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When 1/50th of a vote decided an election

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CRUNCHING THE NUMBERS Former Returning Officer Pádraig Hughes going through the numbers last week from the famous Belmullet Electoral Area count of 1999.

Former Returning Officer Pádraig Hughes takes us through the complex but fascinating count where a fraction of a vote separated Frank Leneghan from Michael Holmes

Edwin McGreal

Could the battle in Belmullet Electoral Area in 1999 be the tightest ever witnessed in Irish politics?
Experienced former Returning Officer Pádraig Hughes cannot recall tighter in his time observing counts and that goes back to 1967. His in-depth knowledge of the system makes for a fascinating examination of an incredible election.
The official count figures will say one vote separated Frank Leneghan from Michael Holmes, 945 to 944.
But the reality is it was a fraction of a vote. The fraction was considered for many years to be 0.54 of a vote but an examination of the figures 20 years on reveals it was much, much less.
Only .02 of a vote, 1/50th of a vote, was between Holmes and Leneghan.
In the quirks of the proportional representation system, this was one election which proved that not only did every vote count, but every preference too.
It is an election and a count that continues to amaze and has been used as a case study by Hughes for training courses for returning officers.
It is not simple to relate just how such a minuscule fraction tipped the balance but it makes for fascinating examination on just how the system works.

Why a fraction?
Fractions can only come into play when surpluses are being distributed.
If someone is eliminated, their vote is distributed and other candidates either get a full vote or no vote from that candidate’s parcels; there can be no fractions here.
Surpluses are not as straightforward but an understanding of how they operate is crucial to understanding how 1/50th of a vote separated Leneghan from Holmes.
If someone exceeds the surplus on the first count, then all of their votes are counted and distributed proportionately. The surplus is divided by the total number of votes from that candidate that are transferable. Then the ensuing ‘transfer fraction’ created is multiplied for each of the candidates remaining by the total number of number two preferences they received from the poll topper.
Example: Quota is 900. John Smith gets 1,000. His surplus is 100. All of his votes are transferable, 1,000. Divide 100 by 1,000 gives you a transfer fraction of 0.1 and multiply that for each of the candidates. If Joe Bloggs receives 200 number twos from John Smith’s vote, then 200 is multiplied by .1, totalling 20 votes. So Bloggs gets 20 actual votes from Smith’s surplus.

Subsequent counts
It is different, however, when a candidate is put over the quota on subsequent counts. Their total vote is not used to calculate the distribution of their surplus.
Rather, it is the bundle of papers that put them over the surplus which is counted and given proportionately to other candidates based on the surplus.
In real terms we can examine the crucial surplus of Pat Kilbane in the Belmullet Electoral Area in 1999. It was the distribution of the Achill councillor’s surplus which left just a fraction of a vote between Leneghan and Holmes.
Before the distribution of this surplus, Tim Quinn’s surplus was distributed and put Leneghan ahead of Holmes for the first time, by just three votes.
Whoever was last after the distribution of Kilbane’s surplus was going to be eliminated. It was down to the wire.
Kilbane was put over the quota in the distribution of Achill Fianna Fáil candidate Michael McNulty’s votes, following McNulty’s elimination.
Of McNulty’s final vote of 622, 284 went to his fellow Achill-man Kilbane to push him over the quota by just 11 votes.
So these 284 votes were what were recounted for next available preferences. Of the 284, 78 votes were non-transferable, that is after Kilbane, the votes did not have any preferences marked for any of the four remaining candidates – Holmes, Leneghan and Fine Gael duo Gerry Coyle and Ian McAndrew.
So 206 votes were transferable and there was a surplus to be distributed of 11 votes. So to start distributing the votes, the surplus of 11 was divided by 206 to get the transfer fraction of 0.053398058. This fraction was then multiplied for each of the candidates by the number of votes each received in transfers from the 206, as shown below.


Election box


So of Kilbane’s 11 votes of a surplus, six went the way of Holmes, four the way of Leneghan and one to Coyle. Holmes had narrowed the gap back to one vote but not close enough.
But how those votes arrived with each candidate shows the intricacies of the system.
The multiplication of the transfer fraction by the transferables in the last parcel for each candidate produced a total of nine full units – that is before the fraction.
Six of these to Holmes and three to Leneghan.
But all 11 votes have to be distributed in order to comply with the requirements of the system.
So how are the other two votes assigned?
Contrary to widespread belief, it is not a case of rounding up anything after .5. In this example, only Leneghan would get rounded up and there would, therefore, be a vote not assigned.
So, rather, what happens is the two votes are assigned to the two highest fractions.
The first goes to Leneghan and the second to Gerry Coyle. Eagle- eyed people will notice that Ian McAndrew had the same fraction as Coyle so how did he not benefit?
Well, in the case of such a tie with fractions, the person with the highest first preference vote gets the nod. Incredibly, both Coyle and McAndrew were level after the first count (1,014 votes each).
It therefore went on to the second count, where Coyle had more votes, so he benefitted and got the extra vote here.
As a very curious aside, the subsequent recount saw this reversed. Both Coyle and McAndrew were on the exact same fractions once more but other anomalies discovered ended up placing McAndrew three ahead after the first count.

Fine margins
But with only one vote between them, every margin was crucial between Holmes and Leneghan.
The gap between them in the fractions here is 0.54368932, the 0.54 of a vote often said to be the difference in this election.
But Holmes did not need to match or beat Leneghan’s fraction. He only needed to outflank the Fine Gael duo to take the second vote that needed to be assigned from the fractions.
He was only 0.01941748 behind them. Less than 0.02 of a vote. The recounts changed none of this – the minuscule margin remained the same.
Had Holmes been in front of McAndrew and Coyle he would have got seven votes from Kilbane’s surplus and four would have stayed with Leneghan.
Had Holmes got 0.02 more of a vote from Kilbane’s surplus, he would have been deadlocked with Leneghan on 945 votes.
That would have eliminated Leneghan on the basis of Holmes having the higher first preference (789 to 615). The prevailing view was Holmes would have overtaken McAndrew thanks to transfers from Leneghan to take one of the last two seats with Coyle.
Leneghan’s victory gave Fianna Fáil a council majority. So 0.02, 1/50th of a vote, tipped the scales for the following five years.
Ahead of Friday’s elections, it is a reminder of just how much every single vote, every preference, can matter.

MORE
Read Holmes and Leneghan’s recollections of a famous election count in our 16-page Election Preview supplement

 

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