Determing the number of possible March Madness brackets

Is there a simple combinatorial explanation to derive the total number of march madness brackets?

Would it be $2*(2^{16}*2^{8}*2^{4}*2^{2}*2)^{2}$ where the final squared takes into account both halves of the bracket?

• What on earth are march madness brackets? – user21820 Jan 24 '14 at 3:02
• How many teams can possibly be in each spot of the bracket? I don't think you've accounted for that. Once the spots in the bracket are assigned, since there are 64 teams there are 63 games, hence $2^{63}$ possible ways for the bracket to unfold. – Jeff Snider Jan 24 '14 at 3:21
• It is a basketball tournament. 64 teams compete for the NCAA championship. – lord12 Jan 24 '14 at 3:21
• @Jeff I have now. I used a "binomial tree" approach. I feel this is definitely more intuitive. From the 32 teams that start, there are $2^{16}$ outcomes, then $2^{8}$ outcomes, then $2^{4}$ then $2^{2}$ then $2$. I square that quantity to take into account the other half. Then I multiply by 2 since there are two possible teams that can win the championship. – lord12 Jan 24 '14 at 3:23
• What I mean by "each spot of the bracket" is which teams make it into the bracket in the first place, and what are the match-ups. For me the "there are 63 games so $2^{63}$ ways for the bracket to unfold" is more intuitive (and easier to read). – Jeff Snider Jan 24 '14 at 3:35

Note that at the end of the tournament, $63$ teams have been eliminated. Since one team is eliminated per game, there are $63$ games. Since each game has two possible outcomes, and the number of possible outcomes given a single game result is independent of that result, we see there are $2^{63}$ possible tournaments.

Since I don't know anything about NCAA's March Madness, I'll assume that you're talking about a 64-team single-elimination tournament.

In the first round, there are $32$ games, each with $2$ possible outcomes. This provides for a total of $2^{32}$ possibilities.

In the next round, there are only $16$ games, again each with $2$ possible outcomes, for a total of $2^{16}$ possibilities.

You keep going, eliminating half of the players each round and taking half as long as the previous round to play the games out, until the finals, at which point there is $1$ game with $2$ possible outcomes.

This is a total of $2^{32} \times 2^{16} \times 2^8 \times 2^4 \times 2^2 \times 2^1$ possible outcomes.

Your answer of $2*(2^{16}*2^{8}*2^{4}*2^{2}*2)^{2}$ is equivalent to this, which means that it's correct. It's a rather roundabout way of coming to the same answer (you don't need to compare each bracket separately), but they are equivalent.

• It's true that any 64-team single-elimination game will have exactly 63 games and therefore $2^{63}$ outcomes; however, the way those 63 are calculated can be slightly different depending on the way the brackets are organized. – Joe Z. Jan 28 '14 at 8:17