16-0 X2 or a Colts/Saints SB?

Which is more likely, that we have two 16-0 teams or that we have a Colts Saints SB?

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
10000	4662	5760	2655	4738	4684	2212	584

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.