Most Vulnerable House Seats, Revisited

Based on a recent critique from Nate Silver, I have decided to conduct a seat-by-seat analysis of the U.S. House. In the tables below, I assign probabilities of Republican gains in the 88 most vulnerable Democratic seats. I also list probabilities of Democratic gains in the 10 most vulnerable Republican seats.

As the reader may notice, the Republicans have a much greater than 65 percent probability--the estimate that Nate Silver asserts--of winning the House. The Democrats are likely to win four or five Republican seats. However, the Republicans have at least a 92 percent probability of winning each of 31 Democratic seats. They have at least a 50 percent probability of winning each of the next 24 most vulnerable seats. They even have over a 10 percent probability of winning each of the 56th through 75th most vulnerable seats. This analysis implies that the Republicans are expected gain approximately 31 + 1/2*44 = 53 seats and lose about five seats. Thus, the Republicans are poised to net about 48 seats, while they only need to gain 39 seats to take control of the House.

What's the probability the the Republicans could only gain 38 or fewer seats? Using a back-of-envelope calculation, I would say that the Democrats only have about a two percent chance of retaining the House. Now, I am assuming independence between the individual seats and zero uncertainty in my seat-specific probability estimates. So, in reality, the Democrats may have as much as a 10 to 15 percent chance of retaining the House (coincidentally, my analysis on the generic ballot question yields about this probability). However, I do not think that the Democrats have anywhere near one-in-three odds of maintaining the majority.


Competitive Democratic Seats


Competitive Republican Seats