With less than three weeks to go before Election Day, prognosticators are all over the map on how many seats the GOP will gain. Everyone has his own proprietary model, but I’m going to let you in on the simple model that has proven to work with great regularity in recent years — the 70 percent solution.
This approach relies on the observation that, in wave years like this one (i.e., years in which one party wins by a large margin), approximately 70 percent of the seats in play on Election Day are usually carried by the victorious party. This holds true whether one looks at the Senate or the House.
For the House, look at Charlie Cook’s listing of seats in play. As Chris Cillizza noted yesterday in his “Morning Fix” newsletter, Cook currently has 92 House seats in play, only seven of which are held by Republicans. Consider history: In 2008, Cook had 53 seats in play, 13 of them Democratic; in 2006, he had 54 seats in play, nine of them Democratic. In 2008, the Democrats gained 21 seats, which means they won 34 of the 53 — 64 percent. In 2006, the Democrats gained 30, which means they won 39 of the 54 — 72 percent. These numbers match historical Senate patterns.
Applying the 64–72 percent pick-up range to Cook’s current list of House seats in play yields between 59 and 64 Republican wins. Subtract the seven seats already held by the GOP, and you’re left with a net pick-up of between 52 and 57 seats.
Looked at from this approach, the House GOP’s recent decision to start buying ad time in many Democratic-held districts not previously thought to be targets makes sense. Putting more seats in play increases the odds for a really big payoff. Expect to see more House “sleeper” seats pop up in the next ten days, and don’t be surprised if one more Senate seat gets placed on your radar with some intense, late ad buys.
— Henry Olsen is vice president of the American Enterprise Institute.