Friday, May 7, 2010

U.S. House of Reps Electoral Math

Jim Geraghty of National Review provides an invaluable breakdown of 99 Congressional seats that are in play this fall. Let's start with that and apply some math.

The current breakdown of the House is 253 Democrats, 178 Republicans, and 4 open seats. So either party needs 218 seats to secure a majority. According to Geraghty, all the Republican seats except one (Joseph Cao of Louisiana) are "safe." And 158 Democratic seats are also "safe." Let's assume that all of those safe seats remain unchanged (or at least that if one or two switch sides, they will be balanced by other switches so that the total numbers remain unchanged). What is the probability that the Republicans gain control of the House?

I assumed that a "code blue" in Geraghty's construction was an 80% chance of winning, a "code green" was 65%, a "code yellow" 50%, a "code orange" 35%, and a "code red" 20%. I also placed Joseph Cao in the "code orange" category.

The Republicans need to pick up 41 seats to get a majority. If Geraghy is right and my probabilities are reasonable, that gives them a 99.5% chance of doing so. Wow! It's worth looking at how the odds break down for some other possible splits.

Split# of GOP PickupsProbability
DEM +1 (or more)40 or les0.5%
GOP +1 (or more)41+99.5%
GOP +5 (or more)43+98.5%
GOP +11 (or more)46+93.7%
GOP +15 (or more)48+86.5%
GOP +21 (or more)51+68.0%
GOP +25 (or more)53+51.8%
GOP +31 (or more)56+27.8%

It's certainly looking very good for the GOP right now. Just to check myself, I re-ran the study with more conservative probabilities (thus reducing the likelihood of taking over any seat), and even then the GOP has an 81% chance to end up with a 15-seat (or better) majority. It's the GOP's race to lose.

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