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On Caucus Math 
26th-Jun-2014 01:57 am
party politics
(... The curse of being a math geek living in a state where they have caucuses instead of primaries (not to mention having spent some time observing party rules committees) is I end up thinking about this stuff...)

So here's a fun issue with caucuses:

First, a quick review of the basic rule for awarding delegates at caucuses.

Your precinct is allocated some number of delegates, D, to elect. Some number of people attend your caucus and each declares a preference for a particular candidate. Twenty minutes later, once the blood has been mopped off of the floor, the battle lines have hardened, and everyone who might have been inclined to change his/her mind has been talked to death, you then compute for each candidate the following number
(# of votes for that candidate)
—————————————  × D
(# attendees)
I'll note first of all that every attendee will in fact be included here, because if you don't ever actually declare a preference, that's treated as equivalent to declaring a preference for "Uncommitted," this extra fake candidate that's always added to the mix. So it's guaranteed that all of these "delegate-share" numbers will indeed add up to D.

The next step is to split each delegate-share into a whole number plus a fractional part. For each candidate, the whole number gets awarded directly, and, if those numbers by themselves don't sum to all of the available delegates, you then rank the fractional parts and distribute any remaining delegates, one each, to the highest candidates in that fractional ranking.

(...And yes, for those of you who know about this, I'm skipping the 15% threshold rule, which some states apply at the precinct level. Thankfully, in Washington, we got rid of that 8 years ago, since it's a complete waste of time at the precinct level [also has any number of bad effects, but that's a whole 'nother discussion].)

So first, I'll present Survival Trick Number One, so that you can survive in a chaotic caucus environment without having to do long division in your head. It goes like this: we rewrite the formula above as follows:
(# of votes for that candidate)
((# attendees)/D)
i.e., just divide numerator and denominator by D, which works because multiplication is commutative (...except that the DNC stupidly ruins the commutativity by including a 3-decimal rounding rule in the process, but, as it happens, this doesn't affect things very often at the precinct level, and in any case this still works fine as a rule of thumb so that you can wrap your head around what's going on...)

Bottom line is there's a certain magic number of votes ((# attendees)/D) that you need to get a "whole delgate".

Meaning you can take your caucus, divide it up into blocs of that many people who are all voting for the same candidate. For each such bloc, that candidate gets a "whole delegate", and then whatever votes you have left over, you rank those, and the candidates that are highest on that ranking get the remaining delegates. The advantage of doing things this way is that you're just counting votes without having to do any long division in your head.

Concrete example:

You're in a precinct that's been allocated two delegates.
Twenty people show up.
14 are Kerry supporters.
6 are Dean supporters.
If you follow the worksheet, then it's (14/20)*2 = 1.4 vs. (6/20)*2 = 0.6.
Kerry gets 1 whole delegate and then, because 0.6 beats 0.4, Dean gets the other one.

Or you can do it my way, seeing that (20 attendees)/(2 delegates) = 10 votes needed to get a whole delegate. Thus, Kerry's 14 votes produce one whole delegate (10 votes) with 4 left over that then lose to Dean's 6 leftovers, and so Dean gets the other delegate out of the "fractional ranking", never mind that we're not having to rank fractions any more.

Now for the problem.
It turns out that the number of votes that you need to get a whole delegate is NOT the same as the number of votes you need to win a delegate out of the fractional ranking.
In fact, if your candidate is getting awarded any whole delegates at all, there's a fair argument that some of your votes are being wasted.

What do I mean by this?

Back to our example: Thus far, it's 14 to 6 with each candidate ending up with one delegate. But then the Kerry folks wonder if they can do better. And it turns out, they can!

After a brief strategy session, 7 of the Kerry voters change their preference to Uncommitted. Which now means the totals are 7 Kerry, 7 Uncommitted, and 6 Dean. Since you need 10 to get a whole delegate, there are now zero whole delegates, the fractional ranking then has to award two and they go to Kerry and Uncommitted.

Except,... since the "Uncommitted" folks are really all Kerry supporters, it's a good bet that "Uncommitted" delegate will be signing in for Kerry at the next caucus level.

Which means Kerry has just effectively cleaned up and claimed both delegates.

WTFF? How did that happen?

