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#1
03-16-2005, 04:07 PM
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Bracketology
I was asked in the office today what the probability of getting all the games right in the NCAA tourney would be. Of course, my first thought was to assign a .5 probability to each game and answer with .5^32. However, we know that .5 is too low and .7 might be too big. Of course .7^32 is much much larger than .5^32 so averaging around does not help much. Anyone have an idea of how to best do this? I think the problem is that the nature of bracketology is so conditional and that as the probabilities of success converge to .5 at the end of the brackets, potential upsets along the way can skew the results greatly....
Help? Jason, gaming mouse, anyone??? Indiana 
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#2
03-16-2005, 06:55 PM
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Re: Bracketology
"There are over 1.8•1019 different combinations of NCAA brackets. To put these odds in perspective, you have a much better chance of winning the lottery and getting struck by lightning in the same day, than you have in correctly predicting the entire bracket."
http://www.harvardindependent.com/ne...b-216890.shtml
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#3
03-16-2005, 07:31 PM
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Re: Bracketology
The ease in picking the winner varies according to opponent rankings. In a game where the 1 seed plays the 16 seed... your chances are very good... in the 8vs9 game you have almost 50-50 chance...
Lets make each choice have a chance = .5 + .035*(ranking difference)... max 1 obviously 1vs16 = 1 15-2 = 0.955 14-3 = 0.885 13-4 = 0.815 12-5 = 0.745 11-6 = 0.675 10-7 = 0.605 9-8 = 0.535 There are 4 games for each of these match ups so... Correctly picking the first 32 games = 1^4 * .955^4 * .885^4 * .815^4 * .745^4 * .675^4 * .605^4 * .535^4 = 0.000158013 or about 1 in 6329 Why did I pick .035 you ask? Well... a 1 seed has never lost to a 16 seed (atleast that's what they said on the radio today... dont beat me if Im wrong here) and one of the 5 seeds always loses. notice the 12-5 match up comes in right around .75 You can extrapolate onward, the next rounds will have fewer match ups with large differences in ranking. Worst case: 8v1 = 0.745 7v2 = 0.675 6v3 = 0.605 5v4 = 0.535 Again 4 of each of these games... result = 0.000701903 and onward.... If all the number 1 seeds reach the final four... then the last 3 games would be .5 each This method yields a worst case of getting all 63 games correct = .0000000000124661511948 or 1 in 80217220566 Good Luck
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#4
03-16-2005, 07:39 PM
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Re: Bracketology
what is "1.8•1019 different combinations"....
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Linear Mode
