David Hill is a retired statistician. He is a member of Council and Chairman of the Technical Committee of the ERS.

The Meek system for counting an STV election overcomes most of the troubles encountered in using older systems designed for counting by hand, but the problem of premature exclusion remains. Premature exclusion of a candidate occurs when someone is the lowest because hidden behind another who, in the end, is also not going to succeed. If A, who would otherwise have been elected, fails because B stood and was elected instead, it is bad luck for A but there is nothing disturbing about it in principle. If, however, A fails because B stood, but then B does not get in either, that is disturbing.

Exclusion of the lowest candidate, when an exclusion is necessary, is the trouble. After all, if the so-called first past the post is not necessarily the right person to elect, then neither is the last past the post necessarily the right one to exclude. Is there some other way of handling things that would do better? What is needed is a mechanism to discover initially which candidates have some hope of election and which have virtually none, and to get rid of the 'no-hopers' at the start of the count. Others cannot then suffer from their presence.

Let the election be to fill *k* seats from *n* candidates, and let
*m* = *n* - *k*. Sequential STV then consists of a number of
main-phases and sub-phases, each being an STV election for *k* seats
but with varying selections of candidates. The STV elections are preferably
conducted using Meek-style counting but other rules could be used.

**Main-phase 1**. All *n* candidates, but instead of dividing into
elected and excluded, divide them into probables and others respectively.
Set all *n* candidates to unmarked.

**Sub-phase 1.1**. The *k* probables plus any other one candidate.
Set the winners to marked.

**Sub-phase 1.2**. The same *k* probables plus any other one
candidate not yet tested. Set any unmarked winners to marked.

etc.

**Sub-phase 1**.** m**. The same

If at any sub-phase there is a tie that has to be settled using random
selection, then all *k* + 1 of the candidates involved are set to
marked.

**Main-phase 2**. All marked candidates, dividing into probables and
others. If the resulting set of probables is the same as a previous set,
those candidates are elected and the process finishes. Otherwise reset all
*n* candidates to unmarked and continue.

**Sub-phases 2.1 - 2.****m**. As 1.1 - 1.m but using the new
probables.

**Main-phase 3**. As main-phase 2.

etc. etc.

It may be noted that anyone getting a quota of first preferences on the original count is, in fact, certain to be elected in the end, but to be classified for the time being as probable does no harm.

The process must terminate because there is only a finite number of sets of
*k* that can be formed from *n*. Usually it will terminate with
two successive main-phases showing the same set of *k* probables. In
that case the result is firmly established. If, however, the two showing
the same set are not successive it will mean that the system is cycling in
Condorcet-paradox style. In that case it may be that a better rule could be
devised than taking the first set to occur twice but it has to be recognised
that a totally satisfactory answer is impossible.

Each candidate is given a fair chance by being tested against each new set
of probables and since each sub-phase consists of only *k* + 1
candidates for *k* seats, exclusion is never necessary during the
sub-phases so the 'exclude the lowest' rule is not operative there.

104 AEBCD 103 BECDA 102 CEDBA 101 DEBCA 3 EABCD 3 EBCDA 3 ECDBA 3 EDBCAIt is evident that E is a strong candidate, in that if any one of A, B, C or D were to withdraw, E would be the first elected. Yet under simple STV the first action is to exclude E, and B and C are elected. Under sequential STV we find

Phase Candidates Winners Probables Marked 1 ABCDE BC BC 1.1 BCA BC BC 1.2 BCD BC 1.3 BCE BE E 2 BCE BE BE 2.1 BEA BE BE 2.2 BEC BE 2.3 BED BE 3 BE BE BEB and E are consequently elected. It will be noted that some elections may be repeats of ones already done (main-phase 2 and sub-phase 2.2 in the above example are both repeats of sub-phase 1.3). The result may of course merely be copied down without actually repeating any calculations.

The following example to demonstrate this trouble is derived from those that Douglas Woodall devised to prove his 'impossibility' theorem. Let there be 3 candidates for 1 seat and votes

25 A 17 BC 16 C

Phase Candidates Winner Probable Marked 1 ABC A A 1.1 AB A A 1.2 AC C C 2 AC C C 2.1 CA C C 2.2 CB B B 3 BC B B 3.1 BA A A 3.2 BC B B 4 AB A ASo A is elected. But if the A voters had put in C as a second preference, we get

25 AC 17 BC 16 C Phase Candidates Winner Probable Marked 1 ABC A A 1.1 AB A A 1.2 AC C C 2 AC C C 2.1 CA C C 2.2 CB C 3 C C Cand C is elected. So the A voters have failed to elect A because they gave C as a second preference.

Even if this is a rare event, it still means that we cannot assure voters that their later preferences cannot upset their earlier preferences. I believe that this is too high a price to pay. There is not much point in reducing the frequency of one type of fault if, in doing so, you introduce another fault as bad.

For example, let there be 4 candidates for 1 seat and votes

98 ADCB 98 CDBA 99 BDAC 3 ACBD 2 CBAD Phase Candidates Winner Probable Marked 1 ABCD A A 1.1 AB B B 1.2 AC A A 1.3 AD D D 2 ABD B B 2.1 BA B B 2.2 BC C C 2.3 BD D D 3 BCD C C 3.1 CA A A 3.2 CB C C 3.3 CD D D 4 ACD A ASo A is elected, even though D would be the Condorcet winner (for the results of AD, BD and CD are all D). It should be emphasised, though, that this is not likely in practice but only with carefully devised artificial examples.