Voting mathematics: 101

A vote. What is it? You can define it one of two ways. It is either the process of performing a vote, or it is the action of your choice being counted. I prefer the terminology in the first case (as do those that tend to describe voting systems... one man, one vote refers to only having to make one trip to the booth, not that they only put one "X" down in general), but it is irrelevant which you prefer in reality; the outcome for both is the same.

Under AV you carry out your vote, by putting as many preferences down as you want. Your vote is carried down the preference list you have stated as your candidates are eliminated. One vote, transferred as necessary.

Or perhaps you look at it like you put multiple votes down against multiple candidates, with only one vote counted per round regardless of which candidate is currently determined to have your support.

Take the following rows of "votes" for several candidates, A, B and C respectively:

|||||||||||||||||||| = 20
|||||||||||||||| = 16
||||||| = 7

You can see here, there's a total of 20 + 16 + 7 votes, or 43 votes.

The lowest candidate is eliminated, the votes stand as such...

|||||||||||||||||||||| = 22
||||||||||||||||||||| = 21

The total votes counted here is, again 43.

So who has had their vote counted more than once? If you follow the simple maths above, you'll see that no-one has had their vote counted more than once, or rather that everyone has had their vote counted more than once. Either analysis can be seen as true

If you feel that a vote is the process of voting, then those 7 votes are "re-counted" for different candidates as a second preference. Mathematically, if they were counted twice then we would have 50 votes on the table, not 43...but we don't, with the elimination of the last candidate the votes they accrued are nullified. The process is this:

A = 20 + 2
B = 16 + 5
C = 7 - 7

However if you feel that votes are choices you cast, then every round your vote is cast once more. The original votes of 20, 16 and 7 are counted, and then the second set of votes, 22 and 21 are counted. Two sets of 43, every single voter having their vote counted equally to everyone else in the process.

If someone had their votes counted more than once, how would it look?

Well round one would look like above, 20, 16 and 7. Round two?

|| = 2
||||| = 5

If you are not counting the votes of those who have not moved their vote, or are having a vote counted for a different candidate, then the election solely comes down to those who transfer. But we can see from above that's not the case. The actual result is that Candidate A wins by 1 vote, yet if we follow basic logic and mathematics, that only those who move their votes are counted again, then candidate B should win, 5 votes to 2. It's impossible for the result to stand as it does where candidate A wins by one vote without accepting that everyone's vote is counted equally, regardless of vote transference.

Those who voted for the most popular two candidates, and therefore don't have to change preferences, aren't forgotten and only counted the once under AV. Either their vote counts still because it is not yet nullified and redistributed, or it is counted equally because every round means each current preference is counted as a new vote for everyone involved.

Simple.