Update: The numbers changed a bit since this entry was posted the other day. The Democrats got 55% of the average district vote, not 56%. (The confusion came because we used numbers from the New York Times that counted third-party votes in a different way than we did, I think; see note at the end of this entry.)
Back to the main story:
The Democrats’ victory in the 2006 election has been compared to the Republicans’ in 2004. But the Democrats actually did a lot better in terms of the vote. The Democrats received 54.8% of the average district vote for the two parties in 2006, whereas the Republicans only averaged 51.6% in 1994.
There was a big jump in 2006. Here’s the time series:
(Bigger version is here.) The shaded areas on the graph show the periods where Republicans have controlled the House. The 2006 outcome of 55% for the Democrats is comparable to their typical vote shares as the matjority party in the decades preceding the 1994 realignment.
54.8% of the vote, 53.3% of the seats
Even with their large vote majority, the Democrats only received 53.3% of the seats in the House. This is as we and Bafumi et al. anticipated. More info on the seats-votes relationship is in our recent paper. (For example, had the Republicans received 54.8% of the vote in 2006, we estimate they would’ve won about 245 seats.)
By the way, the Democrats’ 54.8% share of the two-party vote tracks closely with the “generic congressional vote” in which they were getting 56% in the polls (that is, 52.1%/(52.1% + 40.6%)).
We actually calculate average district votes by imputing 75% for uncontested races (to represent the strength that the party might have had if the district had been contested; the 75% comes from computing the average vote in districts just before and just after being uncontested, based on historical data), so we needed to make corrections for uncontesteds. In 2004, we have 31 uncontested Democrats and 37 uncontested Republicans: the average district vote for the Democrats was 50.5% using our correction (or 50.0% if you simply plug in 100% for uncontested races). In 2006, there were 45 uncontested Democrats and 10 uncontested Republicans, yielding an average district vote of 54.8% using our correction (or 56.8% if you simply plug in 100%). The 56.8% number is more dramatic but I think it overstates the Democrats’ strength in giving them 100% in all those districts.
Another option is to use total vote (see here) rather than average district vote. We discuss this in Section 3.3 of our paper (in particular, see Figure 4). The short answer is that we use average district vote because it represents total support for the parties across the country. The Democrats tend to do better in lower-turnout districts and so their total vote is typically slightly lower than their average district vote. See the scatterplot here.
P.S. Here’s a picture of the estimated seats-votes curve for the 2006 election:
More discussion here.
P.P.S. The first version of this blog entry said that the Democrats received 56% of the votes. When doing this calculation, I didn’t have a spreadsheet with all the 2006 data, but the New York Times reported that the average swing was 7.6%. We then corrected for uncontesteds (using the NYT numbers on how many districts were uncontested for each party) to get an average district vote of 56.1%.
Actually, though, it appears based on John Kastellec’s district-by-district tabulations that 56% wasn’t right either. We think the discrepancy arose because of how the media’s count reflects districts in which only one major party candidate ran but third party candidates received some votes. We just want to work with the two-party vote, and we consider an election as uncontested unless it is contested by both the Democrats and the Republicans. The 7.6% average swing reported in the press was counting third party votes in some way, we think.