Jens 'n' Frens
Idle thoughts of a relatively libertarian Republican in Cambridge, MA, and whomever he invites. Mostly political.

"A strong conviction that something must be done is the parent of many bad measures."
  -- Daniel Webster

Friday, January 04, 2008 :::

Does anyone have access to the Iowa Democrats' "Caucus Mathematics Worksheet and Reporting Form"?
The caucus mathematician, a gray-haired grain and livestock farmer named John LaFratte, sat beside Mr. Ryner and inked the final tally onto a broad sheet of paper titled "Caucus Mathematics Worksheet and Reporting Form": 45 votes for Edwards, 26 for Obama, 24 for Richardson, and 21 for Clinton.

To divide the precinct's nine delegates among the candidates, Mr. LaFratte punched numbers into a calculator, using a mathematical formula supplied by the party that factors in the size of each preference group and the number of caucusgoers. Three delegates would go to Edwards and two each to Obama, Richardson, and Clinton.
In this case, if you do the calculations naively and round off, these are the numbers you get. If one more person had come to vote for Edwards, though, it would push Edwards above 3.5, and you would find yourself trying to give out ten delegates when you only have nine. Does anyone know what the apportionment scheme is? (If they were states being allocated nine congressmen by population, Edwards would need 52 votes before he took a delegate from Clinton. Other systems I know of would already give him one of her delegates with the actual data.)

I've previously thought that I like single transferable vote for this sort of thing, but that it would be hard to explain or get people to do. Reconsidering this, though, I would think if you said "get into groups and the nine largest groups each get a delegate", it would allow you to do essentially the same thing in a way that wouldn't be too unintuitive. In this case Richardson's supporters would divide into two groups of 12 and he and Obama would get 2 delegates each; Clinton would get one and Edwards three, while Clinton's supporters and Edwards are each trying to convince two of Obama's most pragmatic or tenuous supporters to come help them lay claim to the last delegate. The problem with this is the likelihood of a number of groups tied for the ninth and tenth largest. How would a precinct with 3 delegates and two equally sized camps handle the analogous situation today?

::: posted by dWj at 6:38 PM

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Idle thoughts of a relatively libertarian Republican in Cambridge, MA, and whomever he invites. Mostly political.

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