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Author Topic: The Delegate Fight: 2012  (Read 27503 times)
Erc
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« Reply #75 on: January 25, 2012, 10:50:02 am »
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The 'Uncommitted' delegates are indeed very much like the Democratic superdelegates, except there are far fewer of them, fewer than 150 in total (there's only 3 per state and many states, like Nevada, do pledge them to the winner of their state's primary).  While they certainly might make a difference if this actually goes to the convention, they aren't nearly as large of a factor as they could have been on the Democratic side in 2008.
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« Reply #76 on: January 25, 2012, 12:17:38 pm »
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A lot of these "super-delegates" have to pledge themselves to the winner of the state primary/caucus, though if there is indeed a brokered convention, they're presumably allowed to vote for whoever they want on the second ballot.
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Erc
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« Reply #77 on: January 25, 2012, 01:24:47 pm »
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A lot of these "super-delegates" have to pledge themselves to the winner of the state primary/caucus, though if there is indeed a brokered convention, they're presumably allowed to vote for whoever they want on the second ballot.

None of the "superdelegates" from NH, SC, FL, AZ, and MI will have voting privileges due to RNC sanctions.

The "superdelegates" from DE, GA, KS, NJ, and NV are bound by the vote in their state, effectively acting like regular pledged delegates. (Demconwatch says the same about VT & MO, but I think this is outdated information from 2008).

This means that there are 46 states and jurisdictions (remembering the insular territories & DC) with 3 "superdelegates" a piece, plus Huntsman's two delegates from NH, for a total of 140 "superdelegates," or 6.12% of the total number of delegates.

This is a sizeable chunk (comparable to Texas' 152 delegates), but it's unlikely that they'll break so overwhelmingly for one candidate or another to make a difference.
« Last Edit: January 25, 2012, 01:27:15 pm by Erc »Logged
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« Reply #78 on: January 25, 2012, 05:35:31 pm »
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well these GOP superdelegates are enough to give the current delegate lead to Romney
http://projects.wsj.com/campaign2012/delegates

Lets face it, the superdelegates will go overwhelmingly for Romney given a choice. If this thing is close, they could make the difference
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Mr. Morden
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« Reply #79 on: January 26, 2012, 02:18:28 am »

Even though the RNC seemed to sign off on Florida's WTA rule, there could still be a legal challenge:

http://www.tampabay.com/news/politics/national/if-gop-fight-drags-on-so-could-argument-over-floridas-delegates/1212342

This would presumably apply to Arizona as well, which is in exactly the same situation.
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Erc
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« Reply #80 on: January 26, 2012, 09:36:46 am »
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well these GOP superdelegates are enough to give the current delegate lead to Romney
http://projects.wsj.com/campaign2012/delegates

Lets face it, the superdelegates will go overwhelmingly for Romney given a choice. If this thing is close, they could make the difference

In general, I will not be counting superdelegates until their states actually vote (sometimes they do something crazy and decide to respect the will of the people).  If I did include them, you are correct, Romney would have the lead right now, as he has 16 endorsements to 1 apiece for Santorum and Gingrich.  Of course, there are still 122 who haven't endorsed anyone.

Information about Maine's caucuses has been updated, many thanks to CARLHAYDEN for the clarifications.  Apparently, some towns caucus as soon as this Saturday, while some caucus as late as March 2nd!  The results of the straw poll will not wait for these stragglers, and will be announced regardless on February 11.
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« Reply #81 on: February 01, 2012, 01:31:48 am »
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According to CNN through last night's Florida contest, the delegate count stands at Mitt Romney 84, Newt Gingrich 27, Ron Paul 10, Rick Santorum 8.  If these numbers are correct, Mitt Romney is roughly 7.5% of the way to the required 1,144 delegates.  Can you confirm these numbers, Erc, or do you show something different?
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Erc
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« Reply #82 on: February 01, 2012, 04:01:31 am »
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Mitt Romney wins all 50 of Florida's delegates with his win in the state.

As was pointed out earlier, if Florida's full delegations is seated, it won't be WTA statewide, but WTA by jurisdiction (by CD and statewide), so Gingrich would receive some delegates.  Note that the new 2012 Florida map doesn't appear to be done yet, which would immensely complicate the process.

