John Armstrong (of The Unapologetic Mathematician fame) asked me about NFL football: assuming all teams are evenly matched, what is the probability that a division sends three teams to the playoffs? For those of you who aren't familiar with the NFL's playoff structure, there are two conferences of sixteen teams each (the NFC and the AFC); each conference is divided into four divisions of four. The winner of each division gets into the playoffs, and the two best teams among non-division winners in each conference get in as "wild cards". (For an example, see this year's final standings at nfl.com.)

This is equivalent to the following question: what is the probability that both wild cards in a conference come from the same division? I suspect that this question is inspired by the fact that this actually happened in both conferences this year, in the AFC South and NFC East respectively. (This must be especially frustrating for fans of the fourth-place team in each of those divisions. The fourth-place team in the NFC East was the Philadelphia Eagles. I think I'll avoid Eagles fans today. But that might be hard.)

Anyway, the probability that this event happens in any given conference in any given year is

**2/11**. Let the teams play out their season. Consider the twelve non-division winners in a given conference, and rank them in order from 1 to 12. (You might think: in the NFL each team only plays sixteen games, so won't there be ties? Perhaps, but the NFL's wild card tiebreakers don't care what division a team is in.) The top two teams get in. The divisions of the twelve teams (from top to bottom) form a word with three N's, three E's, three S's, and three W's, picked uniformly at random. Say without loss of generality that the first letter that the word starts with an N; then the remainder of the word is picked uniformly at random from permutations of NNEEESSSWWW. The chances that starts with an N are 2/11.

Alternatively, there are words which contain three each of N, E, S, and W. The number of these that begin with the same letter twice is -- pick the letter that appears twice at the beginning in one of four ways, then arrange the ten remaining letters however you like. So the probability of the two division winners coming from the same division is 67200/369600, or 2/11. I don't like this approach quite as much, though, because it involves big numbers that cancel out; it seems cleaner to me to not invoke the big numbers in the first place.

By the way, the NFL has had its current structure since 2002, and the years and conferences in which this event has actually occurred are the 2007 AFC South, 2007 NFC East, and 2006 NFC East, making three out of twelve.

The question I asked in September was not the analogue of this one (MLB only has one wild card, so there is no analogue); there I found the probability that the four MLB teams making the playoffs from any league were actually the four teams with the best records. But I don't have a general method to find the solution for some league with arbitrary structure, and the NFL would require slogging through more cases than MLB (two wild cards instead of one) so I won't bother for now. (Or probably ever -- I don't particularly

*care*about that number.)