Yesterday, I took a first crack at trying to figure out if Clinton can win the popular vote. It’s worth investigating because her surrogates keep suggesting that she has a chance to win the popular vote and that that possibility justifies her continued campaigning. Part of the challenge is in predicting turnout in the various states. Yesterday, I used the rather crude method of estimating turnout by comparing states with a similar number of Electoral College votes. Today, I decided I could improve on that methodology by using a different measure. The number of pledged delegates assigned to each state is based on the number of Democratic votes those states provided in the last three presidential elections. Using this measure allows me to differentiate similar sized states based on the differential partisan makeup. For the following chart, I estimated turnout using the following method:
Guam= Virgin Islands, plus 25%
Indiana (72 pledged delegates)= Maryland (70)
North Carolina (115)= Ohio (141), minus 18%
West Virginia (28)= Mississippi (33), minus 15%
Kentucky (51) and Oregon (52)= Alabama (52)
Puerto Rico (55)= Louisiana (56)
Montana (16) and South Dakota (15)= Delaware (15).
This yielded what I consider to be reasonable numbers. If I had to guess, it probably overestimates turnout in North Carolina, West Virginia, Indiana, and Kentucky, but hopefully it’s a wash.
All polling (except Montana and Oregon) is based on the latest Pollster.com polling averages. In all cases I distributed undecided voters in the polls evenly to attain 100%. If there was an odd number of undecideds, I gave the extra percentage point to the candidate that is favored to win in that state. Montana has no polls, so I just gave it the same breakout as South Dakota. Guam has no polls so I left it with a 50-50 split. The Oregon poll is based on the single recent poll, from SurveyUSA.
| State | Date | P. Del | Estimated Turnout | O % | C % | Obama Votes | Clinton Votes | Net Advantage |
|---|---|---|---|---|---|---|---|---|
| Guam | May 03 | 4 | 2500 | 50% | 50% | 1250 | 1250 | Tie |
| Indiana | May 06 | 72 | 760000 | 47% | 53% | 357200 | 402800 | Clinton +45600 |
| North Carolina | May 06 | 115 | 1800000 | 60% | 40% | 1080000 | 720000 | Obama +360000 |
| West Virginia | May 13 | 28 | 350000 | 39% | 61% | 136500 | 213500 | Clinton +77000 |
| Kentucky | May 20 | 51 | 540000 | 34% | 66% | 183600 | 356400 | Clinton +172800 |
| Oregon | May 20 | 52 | 540000 | 55% | 45% | 297000 | 243000 | Obama +54000 |
| Puerto Rico | Jun 01 | 55 | 360000 | 45% | 55% | 165600 | 194400 | Clinton +28800 |
| Montana | Jun 03 | 16 | 96000 | 57% | 43% | 54720 | 41280 | Obama +13340 |
| South Dakota | Jun 03 | 15 | 96000 | 57% | 43% | 54720 | 41280 | Obama +13340 |
| Total | — | — | — | — | — | 2277990 | 2136510 | Obama +141480 |
Based on current polling averages and my best estimate at likely turnout, Obama stands to increase his popular vote lead by 141,480 votes between now and the end of the nominating process.
It would help Clinton if she could tighten the race in North Carolina or expand her lead in Indiana. For example, if she lost North Carolina by a narrow 53%-47% margin and won Indiana by a healthy 58%-42% margin, she would reduce Obama’s projected lead in the last contests to 22,550.
However, as I noted yesterday, Obama currently enjoys an approximate 250,000 popular vote lead even if you give Clinton all her votes from Florida and the 46% of the Michigan votes that the exit polls suggested she would have earned if everyone’s name had been on the ballot.
If Clinton were to win every undecided vote in the current polls, she would reap about 400,000 net votes. That would be more than enough to erase Obama’s 250,000 lead. Of course, that would require Clinton winning Kentucky 72-28, West Virginia 75-25, and Montana and South Dakota by 54-46. Short of scandal, that is out of reach.
To give you another example, if Clinton were to split the projected Obama states of North Carolina, Oregon, Montana, and South Dakota fifty-fifty, and win her projected victories in Indiana, Kentucky, West Virginia, and Puerto Rico with a 60-40 split, she would pick up 272,000 votes. That would be just enough to claim a popular vote victory with Michigan and Florida included in the tally.
As I said above, these turnout estimates are imprecise, and the more they are off, the more these projections are off. But any way you slice it, Clinton’s chances of winning even a disputed claim to the popular vote are more of a pipe dream than a reality. The superdelegates don’t need to wait on the fence to see how this particular measure will turn out.
Update [2008-4-24 19:57:7 by BooMan]: In the 2004 gubernatorial election in Puerto Rico, almost two million voters turned out. That’s half the population of the island (not of registered voters). Based on that, my projection of 360,000 votes in Puerto Rico is probably too low. If one million people vote (using my formula) Obama will see his 140,000 advantage drop to 80,000. If two million people vote, Obama’s advantage will drop to 10,000. In neither case will it put any dent in Obama’s current 250,000 popular vote advantage.
