I read the odds at picking every game correctly in March Madness is 1in15quintillion something. The formula?
Does anyone know the formula for picking every game correctly in the March Madness basketball tournament? I read online that it was 1 in 15.something quintillion. (You have better odds at winning the lottery, megamillion or powerball.)
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June 30th, 2011 at 5:53 am
There are 2^63 possible brackets, since there are 63 games in the tournament (discounting the play-in game) and for each game you could pick either Team A or Team B out of the two that meet for that particular game. So many people say that your chances of getting a perfect bracket is 1 in 2^63 (which is around 9 quintillion).
However, this is a bit simplistic and assumes that all 2^63 brackets are equally likely to occur, which is obviously not true! For all practical purposes we could assume that all four #1 seeds make it to the second round, which cuts out over 90% of the brackets as impossible… although there are still 2^59 brackets that have all 4 #1 seeds winning their first game, and that’s still an astronomical number!
But then again, for most games you figure there’s a favorite that should be more likely to win. Let’s say that for each game one team is a favorite and has a 65% chance of winning that particular game and you pick the favorite. Then your chances of getting a perfect bracket is 0.65^63, or 1 in about 612 billion. Not good, and still a lot worse then winning Powerball, but still a lot better then 1 in 9 quintillion.
June 30th, 2011 at 5:53 am
i think 64! which is 64x63x62x61x60x59x58x57x56x54x53x52………….
June 30th, 2011 at 5:53 am
25%