# Probability question for ATOT math whizzes

#### LordNoob

##### Senior member
Assuming all things are equal for a moment (i.e. that the chance that any given team will win any given game is 50% for the entire tournament), what are the chances of ONE person correctly predicting the outcome of every single game from the first round to the final of the NCAA tournament.

Please provide mathematical explanations for full credit.

Homework?

#### thesurge

##### Golden Member
1/(2^63)=1/(9,223,372,036,854,775,808)=very small.

#### LordNoob

##### Senior member
Originally posted by: Sheepathon
Homework?

No, just curious.

Also, for any of you ATOTers that have not seen the outside world in weeks due to WOW, the NCAA tournament begins with 64 teams. Obviously that would be 32 games in the first round, 16 in the second, and so on.

#### chuckywang

##### Lifer
Originally posted by: thesurge
1/(2^63)=1/(9,223,372,036,854,775,808)=very small.

Yep, in other words, less than the probability of winning the lottery.....TWICE.

#### hellokeith

##### Golden Member
Assuming games cannot be tied, and there are X number of games, then the blind probability of predicting the winner of every game is 2^X.

In reality, it would be much less difficult due to team statistics, matchups, etc., but would still be difficult. Las Vegas could give you a pretty good idea of the realistic chance of picking all the winners.

#### bersl2

##### Golden Member
Never mind, they are equivalent.

#### thesurge

##### Golden Member
Originally posted by: bersl2
Originally posted by: thesurge
1/(2^63)=1/(9,223,372,036,854,775,808)=very small.

Actually, it could be worse than that. Discounting the play-in game, there are two possible outcomes for each game in the first round. However, there are four possible outcomes for games in the second round---just because you got a game wrong in the first round doesn't eliminate subsequent combinations from being counted.

No, I don't think it works like that. After you account for the probability in the previous round, assuming you just want to predict winners, your current placement in the bracket shouldn't matter. Since, due to the linearity of a NCAA bracket thread, there are "only" 2^63 ways of filling out the bracket (discounting the play-in game?).

#### bersl2

##### Golden Member
Originally posted by: thesurge
Originally posted by: bersl2
Originally posted by: thesurge
1/(2^63)=1/(9,223,372,036,854,775,808)=very small.

Actually, it could be worse than that. Discounting the play-in game, there are two possible outcomes for each game in the first round. However, there are four possible outcomes for games in the second round---just because you got a game wrong in the first round doesn't eliminate subsequent combinations from being counted.

No, I don't think it works like that. After you account for the probability in the previous round, assuming you just want to predict winners, your current placement in the bracket shouldn't matter.

I'm not thinking in terms of probability but rather enumeration. I was thinking that just doing the simple 2^(n-1) for n teams might be counting spurious combinations (i.e., a beats b, but b plays in the next game). But it doesn't, and computing things a round at a time yields the same result.

#### Syringer

##### Lifer
It's 2^(n-1) teams, simple.

Take for example a 4 team tourney

A
B
C
D

There are the 2^(4-1) possibilities, or 8.

1) A>B, C>D, A>C
2) A>B, C>D, C>A
3) A>B, D>C, A>D
4) A>B, D>C, D>A
5) B>A, C>D, B>C
6) B>A, C>D, C>B
7) B>A, D>C, B>D
8) B>A, D>C, D>B

Capiche?