This isn’t an attempt to plug AV, nor to speculate on what would happen under it, but just to knock down the notion that it’s fiendishly complex.
(Update 26/4: I'm getting quite a bit of Google traffic to this post, which is nice, but I should say that this video featuring Dan Snow is a much better explanation of the difference between AV and FPTP. It's not neutral, but I'm sure you're smart enough to separate out the 'how it works' from the 'why it's good'.)
How to vote
Anne, Bob, Claire, Doug and Eve all want you to vote for them. So who do you like the most? Pick one (Bob, say) and put a ‘1’ next to their name: that’s your first choice. Then, pick which of the rest of them you’d like the most if Bob were out of the race (Eve, say) and put a ‘2’ next to their name. Then, pick which of the rest you’d like the most if Bob and Eve were out of the race – and so on.
You can go all the way and put ‘1’, ‘2’, ‘3’, ‘4’ and ‘5’ next to each name, or you can stop at the point where you don’t have a view on which of the rest is best. You can just put ‘1’ and leave at it that, if you don’t care for the rest of them at all. Then you’re done.
How to count the votes to see who wins
We put all the votes that mark Anne as the first choice in a pile. And then the ones for Bob and Claire and so on. If one of them has more than half the votes, they win.
If not, then the one placed last (say it’s Bob) gets knocked out and we move to round two. Round two, at root, asks: ‘Who do you like the most out of Anne, Claire, Doug and Eve?’
We look at all the votes from Bob’s pile to see who’s the next choice on each. If you marked Bob ‘1’ and left it at that, then you’ve said you have no view on the rest of them, so your vote gets put to the side. If you marked, say, Eve ‘2’, your vote goes on her pile. We move all the Bob votes that do have a ‘2’ marked to the piles for Anne, Claire, Doug or Eve. The votes these four had from round one still count this time, of course: those who cast them still have the same first choice in a race with no Bob.
And if one of them has more than half the votes, they win.
If not, we do the same thing: say Anne is last this time. She gets knocked out and for round three the votes in her pile get moved to Claire’s, Doug’s and Eve’s piles, as marked. And if one of these three has more than half the votes, they win. If not, we knock out the one placed last and go to round four, in the same way. When we get to the point where one of them has more than half the votes, they win. And that’s that.
(Number three in an extremely occasional series.)
16 comments:
"if one of them has more than half the votes"
You need to specify "votes", as the number of "votes" changes at each round if some are discarded.
In round 1 it is more than half of the votes cast. In subsequent rounds it is more than half of the votes left, which is usually a different number.
Scott: A rather moot point, as if the same round had been a real FPTP election then that person would have most likely spoiled their ballot anyway.
Scott: in every round, including the first, it's more than half the votes expressing any preference for any of the candidates in play in that round. I do say that after the first round, votes showing no further prefs are put to the side, so I'd hoped that was implicit, but maybe not.
If you want monosyllables: "more than half the votes that make a choice from those still in the race".
"more than half the votes that make a choice from those still in the race"
Thanks.
The clarification is important because it has been claimed that AV guarantees the winner more than half the vote, without specifying that not everyone who casts a valid ballot is included in "the vote"
It's on the third round I get lost. If that is when Anne voters' third choice comes into play?If so would the second choices for Anne voters not be overlooked completely?
Clarification welcome!
No, the number preference that you count isn't directly related to what number round it is. If Anne was your first choice, knocking her out means they look at your next - ie second - choice.
If you put Anne 1 Eve 2, your vote then goes to Eve. (Likewise if you put Bob 1 Anne 2 Eve 3, as Bob had been knocked out after the first round and your vote was shifted to Anne.)
If you put Anne 1 Bob 2 Doug 3, your second choice is already out of the game, so your top choice of those left is Doug, so your vote then goes to him.
I think I've lost the will to sit up straight..... Does it basically mean that the winner won't tend to be the first choice of the minority. (Sorry about multi-syllables)
The candidate who's ahead in the first round has a head start for the second. All else being equal, they're likeliest to end up the winner.
It depends on how much support the ones that stay in the race can attract from the eliminated ones.
If it's Con 48 Lab 30 LD 12 UKIP 10, the UKIP 2nd prefs are pretty likely to give the Conservative the win.
But if it's Con 40 Lab 38 LD 12 Green 10, there's a decent chance the Green 2nd prefs would push Labour ahead; then if the LDs split about 50-50, Labour could win.
I think this is good attempt to explain the alternative voting system. It's a pecularity of the system that it sounds more complex than it really is. All it is an attempt to have a run off (multiple round) election all in one go. That's why it's called Instant Run Off in the US.
Another worked example is available here
http://fairervotesedinburgh.wordpress.com/2010/11/30/vote-early-vote-often/
Suppose you wish to just vote for one candidate. Would the vote be included in the first count or would it be a spoilt paper, and never included?
"Would the vote be included in the first count or would it be a spoilt paper, and never included?"
Under the proposed UK system: included in first count. Ditto if you only put 1,2; or if you only put 1,2,3, or whatever.
Yes, you can just put ‘1’ and leave at it that, if you don’t care for the rest of them at all. It'll be included in the first count and in all the following ones as long as your candidate's still in.
But the real quirks in AV arise when the candidates in second and third place are very close together.
So, imagine where it is Con 35, Lab 30, LD 30, others 5
The second and third preferences of the 'others' voters will determine who gets knocked out first among the three leading candidates. Who is knocked out among them determines who eventually wins.
But the second preferences of the eventual runner up are ignored completely.
I'm not convinced that this is any fairer than what happens under FPTP.
You keep mixing up Bob and Doug. Which one is voted out at the start? and is only one of them left to compet to win?
Dammit. Thanks for spotting that: I said Bob in the last paragrpah when I should have said Doug. That's what comes of rewriting something several times and then being too tired to proofread it properly...
I'll just change that.
You are still left with the basic flaw of a low turnout so that there is, in effect, no credence given to abstainers. The only way around this is for a "None of the above" box so that "winners" on a minority vote cannot claim "a mandate from the voters"!
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