While I Thiele PAV doesn’t give what I would consider to be acceptable results in a normal proportional election with individual candidates elected with equal weights, I wondered whether its problems could be overcome with unlimited candidate-weight elections, or party list using these same weights.

One of the problems with Thiele Proportional Approval Voting is that it fails strong PR. For example:

6 to elect

2 voters: U1, U2, U3, A1, A2, A3

1 voter: U1, U2, U3, B1

Under strong PR, the universally approved candidates (U1-3) should not affect the ratios of the other candidates elected. So the correct result would be U1-3, A1-2, B1 (or equivalent).

Thiele PAV elects U1-3, A1-3, failing the criterion. This result gives the A faction 6 candidates and the B faction 3 candidates, and some may still argue that this is an acceptable proportional result.

However, this is not the worst case.

20 to elect

2 voters: U1-10, A1-3

2 voters: U1-10, B1-3

1 voter: C1-6

In this case the strong PR failure leads to a worse PR failing. The correct result seems to be U1-10, A1-3, B1-3, C1-4. The C faction has 1/5 of the voters so should have 1/5 of the candidates, which is 4. The other 16 should be even among the A and B factions.

This result gives the A and B factions each 13 candidates and the C faction 4, but Thiele works to give a 2:2:1 ratio even though the A and B factions aren’t truly separate. Thiele gives the result U1-10, B1-2, C1-2, giving the factions 12, 12, and 6 candidates respectively. However, this causes the C faction to be over-represented in parliament (getting 3/10 instead of 2/10 of the representation), whereas A and B considered together are under-represented getting 7/10 instead of 8/10 of the representation.

Because of this I don’t really consider Thiele PAV a suitable election method for use. And this brings us to the case where candidates can be given different weights in parliament, or where these weights are used in a party-list proportional approval election. In the above two elections, proportionality would be restored.

6 to elect

2 voters: U1, U2, U3, A1, A2, A3

1 voter: U1, U2, U3, B1

In this case the U candidates would win all the weight between them, or for party list, party U would win all the seats.

20 to elect

2 voters: U1-10, A1-3

2 voters: U1-10, B1-3

1 voter: C1-6

In this case A and B would win no seats. U would win 16 (or 4/5 of the weight) and C would win 4 (1/5 of the weight).

So far so good. Thiele’s problem is when factions overlap, but where both factions support a single party/candidate then that party/candidate would win all the weight so the problem of the overlap effectively goes away. But you can have a case like this:

44 to elect

1 voter: AB

1 voter: AC

1 voter: BC

1 voter: D

The correct proportional result is for A, B, C and D to all be elected with equal weight, so 11 seats each. However, this gives the top three voters 22 candidates each and the D voter 11. Thiele here works to give a 3:1 ratio rather than 2:1, and gives a result of 12, 12, 12, 8. There is no “universal” candidate to sort out the problem of the overlap, so this a problem both for a normal election and an unlimited-weight election.

This is also probably the most stark example I’ve seen of a plain proportionality failure with Thiele PAV. The D faction is just a normal faction with no overlap minding its own business, and it should unequivocally get 1/4 of the weight in parliament. However, Thiele PAV gives this faction 8/44 of the weight - approximately 18% instead of 25%.