In this thread, I noticed that a number of my tourney selections mirrored those of another doper. So I asked myself: How many different brackets would I have to fill out to cover every single tournament scenario?
Since high school was so long ago, I could not answer my own question as I have forgotten how to arrive at the answer. Can someone set me straight?
2 to the power of 63 (or 64 if you include the play-in game)
I’m probably missing something here, but isn’t it just 2 to the 64th power?
For a bracket with 2[sup]n[/sup] teams, there are 2[sup]n - 1[/sup] possibilities for the first round, 2[sup]n - 2[/sup] for the second round, and all the way down to 2 possibilities for the nth round. The proper thing to do is to multiply all those numbers, which gives you 2[sup][sub]n[/sub]C[sub]2[/sub][/sup].
For n = 6, you get 2[sup]15[/sup], or 32768.
For a 65 team single elimination bracket, there are 2[sup]64[/sup] ways to fill it out rationally.
I have seen people who fill these things out irrationally, that is, in ways that can’t possibly happen in the real tournament but which these people think might provide a hedge in the betting pool, where the winner is determined by the number of correct game winners named in the bracket. Stay with me here. For example, the guy will pick Oklahoma over Alabama in the first round, but stick Alabama in as a winner in a later round, somehow thinking that that provides a hedge against making a wrong pick in the first round which dogs him in the later rounds. I can’t say whether or not this strategy has any merit.
Anyway, if one allows for this hedging method , there are 2[sup]120[/sup] ways of filling out a 64 team bracket. I decided not to even try to figure it out for a bracket with a play-in.
I don’t see how people are coming up with the 2[sup]64[/sup] figure. Would someone mind explaining?
No, never mind, I get it.