I'm Backing AV 110%

(Sorry to go on about it, but I'm just fascinated by mathematics.)

One of the headline reasons being put forward in favour of AV is that it means that no one can become an MP without winning more than 50% of the vote (assuming every voter uses all of their votes, which is a detail I'm not too fussed about). I'm not sure this is such a great thing, since some of that support might be, to say the least, grudging, coming from voters who could actually think of six or seven candidates that they would rather have had than the winner for whom their vote was finally counted.

However 50% is still 50%. And what's really great about AV is it's not only the winner who gets more than 50% of the vote, so might some of the losers.

Let's return to the constituency I mentioned in my last post, but give it a more definite result. First-choice votes go:

Labour: 15,000

Liberal: 11,000

Tory: 9,000

The Tory drops out and let's suppose 7000 of his second-choice votes were for the Liberal. The Liberal now has 18000, which is more than 50%, and wins. First-choice votes of one candidate plus second choice votes of another gives greater than 50%.

The trouble is, the Labour candidate also got more than 50% if you add the first choice votes of one candidate and the second choice of another. The Labour first-choices plus the Liberal second-choices (assuming a reasonable split) also come to greater than 50% . (It's actually mathematically possible that the Tory could get more than 50% too if the Labour candidate's second choice votes were largely for the Tory - but that's unlikely).

So 50% is a necessary but not sufficient condition to win an AV election. How then is the winner decided? We could go back to a FPTP-style approach, where it's the candidate with the highest number of votes (first and second choice combined), but in this example, that would probably be Labour. Here the Liberal wins with more than 50% of a particular set of first and second choice votes, but still fewer than the combination of votes that Labour got, but didn't get counted.

So the total number of votes cast is more than 100%?

Let's consider the definition of percentage. The percentage of votes for a candidate is:

(n / T) * 100

where n is the number of votes received by the candidate and T is the total number of votes overall.

But that term T could actually have several meanings. It could be the total number of voters, or the total number of votes cast or the total number of votes counted. (It could also be the total number of eligible voters, but turnout is a problem under any system.) Under FPTP those three definitions of T are all the same thing, because FPTP is one person-one vote.

But AV gives several votes to each voter. (You can argue that it's a good thing, but you can't deny it - the voter gets to indicate support for more than one candidate. It may not be that multiple votes are counted, but multiple votes are cast.) If you go with T being the number of voters, then it's clearly true - the winning candidate has more than 50% of the vote. In this case, 51%.

But if T is the total number of votes cast, with three votes per voter (assuming that there were no other candidates and that every voter used all their votes) that gives a total of 105,000 votes - and the winning Liberal candidate gets (18,000 / 105,000) * 100 = 17%.

On the other hand, if T is the number of votes counted, things get more complicated. In the first round, 35,000 votes were counted. The Tory dropped out and so in the second round a further 9000 votes were counted, giving a T of 44,000. So now the winning Liberal candidate's percentage is (18,000 / 44,000) * 100 = 41%. Not a bad result for a winner under FPTP, but this is AV, which supposedly guarantees the winner gets more than 50%. You may think I'm being unfair, counting those second-choice Tory votes into T with the same weight as first-choice votes, but if you do, then you must surely also object to them being counted into n (the votes for the winning candidate) with equal weight - that's one of the main objections that many people have to AV.

So the question that we must all answer, whether we are pro or anti or could not care less, is:

What is your definition of T?

Marathon (and Zeno) Debrief

The news you're all waiting to hear is that I completed the Brighton Marathon in 4 hours 12 minutes 43 seconds.

But I know you're not really interested in how quickly I ran the marathon but, as I asked in an earlier post, what was the distance?

Despite not being able to post diagrams, I suspect Katie had the answer to the question posed (as I guessed she would before I even posted).

The answer is that Mr Tortoise, the slower runner, records the further distance on his GPS.

Suppose the following is the actual route that both runners take on a small fragment of the course. It's a sharp bend, to make things clearer, but the same principals would apply to any bend in the route. (Click the diagrams to enlarge.)

Now Mr Tortoise will complete this section in twice the time of Mr Hare. Given that their GPSs both sample their position once a second, that means Mr Tortoise will have twice as many samples (shown as red crosses) as Mr Hare (green circles).

Given these samples locations, calculation of the actual route taken is done by joining up these locations as shown.

(Note that it's an assumption that the GPS software joins the dot's as straight lines. Conceivably, a much more complex algorithm could be used.)

Clearly the red (Mr Tortoise's) route is going to be longer than Mr Hare's, regardless of the nature of the bend. The only circumstances when this is not the case will be on a straight line, where the intermediate red cross will lie on the green line and the distances will be the same.

So on a perfectly straight route, the total distance will be the same, but if there are any bends at all, the slower runner will always record the further distance.

And of course, both recorded distances will be less than the true route, represented by the black curve.

Which leads me to be puzzled as to why, at the Brighton Marathon, my GPS recorded a distance of 26.41 miles, when an official marathon is only 26.22 miles. Worth further sponsorship, surely?

Zeno’s Paradox Revisited

Two runners, Mr Hare and Mr Tortoise, run a marathon. Mr Hare is faster and completes the course in, let's say, three hours, whereas Mr Tortoise takes six hours.

Each of them wears an identical GPS running watch, which samples the runner's location once every second and uses the information to calculate speed and distance travelled.

The marathon follows a typical street circuit and each of the two runners follows exactly the same path, without cutting any corners or taking any shortcuts.

The question is, at the end of the race, which of Mr Hare and Mr Tortoise is likely to have travelled the furthest distance according to his GPS watch?

If you have a solution, post it as a comment, showing your working. (Note there are no tricks in this, it's purely an issue of mathematics, or possibly physics.)

I'll be posting my solution after the Brighton Marathon on Sunday April 18^th, for which you can still sponsor me at http://www.runningsponsorme.org/jasperkent. I hope that my time will lie somewhere between those of Messrs Tortoise and Hare.