At some point, I'll try to see what the exact counts would be without penalty, but it seems that Gingrich couldn't have possibly won more than 5 CD's (all in or near the Panhandle), which would result in a breakdown of 89 Romney - 10 Gingrich.
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Erc
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« Reply #83 on: February 01, 2012, 04:11:46 am »
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According to CNN through last night's Florida contest, the delegate count stands at Mitt Romney 84, Newt Gingrich 27, Ron Paul 10, Rick Santorum 8.  If these numbers are correct, Mitt Romney is roughly 7.5% of the way to the required 1,144 delegates.  Can you confirm these numbers, Erc, or do you show something different?

My current count is Romney 69, Gingrich 23, Santorum 11, Paul 8.

Why the discrepancies?

1)  Superdelegates.  As stated before, I don't like to include these guys in my count before their state has voted, but it's perfectly legit to do so if you want.  CNN has 19 superdelegates for Romney and 2 for Gingrich which I do not currently have in my count.

2) Iowa.  Iowa doesn't actually select any delegates until June, after a long round of conventions.  My estimates and CNN's estimates differ, as explained earlier in the thread.  They have a 7 Romney - 7 Santorum - 7 Paul - 2 Gingrich - 2 Uncommitted (ex-Perry) split, whereas I think a 10 Romney - 10 Santorum - 5 Paul split is slightly closer to reality.

Taking into account both of these, you recover the CNN count (up to a single Romney delegate, I probably lost a superdelegate in CNN's count someplace)
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Erc
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« Reply #84 on: February 01, 2012, 04:38:51 am »
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Finally decided to cave and am now including a count of all the other superdelegate endorsements (apart from Kim Lehman's, who is already there) on the front page, below the Totals.
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Erc
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« Reply #85 on: February 01, 2012, 02:56:04 pm »
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Arizona: February 28

Overview
29 Delegates (1.27% of total)
Primary
Winner-Take-All

All 29 delegates in Arizona are assigned to the statewide winner.

Arizona violated RNC rules by holding its primary before the first Tuesday in March, and thus lost half of its delegates.  It is possible these may may be restored later by the RNC, in which case the Arizona Primary would be assigning 58 delegates (all still At-Large WTA).  

Since Arizona is assigning its delegates as Winner-Take-All, it is also violating RNC rules forbidding such contests before April.  The RNC may choose to again penalize Arizona, but this seems very unlikely.

RNC Members

In the event sanctions are lifted on Arizona, its RNC members are bound to the winner of the primary, and are included in the 58 At-Large delegates noted above.

Preliminary Results (as of 2/28)

Romney won the state and all 29 delegates.  If sanctions were removed, he would get 55 delegates.

Michigan: February 28

Overview
30 Delegates (1.31% of total)
Primary
28 by CD (WTA)
2 At-Large

Michigan, by holding its primary before the first Tuesday in March, is violating RNC rules and has been penalized half its delegation.  Although Michigan has been through this song and dance before, they have (at long last) adopted plans to allocate delegates given these penalties.

The winner of each CD will receive 2 delegates.  There are 14 CD's in Michigan, for a total of 28 delegates by CD.

The remaining 2 delegates will be awarded based on the statewide vote.  It has been clarified that these 2 delegates are assigned WTA.

RNC Penalties

In the event RNC penalties are removed, the original delegate allocation plans will presumably be followed.  For reference, they are:

42 of these 56 delegates will be chosen by Congressional District, WTA.  14 delegates will be chosen on a statewide basis, proportionally, with a 15% cutoff.  After rounding to the nearest whole delegate, any additional delegates needed will be given to the statewide winner; if there is a surplus of delegates, they will be removed from the candidate with the fewest votes.  In addition, 3 RNC members will be unpledged.

RNC Members

Due to RNC penalties, the 3 RNC members from Michigan will not have voting privileges at the convention.  If the sanctions are removed, they will act as unpledged delegates.

Bobby Schostak
Saul Anuzis - Romney
Holly June Hughes - Romney

Preliminary Results (as of 2/29)

Romney - 16
Santorum - 14

Romney won the state and wins the 2 WTA delegates.

Each candidate won 7 of the states 14 CDs, splitting the CD delegates 14-14.  Discussion here, breakdown from the AP and the MI GOP.

Should sanctions be removed, the delegate allocation would be:

Romney - 28
Santorum - 28
« Last Edit: March 02, 2012, 12:19:37 am by Erc »Logged
Erc
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« Reply #86 on: February 05, 2012, 08:27:05 am »
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I had hoped that by this morning I'd be able to have something other than very preliminary results for Nevada....

As it currently stands, with 70.4% of precincts reporting:

Romney 13
Gingrich 6
Paul 5
Santorum 3

It's not clear how Nevada does the rounding...in accord with most other Republican processes, I'm choosing to round each to the nearest whole delegate for each and then giving the 28th delegate to the statewide winner, Romney.