On the other hand, he did have more than 2/3 of the vote in that precinct, so there's some argument that this isn't actually a totally unfair outcome and perhaps the real question is why should the Kerry supporters have to jump through this extra hoop to get the delegates that are rightfully theirs?

The problem is that, while the number of votes you need to get a whole delegate is
(# attendees)/D,
the number you need to guarantee one out of the fractional ranking is actually
(# attendees)/(D+1),
which, in our example is 20/3 = 6+2/3.

... or, more precisely, if your candidate has at least (# attendees)/(D+1) and there is at least one candidate with strictly more than (# attendees)/(D+1) (even if it's only the slightest ε more), then you are guaranteed a delegate out of the fractional ranking (since in that case there cannot be more than D groups with (# attendees)/(D+1) votes, and you're one of them, so you win)

Dean with 6 votes isn't quite there, and that makes all the difference in the world.

Meanwhile back in the first scenario, where the Kerry supporters are spending 10 votes to get a "whole delegate", this can now be seen as a ripoff, spending 10 when they only needed to spend 6+2/3, thus wasting 3+1/3 of their votes, which costs them a delegate.

More generally, (# attendees)/(D+1) will always be less than (# attendees)/D and, if you're in a close race, chances are you're going to care about that difference.

And apparently, they've even thought about this in Iowa, or, rather, it's the only way I can account for Iowa's version of the threshold rule which not only makes things way more complicated, but also introduces the nastier features of thresholds into the lower-delegate caucuses where they weren't originally a problem.

My fix, which will most likely never be adopted, is much simpler:

Instead of multiplying by D, multiply by (D+1).

That is, for each candidate you instead compute
(# of votes for that candidate)
—————————————  × (D+1)
(# total number of voters)
and then proceed as before, awarding the whole numbers, and again if the total number of delegates awarded in this way is different from D, use the ranking of the fractions to fix it. The only difference now is the (remote) possibility that there will be too many (i.e., D+1) delegates awarded via the whole numbers, in which case, instead of giving out delegates to whoever is at the top of the ranking, we're taking a delegate away from whoever is at the bottom (except that if all of the whole numbers are indeed adding up to (D+1), that means all of the fractional parts will necessarily be zero, so we don't even have to look at any ranking; you just pick somebody at random to lose one).

So in our example, for Kerry the magic number is 14*3/20 = 42/20 = 2.1 and for Dean it's 6*3/20 = 18/20 = 0.9, Kerry gets 2 and we are done; we don't even need to consult the fractional ranking at all.

Or, calculating things my preferred way, you need 20/3 = 6+2/3 votes to get a whole delegate, Kerry has 2 such blocs, Dean doesn't have any, and again we're done. At which point it should be blindingly obvious that there's absolutely no advantage to be had by splitting your voters up over multiple fake candidates; you'll get the same result either way.

Which is the way caucus rules should be (i.e., they just give you an answer and no amount of gameplaying changes it).

But, even though they're never going to adopt this rule, you can still use it, the point of it being that if you're ever in a situation where the Multiply By (D+1) rule is giving you a different answer than the actual Multiply By D rule, that's where you have to watch out.

This effect is most pronounced in the 2 delegate caucuses but it can show up in higher-delegate caucuses as well.

Example 2

4 delegate caucus
20 people show up
17 for Kerry
3 for LaRouche

So now it takes 20/4 = 5 votes to get a whole delegate. Thus, Kerry supporters can spend 15 to get 3 of the 4 delegates. Unfortunately, that means they only have 2 votes left over, which lose to LaRouche's 3 in the fractional ranking and so we get one LaRouche delegate.

What does the Multiply By (D+1) rule say? In this case (D+1)=5, meaning you only need 20/5 = 4 votes to get a delegate out of the fractional ranking, and, with 17 votes, Kerry supporters can produce four such blocs which will all beat LaRouche's 3 votes.

... the only problem being that those blocs need to be for four different candidates. But this is actually doable:

5 for Kerry
4 for Uncommitted (actually Kerry)
4 for Sharpton (actually Kerry)
4 for Kucinich (actually Kerry)
3 for LaRouche

Now Kerry only gets 1 whole delegate while the fractional ranking awards the rest to Uncommitted, Sharpton, and Kucinich, who all change their votes to Kerry at the next caucus, and thus Kerry cleans up all four delegates.
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