Richard Charnin
Nov.4, 2018
Nate Silver calculates an 85% probability the Dems will win the House – if they win the popular vote by at least 5.7%.
As usual, Nate Silver makes the simplistic assumption that the polls accurately reflect voter intent. And as always, he avoids mentioning the fraud factor.
Nate bases his calculation on the latest polls from Real Clear Politics which shows the Dems winning 202 seats and the Repubs 196 with 37 too close to call.
https://www.realclearpolitics.com/epolls/2018/house/2018_elections_house_map.html
Assuming the Repubs have 196 seats, then the probability P that they will win AT LEAST 22 of 37 races that are too close to call and win 218 seats is 1 in 6.
P= 16.2% = 1-BINOMDIST(21,37,0.5,true).
https://projects.fivethirtyeight.com/2018-midterm-election-forecast/house/
But…Given the latest Gallup voter affiliation survey and assuming equal vote shares below, the Dems would need at least 56% registered voter turnout compared to just 46% for the Repubs to win by 52.9-47.1%. Anything less than a 10% Dem turnout edge means the Repubs would win the House.
So the question Nate must answer is this: is it logical to assume that the Democratic turnout rate would be 10% greater than the Repubs? I don’t think so. If anything, the Repubs are more motivated.
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Generic Vote
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| Gallup | Party-ID | Turnout | Votes | Rep | Dem |
| Rep | 27% | 46% | 19573 | 92% | 8% |
| Dem | 28% | 56% | 24711 | 8% | 92% |
| Ind | 45% | 44% | 31204 | 50% | 50% |
| Total | 100% | 47.9% | 47.14% | 52.86% | |
| Votes | 75,488 | 35,586 | 39,902 | ||
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Turnout scenarios
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| Rep | Dem | Ind | Total | Rep | Dem |
| 46% | 56% | 44% | 47.9% | 47.14% | 52.86% |
| 47% | 55% | 45% | 48.3% | 47.65% | 52.35% |
| 48% | 54% | 46% | 48.8% | 48.14% | 51.86% |
| 49% | 53% | 47% | 49.2% | 48.63% | 51.37% |
| 50% | 52% | 48% | 49.7% | 49.10% | 50.90% |
| 51% | 51% | 49% | 50.1% | 49.57% | 50.43% |
| 52% | 50% | 50% | 50.5% | 50.03% | 49.97% |
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