David Hill is a retired statistician, member of ERS Council and Chairman of its Technical Committee.

The Meek system and the Warren system for counting an STV election are very similar, but whereas the fractions of a vote retained by successive elected candidates are multiplicative under Meek, they are additive under Warren. For example, if candidate A is keeping 1/2 of everything received and candidate B is keeping 1/3, a vote reading AB... will, under Meek, give 1/2 of a vote to A and 1/6 of a vote to B (i.e. 1/3 of the remaining 1/2), leaving 1/3 of a vote to be passed on further. With those same fractions under Warren, a similar vote will give 1/2 of a vote to A and 1/3 of a vote to B, leaving 1/6 of a vote to be passed on further. (It should be noted, though, that in any actual case the fractions will not usually be the same under the two systems). The Warren system will often lead to the situation where not enough vote remains for the fraction required; in such a case all that remains is taken and nothing remains to go any further.

There is no difference in the ease of writing a computer program to satisfy the one system or the other; the choice can be made solely on which is regarded as better in principle. It should also be reported that in real examples of STV elections, as distinct from artificially constructed examples, no case has yet been found where the two elect a different set of candidates, so the difference for real life seems to be slight.

There has been much argument over which system is to be preferred. In the end, we have settled on a particular example which demonstrates that each system can be said to suffer from a difficulty that the other one solves. It must therefore be a matter of judgement which difficulty is regarded as the more serious, rather than a firm decision of one always being better than the other.

The Meek rationale is that all transfers from a surplus should be in proportion to the 'votes-worth' put into that surplus. Thus 5 identical votes, each of current value 1/5, should have the identical effect to that of 1 complete vote for the same preferences. The Warren rationale is that no voter should be allowed to influence the election of an additional candidate until having contributed as much as any other voter to the election of each candidate who has already been elected and is named earlier in the voter's preferences. Thus the 5, each of value 1/5, are to be treated as 5, not as the equivalent of 1.

The example that shows the differences has 5 candidates for 3 seats and 32 votes, leading to a quota of 8.0. The votes are:

12 ABC, 12 BE, 7 C, 1 D.Meek supporters can point out the Warren anomaly that A and B each had a substantial surplus on the first count, yet the 12 ABC votes are given by the Warren system entirely to A and B and, in consequence, C fails to get the 1 extra vote needed for election and E takes the third seat. Under Meek, C easily beats E.

Warren supporters can point out the Meek anomaly that if the 12 ABC voters had voted BAC instead, the Meek system would have behaved exactly like the Warren system, and E would have beaten C. It seems illogical that the choice of C or E should depend upon the ordering by those 12 voters as ABC or BAC when A and B were both elected anyway.

Deciding between the two systems must therefore remain a matter of personal preference.

It may be of interest to see exactly how each of the two systems would treat this example. Each would note that A and B are both elected on the first count, each having 12 first preferences for a quota of 8.

The Meek system would calculate that A needs to keep 2/3 of everything received whereas B needs to keep 1/2, these fractions being derived so that each of A and B keeps exactly a quota. The 12 ABC votes would be allocated as 2/3 of 12 = 8 to A, 1/2 of the remaining 4 = 2 to B, the remaining 2 to C. The 12 BE votes would be allocated as 1/2 of 12 = 6 to B, the remaining 6 to E. At the next count the current votes would therefore be A 8, B 8, C 9, D 1, E 6. The third seat is thus assigned to C and no more needs to be done.

The Warren system would calculate that A's amount retained needs to be 2/3 and B's 1/3, again derived such that (under the different counting method) each of A and B keeps exactly a quota. The 12 ABC votes would be allocated as 2/3 of 12 = 8 to A, 1/3 of 12 = 4 to B. The 12 BE votes would be allocated as 1/3 of 12 = 4 to B, the remaining 8 to E. At the next count the current votes would therefore be A 8, B 8, C 7, D 1, E 8. The third seat is thus assigned to E and no more needs to be done.