Consider a single-elimination tournament with 2

^{n}teams entered. Ignoring the play-in games,

*n*is six for the NCAA basketball tournament. Some tournaments have "bys" on the first round, and I am ignoring that aspects of those tournaments too.

Let

*f*(

*n*) be the number of ways of filling out your brackets for the tournament. The problem considered here is to calculate

*f*(6).

Since 2

^{n}teams are in the tournament, 2

^{n - 1}games are played in the first round. Since there are two possible victors for each game, there are 2

^{(2n - 1)}possible ways of filling out your brackets for the first round. The total number of ways of filling out your brackets is then:

One could guess a closed-form solution for this difference equation and prove it correct by mathematical induction. If I recall correctly, generating functions are useful in finding the solution to such recurrence relations. But, since I am only interested in a few terms, I adopt the brute force calculations shown in Table 1.f(n) = 2^{(2n - 1)}f(n- 1)

n | Number ofTeams | 2^{(2n - 1)} | Ways ofFilling Out Brackets |

1 | 2 | 2 | 2 = 2^{1} = 10^{0.301} |

2 | 4 | 4 | 8 = 2^{3} = 10^{0.903} |

3 | 8 | 16 = 2^{4} | 128 = 2^{7} = 10^{2.11} |

4 | 16 | 256 = 2^{8} | 2^{15} = 10^{4.515} |

5 | 32 | 2^{16} | 2^{31} = 10^{9.33} |

6 | 64 | 2^{32} | 2^{63} = 10^{19.0} |

Suppose, by some luck of the draw, you correctly chose the 32 winners of all the first round games. There would still be over one billion (10

^{9}) possible ways of filling out the remainder of the brackets. Even if everybody in the world filled out a bracket for the full tournament, the overwhelming majority of possible brackets would still be unfilled.

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