## Friday, March 25, 2011

### A re-examination of the UK AV proposal - a possible flaw

[Update - I'm leaving this here for archive purposes. Having yet another re-examination there is a specialised case where it can fail. I solve this flaw here]

One of the items I've come across in those opposing AV is that it will cost more money to count and give the minority candidates a greater say. I've explained elsewhere why this wouldn't be the case,but an re-examination of the AV proposal has shown me why this view can be held.

Looking at the Wikipedia page describing the method used to determine the winner it states

If after first preferences have been counted, no one candidate has a majority of the votes cast, then the bottom candidate will be eliminated and their next available preference will be redistributed accordingly. The process continues repeatedly until one candidate reaches a majority and wins.
This is logical, systematic, and absurd.

I'll demonstrate why using the Wyre Forest Parliamentary election results. Firstly I need to make clear that this method can be used with either the actual vote number or the percentage of the vote; I'll show both but use the percentages purely for the sake of clarity.

The results of what would be called the first round under AV are as follows

PartyVote%
Conservative18,79336.9
Health Concern16,11531.7
Labour7,29814.3
Liberal Democrat6,04011.9
UKIP1,4982.9
BNP1,1202.2

Under AV the votes from the BNP will be re-distributed amongst the other 5 candidates. If this doesn't produce a majority then the votes in the UKIP pile (that may include those just placed there from the BNP pile) will be re-distributed; and so on and so forth.

Those with more than a basic grasp of mathematics should be seeing how wrong this is. The current top candidate is the Conservative Party with 36.9% of the vote; even if every single BNP member uses their second preference vote for the Conservatives the highest percentage that can be reached is 39.1%.

There is no way that only re-distributing the BNP vote can result in a majority.

In this instance to guarantee at least the possibility of a majority vote the entire bottom three parties would need to have their votes re-distributed. (36.9+2.2+2.9+11.9=53.9%). By the method proposed we would only reach this point by the fourth round of counting. But the mathematics shows the two intervening rounds to be pointless so why not just jump to it and save all that counting?

But wait, we don't even need to stop there. If we jumped to the point where we reach three candidates we can now see that if even if every single Liberal Democrat, UKIP and BNP supporter voted Labour they still wouldn't be able to reach a majority. So why not just distribute all those votes too?

This may sound complicated, but that's because I've walked you through all the steps. This tedious process can be circumvented by applying the Median Rule - Take the top candidates whose combined total reaches 50% of the total vote and redistribute all the other votes.

Now even this won't guarantee a majority in the second round; if we consider a 20/20/20 split with the remaining 40% split evenly there would still be a requirement for a third round. However as an observation it is possible to state that - the maximum number of rounds required will not exceed that of the number of candidates determined by the Median Rule; i.e. apply the Median Rule and get three parties the maximum number of rounds is three.

So why do it the other way? Well it's possible that not every voter will use their full allocation of votes and thus the number required to hit 50% will change. If every single BNP voter voted only for the BNP the 50% figure will be 24,872 down from 25,432. If every UKIP voter did the same it would become 24,123.Same for the LibDems it becomes 21,103. Repeat for Labour and we're left only with the top two Parties and the Conservatives win simply because no-one from the other Parties voted for anyone else.

But here's the catch - we'd still get the same result if we used the Median Rule:

At random I used the constituency of Kettering.
PartyVote%
Conservative23,24749.1
Labour14,15329.9
Liberal Democrat7,49815.8
BNP1,3662.9
English Democrats9522.0
Bus Pass Elvis Party1120.2

If every single Bus Pass Elvis and English Democrat only voted for their respective parties the 50% majority required would be 23,132 and the Conservative Party win. However if we apply the Median Rule we'd be distributing the LibDem and BNP Vote as well.

Taking our first result, even if every LibDem and BNP vote were added to the Labour one it would still be less than the Conservative result. Conservatives still win.

The systematic count - skim the bottom Party, redistribute, repeat until majority; is simply not necessary.