more often preferred than candidate Y, then X receives a point. (If they tie, they each get half a point.) At the end of the comparisons, the candidate with the most points is selected.

We discuss criteria for a voting system to be fair. One criterion in particular is the "irrelevant alternative criteria" which states:

If an election is held and a winner is declared, this winning candidate should remain the winner in any recalculation of votes as a result of one or more of the losing candidates dropping out.

Can anyone think of an example of when the pairwise comparison method violates the irrelevant alternative criterion?

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Suppose we have three candidates $A,B,C$ and four voters with preferences as below:

$$\begin{array}{c | c | c | c} 1 & 2 & 3 & 4 \\ \hline A & A & C & B\\ C & C & B & A\\ B & B & A & C \end{array}$$

Then $A$ ties with $B$ and beats $C$ so has $3/2$ points, while $B$ ties with $A$ and loses to $C$ so has $1/2$ points, and $C$ loses to $A$ and beats $B$ so has $1$ point, thus $A$ wins. But if $C$ drops out, $A$ and $B$ tie.

This is the worst that can happen with three candidates, since if $B$ beats $A$ head-to-head then $B$ has at least $1$ point and $A$ has at most $1$ point. However, with four candidates $A,B,C,D$ and six voters we can have:

$$\begin{array}{c | c | c | c | c} 1 & 2 & 3 & 4 & 5 & 6\\ \hline A & A & C & C & B & B\\ C & D & D & D & A & A\\ D & B & B & B & D & D\\ B & C & A & A & C & C \end{array}$$ Then $A$ has $2$ points, $B$ has $3/2$ points, $C$ has $1$ point and $D$ has $3/2$ points, so $A$ wins. But if $D$ drops out we have:

$$\begin{array}{c | c | c | c | c} 1 & 2 & 3 & 4 & 5 & 6\\ \hline A & A & C & C & B & B\\ C & B & B & B & A & A\\ B & C & A & A & C & C\\ \end{array}$$

Then $A$ has $1$ point, $B$ has $3/2$ points and $C$ has $1/2$ point so $B$ wins.