don't know the math, but I can google!
don't know the math, but I can google!
from the personal blog of marcus kazmierczak:
The Odds of a Perfect Bracket
It?s March Madness time again which means time to fill out the tournament brackets. So I?m sure we are all wondering what the odds are to pick a perfect bracket, all 64 games picked exactly right.
The straight odds calculations is relatively easy, for each 64 games you must pick the right outcome out of the 2 possible outcomes (win or loss). So our Statisitics 101 class tells us that is two to the power of 64 (2^64) different possible combinations. Only one those will be the perfect bracket.
You have 1 in 18 quintillion chance to pick a perfect bracket
Oh and if you didn?t calculate that, 2^64 = 1.8 x 1019 or more precisely 18,446,744,073,709,551,616. This means you have a 1 in 18 quintillion chance of picking the perfect bracket. You have a better chance of winning the lottery two days in a row then picking the perfect bracket.[1]
However, I don?t buy the straight odds calculation. There has never been a #1 seed beat by a #16 seed, which is 4 games that are practically gimmies. The lowest seed to win the championship is an #8 seed, the lowest seed to make it to the final four is a #11 seed, to the elite eight #12 seed and sweet sixteen is a #14 seed.[2]
So it is relatively obvious that the rankings do give a distinct advantage to be able to pick a perfect bracket.
Here?s how I?m going to divide up the odds for the top seed to win a game in the first round:
#1 vs. #16 = 0.95
#2 vs. #15 = 0.80
#3 vs. #14 = 0.75
#4 vs. #13 = 0.70
#5 vs. #12 = 0.65
#6 vs. #11 = 0.60
#7 vs. #10 = 0.55
#8 vs. #9 = 0.50
Multiplying each of these gives us a 4% chance to pick a perfect round 1 for a regional bracket. There are four regionals so that would be 2.56 x 10^-6 (1 in 390,625 to pick a perfect round 1)
The second round is not quite as easy to give odds to, the #1 seeds would still have a distinct advantage over a #8 or #9, but the odds would be tough for say a #3 vs. #6. Let?s say the favored team for round 2 has a 55% chance of winning. For the 16 round 2 games this would be 0.55^16 = 7.0 x 10-5
Let?s say for the remaining rounds the odds are even, though they wouldn?t be but I want the calculation to be relatively conservative. So the remaining 15 games, actually 16 including the play-in game gives the odds as 0.50^16 = 1.5 x 10-5
Combining all the rounds gives us a total odds of 2.69 x 10-15 which is a 1 in 371 trillion chance, quite a bit easier than the straight odds but still a really really long shot.
So good luck with your picks, don?t feel bad if you miss a couple.
[1] - Assuming a 1 in 30 million chance of winning the lottery; to win the lottery two days in a row would be (3 x 10^6) * (3 x 10^6) = 9 x 10^12, or 1 in 9 trillion chance.
[2] - Seed information from Wikipedia