Gingrich, currently at 22.6%, gains a delegate if he reaches 24.1% and loses a delegate if he drops to 20.3%.

Paul, currently at 18.6%, gains a delegate if he reaches 20.4%, and loses a delegate if he drops to 16.6%.

Santorum, currently at 11.1%, gains a delegate if he reaches 13.0% and loses a delegate if he drops to 9.2%.  He will also lose a delegate if there is a net gain of two delegates by other candidates, or if Romney reaches 50% without any other candidates losing delegates.

Romney, currently at 47.6%, gains a delegate if he reaches 50.0%.  If not, he can gain a delegate if the other candidates net lose one delegate.  He loses a delegate if there is a net gain of one delegate among the other candidates.
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Erc
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« Reply #87 on: February 05, 2012, 10:05:27 pm »
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I can't count.  Romney wins 14 delegates out of Nevada, not 13.  (A 15th delegate would require him getting nigh on 54%, which he isn't going to manage at this point).

This means the count out of Nevada, barring major swings in the last 10% of the precincts, is
Romney - 14
Gingrich - 6
Paul - 5
Santorum - 3

Again, the rounding rules are not firmly established...but most reasonable rounding methods would come up with the same count.  Romney may have enough support at the convention to push through obscene rounding rules (all fractions rounded down, or something) which might net him another 0-2 delegates, but I doubt it'll come to that.
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Erc
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« Reply #88 on: February 07, 2012, 10:04:03 pm »
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With Santorum now projected to win in Missouri, I estimate he will receive 0 (out of a possible 0) delegates.  Wink

Delegate estimates from MN and CO will be forthcoming pending more results.
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« Reply #89 on: February 07, 2012, 10:46:24 pm »
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Santorum is currently at 46% of the straw poll vote in MN. If he wins a majority of delegates to the state convention, could his delegates make it a winner-take-all state, by changing the delegate allocation rules by a majority vote?
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Erc
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« Reply #90 on: February 07, 2012, 11:16:06 pm »
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Santorum is currently at 46% of the straw poll vote in MN. If he wins a majority of delegates to the state convention, could his delegates make it a winner-take-all state, by changing the delegate allocation rules by a majority vote?

This isn't the Democratic Party, where proportionality is required even at conventions---or even Nevada, where the straw poll is binding on the delegate allocation.  If Santorum supporters do form a majority at the convention, they can elect a completely Santorum slate if they want to.

This is entirely plausible, especially since a lot of Gingrich/Romney support may be winnowed out by rounding errors at the precinct caucuses and the BPOU conventions.

If I can find any detailed data on how many delegates each precinct/BPOU elects, I may be able to find out how likely this is.
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Erc
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« Reply #91 on: February 07, 2012, 11:26:24 pm »
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Complicating the process in Minnesota is the fact that they have apparently not yet completed redistricting (and might not before the first BPOU conventions).  As Minnesota neither gained nor lost delegates with the census, they may just use the old boundaries.
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« Reply #92 on: February 08, 2012, 12:11:41 am »
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I'm not well versed in the nuts and bolts of how the delegates are selected for each state, but it's looking fairly likely that nobody is going to have enough delegates going into the convention. Which wouldn't be such a big deal except that all the modern convention is supposed to be is a giant televised, impeccably choreographed pep rally. With cameras rolling, is it really going to get down and dirty in Tampa this summer? Cause that would be hilarious.
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« Reply #93 on: February 08, 2012, 12:18:39 am »
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Complicating the process in Minnesota is the fact that they have apparently not yet completed redistricting (and might not before the first BPOU conventions).  As Minnesota neither gained nor lost delegates with the census, they may just use the old boundaries.

The Courts will release new redistricting maps two weeks from today. It will be a mess getting it organized but I do believe they will be using the new lines.

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Erc
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« Reply #94 on: February 08, 2012, 01:45:28 am »
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Very, Very Preliminary Delegate Estimates:

Colorado: 
Santorum - 12
Romney - 11
Gingrich - 6
Paul - 4

Minnesota
Santorum - 18
Paul - 10
Romney - 8
Gingrich - 1

This is pure guesswork at the moment; I'll have a chance to do a more thorough analysis by CD tomorrow for something slightly closer to the truth.  For reasons discussed above, I am probably well underestimating Santorum's performance in Minnesota.
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Erc
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« Reply #95 on: February 08, 2012, 02:04:33 am »
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Of note tonight:

Santorum appears to have passed Gingrich in delegates, at least after a first pass of tonight's results.  As a result, I've moved him back to the second column position he held before South Carolina.

