For the true hardcore…A version of the MATLAB code I used is now available. I used gerry.m for Part 1, and delegations.m for Part 2. Voting dat...
Gerrymandering notes: how many votes to take House, 1992-2012
Dear PEC Readers, we’re going casual tonight. This is for frequent visitors and those of you on the RSS feed. I’ll explain…
I’m preparing a long-form piece (for elsewhere) on the topic of partisan House gerrymandering. We’re cooking up some graphs to drive home some basic points. Your immediate reactions and critical questions will be welcome.
This graph shows what fraction of the two-party vote would have been needed for Democrats to control the House of Representatives.
The procedure was:
- Calculate the % two-party vote for all 435 districts.
- Calculate the shift in vote needed to make an outcome of exactly 218 Democratic seats.
- Add this shift to the national % Democratic vote.
The colored horizontal line segments indicate which party was in control. Generally, the out-party needs a bit more than 50% of the two-party vote to gain control. This extra barrier is an advantage for the incumbent party.
Note 1: Dealing with uncontested races is a challenge. For instance, the 2006 data point is distorted by the fact that there were 47 uncontested races won by Democrats (versus only 10 won by Republicans). Forty-seven is an unusually high number. With other definitions, this data point is more comparable to 1996-2004.
Note 2: I came into this analysis expecting the 2012 value to be unusually high because of partisan gerrymandering. It is indeed high – but it is only on a par with 2004. I am pondering if there is a problem I am missing.
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