Comments on: Odds of a perfect NCAA bracket

By: michael

michael — Mon, 21 Apr 2008 03:33:51 +0000

in response to clay, I do not think that the calculation you proposed is valid. Even though half the field gets eliminated every round, that does not change the fact that you have still have a 1 in 2 chance of picking every game, and you must do it 63 times in a row. I have no clue why you would add every round the way you did, that makes absolutely no sense to me.

I believe it will be a very long time before anyone gets a perfect bracket. Not only are the odds astronomical, you have to take into account the fact that out of the millions of brackets, most people take fairly obvious picks, which are likely to be the same as other peoples lowering the chances of hitting. Also, the people crazy enough to try and call the cinderellas (Davidson or George Mason) often are crazy enough to pick other huge upsets that don’t happen. It may happen eventually, but i haven’t found any legitimate cases of a perfect bracket.

By: corey

corey — Sun, 30 Mar 2008 03:29:43 +0000

has anyone ever gotten it all right

By: kyle

kyle — Sat, 29 Mar 2008 19:47:34 +0000

well the 1’s always win and the 2’s and 3’s almost always win so you can just pick those and change all the other ones. then the odd go way up. they still aren’t good but they improve significantly.

By: Clay

Clay — Fri, 28 Mar 2008 18:06:33 +0000

Actually, the 18 quintillion number doesn’t take in to account that half of the field gets eliminated every round…so your calc really should be.
2^32 + 2^16 + 2^8 + 2^4 + 2^1= 4,295,033,106….still a large number but 4 billion is better than 18 quintillion.

In the championship game, there can be one of two winners, team a or team b. so you get 2^1. in the round before, there are 4 possible out comes. Team a beats team d, team d beats team a, team b beats team c, or team c beats team b. So we get 2^2. double this to move back a bracket for 2^4, then 2^8, 2^16, then 2^32. Add the sum of these squares and you get 4,295,033,110.

By: SPENCIE

SPENCIE — Wed, 26 Mar 2008 19:37:17 +0000

well actually the correct number by the mathematical equation:
B = (2A*9) k_i/64^2
where A = the locations of the final four
where K = number of rounds ever played in this NCAA tournament
where i = the imaginary number which is equivalent to a/-1
where _ = the exact odds of getting a perfect bracket
(the star is where you insert your middle name)
The answer therefore is q^-5^10
which equals 1:4096

By: matt

matt — Sun, 23 Mar 2008 19:07:32 +0000

I have had 5 perfect brackets in the past 5 years. I fill out 18.5 quintillion each year. My hand is tired…. From winning.

By: scott

scott — Sat, 22 Mar 2008 21:37:03 +0000

There has never been a perfect bracket and there never will be. The odds are astronomical. Even if you asssign weights to each game like a 1 seed has a 98% chance of winning the first game. The 1 seed is a 85% chance of winning thr first game. The odds of picking a perfect bracket is way out there I canot fathom it occurring.

By: Jake

Jake — Sat, 22 Mar 2008 03:24:46 +0000

i got a perfect bracket before.

not really, but it would be cool if i did

By: Brian Flick

Brian Flick — Thu, 20 Mar 2008 03:58:25 +0000

i love bowling!!!!

ill be in the press for making all conference bowling thanks you..!!! h
and for this bracket… calm down brothers and sisters… its just a bracket:)

Peace and Love

By: Jack Upton

Jack Upton — Thu, 20 Mar 2008 02:23:51 +0000

what about odds of getting all picks incorrect? we are thinking 2^32 power, since you only need to get all games in the first round wrong…much easier!