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CA Delegates

05 Feb 2008 09:46 am

As Marc Ambinder lays out here, it's extremely difficult for one candidate to gain a large delegate advantage over her opponent as long as there are only two candidates in the field and both have substantial bases of support. In particular in a congressional district that's been allocated an even number of delegates, you need to win a big supermajority in order to avoid an even split of the delegates.

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By this logic, Clinton would have had a delegate lead in FL smaller than her popular vote lead (19%). But it was actually larger.

Now, it's true that if it's a close election -- 5%, the chances are the delegate margin will be zero. But if it's 10% or so, it will probably be a 10% or so delegate margin. The only way it can't be is for Obama to keep just above the "magic number" (38%) in all the even districts, while splitting the odd districts with Clinton (i.e. getting 50.1 in one, and 30.1--enough for 2 of 5 delegates--in another). But that's just unlikely.

Clinton had no delegate lead in Florida at all as Florida had no delegates.

This is just spin from the Clinton camp (would you expect anything else from Ambinder?) in order to lower expectations.

What he doesn't tell you is that Obama has the demographic advantage in many of these even districts, while Clinton has the advantage in the odd districts. So this system helps Clinton.

Ambinder also fails to note the delegates that will come from the overall statewide victory. All told, Hillary should still get a healthy margin in the delegate count.

Okay, but in a congressional district that's been allocated an odd number of delegates, you need to win by 1 vote to get an extra delegate. So in a CD that awards 5 delegates, you can win 50.0001% of the popular vote and get 60% of the delegates.

So Clinton should be wracking up delegates disproportionate to her popular vote lead in this districts.

The question, which Ambinder leaves unanswered, is which effect is stronger.

It seems likely that CA (my home state) will be a virtual push in the delegate count, and with Obama's home state strength in IL somewhat offsetting Clinton's in NY, SuperSpectacular Tuesday will probably not result in a clear winner in terms of delegates. The media may anoint a winner of the day, but even that hasn't proved lasting this cycle.

Over at
Open Left
, Chris Bowers argues that this means the nomination will be determined by the super-delegates. My guess is that those super-delegates will be closely watching the contests through March 5 to see which way the wind is blowing.

In California there are 2 districts with three delegates, 26 with four, 19 with five, and 6 with six, so that's 21 odd districts to 32 even, for what that's worth. The question, of course, is who's ahead in which specific districts.

As the Boston Globe points out Obama could get a big boost in delegates by winning a supermajority in the 20 predominantly black congressional districts that are voting today. Because these districts consistently vote Democratic, they are given more delegates. So it's a potential double win for him.

This diary on MyDD gives predictions for each district:
http://www.mydd.com/story/2008/2/2/212343/0037


As you can see, Hillary has the advantage in 17 of 21 of the odd districts. She will probably get a supermajority in a few even ones as well. Obama will get a majority in a number of even districts, but won't reach the supermajority needed to get an advantage from that.

All this adds up to a Hillary win of about +30.

California awards only 241 delegates proportionaly by district and the other 129 at large delegates go to the state vote winner. A total of 370 delegates are in play today.

The other 71 California delegates are 'super delegates' and round out the California delegation at 441 delegates to the convention, the largest in the nation.

California awards only 241 delegates proportionaly by district and the other 129 at large delegates go to the state vote winner. A total of 370 delegates are in play today.

The other 71 California delegates are 'super delegates' and round out the California delegation at 441 delegates to the convention, the largest in the nation.

To continue with what KC said, if there are three delegates, a candidate earns one each for surpassing 15, 50, and 85 percent of the vote.

If there are four delegates, you earn another delegate when you exceed 15, 37.5, 62.5, and 85 percent of the vote.

If there are five delegates, the additional delegate is earned at 15, 30, 50, 70, and 85 percent.

If there are six delegates, they are earned at 15, 25, 41.7, 58.4, 75, and 85 percent of the vote.

There is some chance a candidate could get to the 58.4 percent amount in a district to win an even one 4-2. But if you assume everything breaks close to 50-50, then you will see each candidate win a minimum of 110 delegates and a maximum of 131.

That is using KC's numbers, which total 241 delegates. The number reported for California is 370. I am not sure if it is super-delegates or something else causing the difference.

Curtis, the difference, as ken says, is the at-large delegates. But the DNC rules say those at-large delegates are divided proportionally according to the popular vote in the state, not all awarded to the state vote winner.

There was an article on Slate about this yesterday, and the fallacy there is thinking that because the race overall is expected to be close, extrapolating from that to think that the race will be close in each specific district.

The reality is that the congressional districts are smaller sample sizes to begin with, and in many cases they were specifically created to favor one demographic group or another. What we should expect in the popular vote in California is a big range of victory margins some close, some not so close, by the two candidates, creating a very close range only in the aggregate.

In other words, it's very possible that there will be significant numbers of districts where the 62.5% supermajorities come into play.


Comments closed February 19, 2008.

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