Posted by Neil Paine on December 16, 2009
"Q: With just three games left, do you think both the Saints and the Colts will go undefeated the rest of the way and play each other in the Super Bowl? -- Logan (Reno, Nev.)"
"A: Logan, I think there is a better chance that each team will go unbeaten in the regular season than there is that these two teams will play in the Super Bowl. The Super Bowl matchup many expect on Dec. 16 is seldom the one that materializes in late January or early February. Don't know which team it'll be, don't know when it'll be, but one of these teams will stub its toe in the postseason. If memory serves me correctly, the last time two No. 1 seeds met in the Super Bowl was in 1993, when the Cowboys beat the Bills. Rarely happens."
My stat-sense was tingling, so I decided to set up a Monte Carlo simulation to see if what Schefter claimed, that Indy and New Orleans both going 16-0 was more likely than them eventually meeting up in the Super Bowl, was true. Using each team's current Pythagorean W% (and a 58.7% home-field advantage, the league-wide rate in 2009) to predict game-by-game outcomes, I simulated the rest of the regular-season and the playoffs 10,000 times, using authentic seedings, matchups, etc. (except that all ties in the standings were broken by Pyth%). Here's what I found:
|Simulations||Colts go 16-0||Saints go 16-0||Both go 16-0||Colts in SB||Saints in SB||Both in SB||Both 16-0, in SB|
So it turns out that Schefter was, in fact, correct -- provided both teams try their hardest to win all of their remaining games, there's a 26.6% chance both the Saints and Colts finish the regular season 16-0, compared to just a 22.1% chance that the two teams meet up in February. And, of course, there's always a 5.8% chance that both teams go into the big game with no losses, in which case there's a 100% probability that Mercury Morris' head would explode.