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Monday, March 25, 2013


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.


Vance Maverick said...

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.

Jonathan said...

I feel the need to use my junior high school math once in a while.

Jamesio said...
This comment has been removed by the author.
James said...

Your 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))]...


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.

Jonathan said...

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...

(not 2, 4, 8, 16, 32, 64)...

So it is more exponential than I thought.