Romney has probably lost his delegate majority (even including his declared supers), depending on the final fallout from CO/MN.
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Erc
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« Reply #96 on: February 08, 2012, 02:27:40 am »
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I'm not well versed in the nuts and bolts of how the delegates are selected for each state, but it's looking fairly likely that nobody is going to have enough delegates going into the convention. Which wouldn't be such a big deal except that all the modern convention is supposed to be is a giant televised, impeccably choreographed pep rally. With cameras rolling, is it really going to get down and dirty in Tampa this summer? Cause that would be hilarious.

Even with the simply ridiculous nature of this race so far, I would be very surprised if this does end up going to the convention floor, for a variety of reasons.

You obviously need three (or more) candidates with a sizeable number of delegates in order to have a brokered convention, barring a knife-edge outcome.  I sincerely doubt that both Gingrich and Santorum will stay viable; even without dropping out, one or the other will likely fade into irrelevancy at some point in the coming weeks (at this juncture, presumably Gingrich).  If neither does, then you run the risk of Romney running away with a lot of delegates against a divided vote (remember, very few states really award delegates proportionately).  There's a slim possibility that Gingrich could maintain viability (and grab a bunch of delegates on Super Tuesday) as a purely regional candidate (playing well in the South while Santorum doesn't)...though this seems unlikely for a number of reasons.

Of course, there's also the Paulite fantasy where the Paul delegates hold the balance.  In primary states, Paul is likely to receive very few delegates (given his low ceiling, and given how few primary states actually award delegates proportionally).  He has a better chance in caucus states, and not only due to his vaunted 'caucus organization'---caucuses and their ensuing conventions are messy affairs, and if it really does come down to a long slog of a battle between Romney and Santorum (or Gingrich), Paul may pick up delegates in extremely divided conventions (rather than either side ceding any to the 'enemy').  However, unless Paul really does end up controlling any state conventions by entryism or by playing the caucus game really well, it's going to be a modest number of delegates that he wins overall.  And there's always the very likely possibility state conventions not controlled by Paul understandably decide to vote for a slate of delegates excluding Paul supporters.

And if it does come down to mainly a one-on-one slog in the later primaries, very few of the contests are truly proportional, and are likely to give far more delegates to the winner than to the loser.  This makes the sort of evenly-matched delegate-by-delegate Obama-Clinton fight extremely unrealistic.  Of course, it could in the end come out balanced close enough for a combination of Paul, Gingrich (or Santorum) remnants, and superdelegates to make a difference...and boy that would be interesting.

Or, this could be the equivalent of Huckabee's early-mid February successes (a vaguely similar reaction against the presumed nominee), and could fizzle out by the end of the month, with Romney running away with this thing.  Of course, this time around, Romney doesn't have that lovely Super Tuesday delegate lead McCain had, hadn't seen his biggest rival drop out, and didn't do nearly as well as McCain did in his losses.
« Last Edit: February 08, 2012, 02:34:03 am by Erc »Logged
Minnesota Mike
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« Reply #97 on: February 08, 2012, 06:31:09 pm »
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Very, Very Preliminary Delegate Estimates:

Colorado: 
Santorum - 12
Romney - 11
Gingrich - 6
Paul - 4

Minnesota
Santorum - 18
Paul - 10
Romney - 8
Gingrich - 1

This is pure guesswork at the moment; I'll have a chance to do a more thorough analysis by CD tomorrow for something slightly closer to the truth.  For reasons discussed above, I am probably well underestimating Santorum's performance in Minnesota.


If the race is still being actively contested I don't see how Romney gets more than 1 or 2 delegates from Minnesota, In fact I bet he gets shut out.  My very very rough guess would be Santorum 20, Paul 17

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« Reply #98 on: February 08, 2012, 06:32:04 pm »
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Could you do another count with Paul/Romney added up and Santorum/Newt added up, because I think it's clear that Paul would eventually support Romney and Newt would support Santorum.
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« Reply #99 on: February 08, 2012, 06:46:47 pm »
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Example districts in Colorado:


This didn't happen everywhere, but Paul probably won Minnesota and is a lot stronger in Colorado than the straw poll shows.  Caucus states are great for Paul, primaries are black and white straight popular voting (MO didn't count - they have a caucus that actually matters in a month).
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