NFL futility (in the wild)

The Shutdown Corner, a Yahoo! sports blog notes that the redskins will play their sixth consecutive game against a winless opponent. Quite impressive. In the post they note that the chances of this happening are 1 in 32,768. I have some comments:

I think what they mean to say is that given the Redskins schedule this year and assuming each of their opponents has a 50% chance of winning each of their games the chances of this happening are 1 in 32,768. (That calculation checks out.)

However, the assumption that any of the redskins opponents has a 50% chance of winning their games is laughable at best (i.e. Lions, Bucs, Chiefs). If we re-do this calculation with last years respective winning percentages for these teams the chances are only about 1 in 1,703. ((14/16)*(7/16)^3*(4/16)^3*(14/16)^5=0.000587). But the bucs and panthers are much worse than last year, so the chances are probably a bit larger than 1 in 1,703.

On the other hand, for this to happen to a random team with a random schedule is very unlikely. Evidence of this is that it has never happened in NFL history before, as noted in that blog post.

So while the Redskin’s feat is impressive (or depressing?), I think they got the numbers wrong.

Cheers.

Posted on October 14, 2009, in Uncategorized. Bookmark the permalink. Leave a comment.

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