This question was asked on 610 AM today. Here is my theory: Look at each round is a binary number, it's length determined by the number of games played. Final round, 1 game 0 or 1 2^1 = 2 Final four, 2 games 00, 01, 10, 11 2^2 = 4 Elite Eight 4 games 0000, 0001, etc 2^4 = 16 Sweet Sixteen 8 games 00000000, 00000001, etc 2^8 = 256 Second Round 16 games 0000000000000000, 0000000000000001, etc 2^16 = 65,536 First Round 32 games 00000000000000000000000000000000 2^32 = 4,294,967,296 If it was just a 4 team tourney it would have 2^1 * 2^2 = 8 possible outcomes 8 teams 2^1 * 2^2 * 2^4 = 128 16 teams 2^1 * 2^2 * 2^4 * 2^8 = 32,768 32 teams 2 * 4 * 16 * 256 * 65,536 = 2,147,483,648 64 teams 2*4*16*256*65,536*4,294,967,296 = ***9,223,372,036,854,775,808*** This is the magic number. Is my math right? I've sent this into 610 and they haven't read my answer, but I'm pretty sure that number is the total number of possibilities. B
Yeah, it's 2^63 because 2^1*2^2*2^4*2^8*2^16*2^32=2^63 When you multiply different powers, you add the powers together. B
Since 610 doesn't seem to believe the MIT guy... 1 game (Final round) A vs B, A or B wins - 2^1 = 2 2 games (Final four) A vs B, C vs D - 2^1*2^2 = 8 A vs C, A wins A vs C, C wins A vs D, A wins A vs D, D wins B vs C, B wins B vs C, C wins B vs D, B wins B vs D, D wins 4 games (Elite Eight) A vs B, C vs D, E vs F, G vs H 2^1 * 2^2 * 2^4 = 128 8 games (Sweet sixteen) 2^1 * 2^2 * 2^4 * 2^8 = 32,768 16 games (Second round) 2^1 * 2^2 * 2^4 * 2^8 * 2^16 = 2,147,483,648 32 games (First round) 2^1 * 2^2 * 2^8 * 2^16 * 2^32 = 2^63 2^63 = 9,223,372,036,854,775,808 B
There's the playin game, making the total number of games 64 and thus the total number of outcomes 2^64.
I found out I actually made it on the air last night. Does anyone know it my reference to the Clutchcity BBS made it on the air? B
