The odds of picking the winner of a basketball game blindly (without knowing which team is better, more highly ranked) is 50%
The odds of picking (in advance) winners of a four-team tournament is 12.5%. You first have to guess right in the first two games, so you are at 25% there, and then also have picked in advance the winner of the final game.
Now with 8 teams, you have to pick four winners in the first round correctly. You have 1/16 of a chance to get that right. If you do, then you have a 1/4 chance of picking winners for the next round, and, again, a half and half chance of getting the final game right. So 1/16 x 1/4 x 1/2. I would guess, then, that with every round you add onto the tournament, you multiply the difficulty exponentially by 2 to the power of y, where y is the number of the round you added.
You're right. Assuming N, the number of teams, is a power of two, the number of games played is N - 1. (Three games in the case of four teams, etc.) Assuming a 50% chance of picking each game right, your chance of getting it right overall is 1 / (2 ^ (N - 1)), or 2 / (2 ^ N). And of course the relationship of the number of (complete) rounds to the number of games is exponential already, so pretty soon you're talking about real money.
ReplyDeleteI feel the need to use my junior high school math once in a while.
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ReplyDeleteYour conjecture was already disproven in round 3, when you multiplied by 1/16 (16 is 2^4 not 2^3). I'd guess you multiply each time by 1/[2^(2^(y-1))]...
ReplyDeleteProof:
let y be the round number
this has 2^y teams
hence 2^y / 2 = 2^(y-1) games
so chance of guessing that round entirely correctly is 1/2^(2^(y-1)), which you have to multiply onto the previous chance.
Vance Maverick's error was using the number of teams instead of the round number, as you did.
1 Round: 1/2
2 Rounds: 1/2 * 1/4 = 1/8
3 Rounds: 1/2 * 1/4 * 1/16 = 1/128
4 Rounds: 1/2 * 1/4 * 1/16 * 1/256 = 1/32768
Seems like the multiplier to add a new preceding round is half the chance of guessing all the rounds that come after it.
So that means the chance of guessing n rounds correctly is 1/2^(2^n - 1)
Oh, that's obvious... there are always 2^n - 1 games in n rounds.
Yes. You are right. It is not two to the power of whatever round it is (counting backwardds from the end), but rather a sequence of 2, 4, 16, 256...
ReplyDelete(not 2, 4, 8, 16, 32, 64)...
So it is more exponential than I thought.