Odds of Perfect Bracket

[Eric]

Has anyone wondered how many possibilities you have for a perfect bracket? I have, and I hurt my brain by looking at these numbers:

Games played:

32 + 16 + 8 + 4 + 2 +1= 63 games played

Since you have 2 choices per game, you put 2 to the 63rd power.

2^63 = 9,223,372,036,854,775,808

That is 9 quintillion, 223 quadrillion, 372 trillion, 36 billion, 854 million, 775 thousand, 808 posibilities. So, uh.......I think I'll get a perfect bracket next year; the odds are on my side!

[Spence]

Has anyone wondered how many possibilities you have for a perfect bracket? I have, and I hurt my brain by looking at these numbers:

Games played:

32 + 16 + 8 + 4 + 2 +1= 63 games played

Since you have 2 choices per game, you put 2 to the 63rd power.

2^63 = 9,223,372,036,854,775,808

That is 9 quintillion, 223 quadrillion, 372 trillion, 36 billion, 854 million, 775 thousand, 808 posibilities. So, uh.......I think I'll get a perfect bracket next year; the odds are on my side!

We really need football to get started.

By the way, Eric, nice run in the NIT. You guys had a good run in the tourney, probably the most successful of any b-10 team.

[openSkies]

Hey Eric,

Run the numbers for predicting the national championship teams for football.

[Eric]

Ay, ay, ay, I can't do everything! Okay, I'll see what I can do.....(meaning I'll just go on ask.com)

[Eric]

Okay, I looked up the odds of picking the correct 2 football teams and, if I did my math correctly, came up with 7,021 possibilities. I forgot that formula, but you'd start with 118 matchups, then add 117 matchups, then add 116 matchups, and so on.

For example:

Team 1 vs. Team 2
Team 1 vs. Team 3
Team 1 vs. Team 4
Team 1 vs. Team 5
Team 1 vs. Team 6
Team 1 vs. Team 7
Team 1 vs. Team 8
Team 1 vs. Team 9
Team 1 vs. Team 10

This totals 9 games.

Then:

Team 2 vs. Team 3
...........................
Team 2 vs. Team 10

This equals 8 matchups. I think this is correct, although knowing my math skills, I always tend to screw up one tiny thing then the whole problem goes awry.

[openSkies]

Where'd you come up with that formula? I think it's flawed...

[Eric]

Look:

Texas vs. USC
Texas vs. Georgia
Texas vs. West Virginia
Texas vs. Ball State
Texas vs. Washington...........

Then you do another column, like "team 2". Since there are 119 teams, and if you selected one team and matched them up with every other team, you'd have 118 games. Then you take another team, but you've already covered ONE game with "team 1". Making it 117 matchups. Then when you get to "team 119" there are no matchups left, so actually it should be 7,020 matchups. Because the final matchup would be "team 118" vs. "team 119".

[openSkies]

I meant the basketball one. But I misread.

My Calculas-loving girlfriend and I are going over the numbers now for the Nat'l Championship match-up. Thought I had it, but I messed up one thing

[Eric]

I knew that, your question, at least the way I interpreted it, was the correct national championship PAIRING. So that's where I get 7,020 possible matchups.

[openSkies]

Chance of guessing Team 1:
1 / 119
0.840%

Chance of guessing Team 2 (out of remainders):
1 / 118
0.847%

Team 1 % x Team 2 %:
0.840 x 0.847

Outcome:
0.007121%

Are we thinking this through correctly?

//

Or more easily:

( 1 / 119 ) ( 1 / 118 ) = 0.007121%

[Eric]

Right, that's the precentage of possibly predicting the correct matchup. My mathematics was the amount of possibilities.

[openSkies]

K, cool. This year by the way...

104 teams are playing 12 games, while 15 are playing 11 games. Shouldn't that be factored into the equation somewhere?

[openSkies]

The computer has predicted the national title match-up three times (1993, 1998 and 2005), and at least one contestant every year except 1994, 1997, and 2001. The team it picked to win the title did so three times (1993, 1998, and 1999), and lost in the title game four times (2000, 2003, 2004, 2005).

3 years correctly guessing, 13 years of predicting. 0.007121% chance of correctly guessing.

3 / 13 = 0.23%

0.23 x 0.007121 = 0.001637943313%

//

I think that number right there show's how darned amazing this CFP thing really is haha.

[Spence]

Your odds would go way up if you used the teams with the highest winning percentage all time.

[openSkies]

Don't make me hit you, Spence

After checking with the government websites, the computer equation used for the CCR 119 has almost EXACTLY the same odds of predicting the title match-up three our of thirteen years as...

Your odds as dying while walking.

1 in 611 : CFP's odds
1 in 612: Your odds of dying on your midnight stroll