Dumb Jocks

I had been mulling over writing this all up, under the title "What's a Quintillion Among Friends", but it's only a bunch of college basketball games and not worth the effort to get the number exactly right. Nah, ain't worth it. BUT...

Two things happened.
1) I saw Nate Silver bloviating, on Olbermann IIRC, and he didn't scotch the 9.2 quintillion number
and
2) I caught a bit of the intermission shows on one of the networks, IIRC it was TNT, and one of the jocks, again IIRC, Kenny Smith made the salient point

That salient point is that the 9.2 quintillion numbers results from, as Kenny (I think, as they were all screaming at each other at once) said, coin flips across all 64 teams. And that's baloney. That number is 2^63: all permutations all games. So not happening. But all the reporting repeats that number. If Silver gainsayed it, I missed it in his interview.

By sometime last night, all the ESPN and Buffett brackets had been torpedoed. Lots of double digit seeds had won their games. Not going to buy my Caribbean Island. And so the discussion ensued.

Not all permutations can occur. There are 4 sets of 16, which if you treat each as permutations of coin flips (2^15) is 32,768 outcomes. Now, if you treat the sets as independent, that would be to the fourth power, and thus about 1.1 quintillion. Again, too many possibilities, since what one wants is the distinct number of binary trees.

Stackoverflow has discussion of binary trees, this one, at least not couched in March Madness terms. What a relief. But this could easily be wrong, too, since it is a calculation of root to branch tree building, and would thus have to account for each region having 16 different starting nodes (in this upended view).

If you do Catalan numbers, for 16 seed region, pivot on 16, then multiply be 4 regions, you get 615,506,319,360. Not sure that's quite right, since it doesn't account for the Final Four? Anyway.

Too early in the morning, and only because the dogs insist they have to poo, to work out such a meaningless number.