IRV failing monotonicity criterion

Question

I am looking for the simplest possible example of instant runoff voting failing the monotonicity criterion. By “simplest possible” I mean the scenario with the fewest number of candidates $(3)$ and the fewest number of votes switched $(1?, 2?)$ that still demonstrates the effect. I'm guessing that the result is the same, but for the sake of specificity, let's say we want the case where increasing support for a candidate causes them to lose the election in round two (as opposed to decrease support causing them to win).

The wikipedia article (https://en.wikipedia.org/wiki/Monotonicity_criterion#Instant-runoff_voting_and_the_two-round_system_are_not_monotonic) uses an example with 3 candidates and 2 votes being switched, but with $100$ total votes. I’m wondering what the minimum number of total votes needed is. Also, is there a general proof to show that the solution really is the minimum?

Thanks in advance for your insights! -James

Answer 1

There's a simpler one on RangeVoting.org's (non)Monotonicity and Instant Runoff Voting :

The original ballots are:

• C > B > A (5 voters)
• A > C > B (4 voters)
• B > A > C (8 voters)

As shown, A gets eliminated first, all of A's 2nd choices go to C, and C wins with 9.

If two of the B > A > C voters cross the line and switch to A > B > C, decreasing their support for B:

the ballots become:

• C > B > A (5 voters)
• A > C > B (4 voters)
• B > A > C (6 voters)
• A > B > C (2 voters)

then C is eliminated, all of C's 2nd choices go to B, and B wins with 11.

So 2 voters decreasing their support for a candidate caused that candidate to win.

For this cyclic example to work, these criteria must be met (where the numbers are the first or second scenario):

• CBA > ACB
• BAC1 > ACB
• ACB + CBA > BAC1
• BAC1 = BAC2 + ABC2
• BAC2 > CBA
• ACB + ABC2 > CBA
• CBA + BAC2 > ACB + ABC2

I wrote a quick computer program to try every combination and it didn't find any simpler solution (for these conditions).