Sam Wang: Trump is Romney Lite
However, there is a quantitative difference: Trump’s underperformance is substantially worse than Clinton’s, in a way that differs from random expectations (two-sample t-test, p=0.0004). Electorally speaking, Donald Trump is Romney Lite. His weakness is spread across many states, most notably Utah and Kansas, states that voted for Ted Cruz in the primaries. Together with the increased number of undecided voters, this suggests that a lot of Republican voters have gone missing.
If Trump does not bring enough Republican voters home, there could be unexpected wins for Clinton. If all states within 5% went Democratic, the electoral total would be Clinton 381 EV, Trump 157 EV. This is the downside risk for Republicans. On the flip side, Clinton’s biggest major weakness is Pennsylvania, where she leads by a median of only 1%. If all states within 5% went Republican, the electoral total would be Trump 318 EV, Clinton 220 EV. So it is not crazy to imagine a Trump victory…if he could somehow become a candidate that did not repel members of his own party.
The November win probabilities in the banner quantify how likely these outcomes are. The probabilities are based on evaluating how likely it is that today’s Meta-Margin, which is Clinton +3.6%, would move to zero between now and November. Recall that the Meta-Margin is defined as how much state polls would have to narrow, on average, to create a perfect toss-up. So if Trump makes up 3.6% on average (for instance by making a net gain as undecideds become committed), then he would have an even-odds shot of winning the Presidency.
That doesn’t sound like very much change. However, the last five Presidential elections (1996, 2000, 2004, 2008, 2012) have not shown much movement, a standard deviation (SD) of only 2 or 3 percentage points. This is a feature of intense political polarization, in which opinions are entrenched and don’t move much. If the SD is only 2 percentage points, then Hillary Clinton’s November win probability would be 90%.
The win probabilities in the banner are more cautious. They assume greater uncertainty based on historical standards: from 1952 to 2012, the average July-to-November movement has been 7 percentage points, which I use as the SD. That gives a random-drift probability of 70%. For the Bayesian prediction, I assume that this year’s up-and-down poll movement sets a range for the final likely outcome. This re-sharpens the prediction a bit, to give a November win probability of 85%.
http://election.princeton.edu/2016/07/07/is-85-an-underestimate/
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