In approval Monroe, voters approve however many candidates they want, and each voter is assigned to a candidate, with equal numbers of voters assigned to each elected candidate. The winning slate of candidates is the one where the most voters can be assigned to a candidate that they approved.

It’s been discussed previously that Monroe has problems because it ignores how popular a slate of candidates is other than the one candidate that each voter is assigned to. For example, in this 2 candidate, 2 voter election:

1 voter: A, B, C

1 voter: A, B, D

Any set of two candidates would be considered as good as any other. AB has unanimous support but is considered no better than e.g. CD, where each candidate has 50% support each. But this is well known about and not what this new thread is about.

I was looking at what sort of results Monroe would actually give in certain scenarios. I think I have proved that where voters vote purely for candidates of their favourite party and there are just two parties, Monroe is equivalent to Sainte-Laguë party list. I won’t give the proof here, but I’ll give a couple of examples:

4 to elect

5 voters: Party A

3 voters: Party B

Under Sainte-Laguë party list, it would be a tie between

A: 3 seats

B: 1 seat

and

A: 2 seats

B: 2 seats

This is also the case under Monroe. There are 8 voters and 4 seats, so each elected candidate has 8/4 = 2 voters assigned to them.

Under the 3:1 result, all 5 A voters are satisfied (as they are over-represented), and 2 B voters are satisfied. 5+2 = 7

Under the 2:2 result, 2*2 = 4 A voters are satisfied along with all 3 B voters. 4+3 = 7. This gives a tie.

Next example:

3 to elect

5 voters: Party A

1 voter: Party C

Under Sainte-Laguë party list, it would be a tie between

A: 3 seats

C: 0 seats

and

A: 2 seats

C: 1 seats

The maths is as simple as in the first example, but to cut to the chase, Monroe gives the same result.

Next example:

2 to elect

3 voters: Party B

1 voter: Party C

Under both Sainte-Laguë and Monroe, there would be a tie between

B: 2 seats

C: 0 seats

and

B: 1 seat

C: 1 seat

Again, this is easy to verify. However, it starts to go wrong when there are three parties. You can combine the three examples above as follows:

4 to elect

5 voters: Party A

3 voters: Party B

1 voter: Party C

Under Sainte-Laguë party list, there is a three-way tie between 3:1:0, 2:2:0 and 2:1:1. But the equivalence with Monroe no longer holds. It starts to go wrong when we look at the 3:1:0 and 2:2:0 cases. Comparing these two is exactly the same as the first example above except that there are more voters. There are now 9 instead of 8. This means that each candidate has 9/4 voters assigned to them instead of 8/4.

The number of satisfied voters under 3:1:0 is 5 + 9/4 = 7.25.

Under 2:2:0 it is 2 * 9/4 + 3 = 7.5

Of course, we could consider only the voters who approved at least one candidate in the slate. That way we would be back to 8/4 voters per candidate, and 7 satisfied voters under either result, as in the first example. But is it right to ignore voters in this way? By not ignoring these voters, we would get failures of independence of irrelevant ballots, but by ignoring them, it doesn’t seem like the right thing to do. Plus it’s a weird discontinuity to consider at 100% weight all voters who have approved 1 or more candidates in a slate (so it could be 2, 3, 4… candidates), and those who have approved 0 at 0%. But that’s enough philosophising, because the final possible result kills it off anyway: 2:1:1.

In this case, every voter has approved at least one candidate, so the number of voters assigned to each candidate is 9/4 however you cut it.

And the number of satisfied voters is 2*9/4 + 9/4 + 1 = 7.75. This matches none of the numbers above.

I think Monroe has been let to exist as some sort of ideal of proportional representation without ever really being tested. But different numbers of voters approving at least one candidate in a slate breaks it quite badly.

I don’t really follow this forum that closely but I know that Monroe has been discussed in other topics. I probably wouldn’t touch it with a barge pole though.