The Magic Cafe Forum Index » » Magical equations » » Math Smaries to the rescue! (0 Likes)

 Slim King Eternal Order Orlando 17672 Posts Posted: Dec 16, 2005 07:45 am    0 Thanks everyone for your help on the last question. The new one is just the same but the participants are allowed to choose any numbers ( 2thru 9), even if it is the same number four times, or any combination. Will the odds change? I know that choosing 3 or 6 twice would be a good thing, right? This allows a more random selection of numbers, but how does it affect my odds? Thanks Dave THE MAN THE SKEPTICS REFUSE TO TEST FOR ONE MILLION DOLLARS.. The Worlds Foremost Authority on Houdini's Life after Death..... stanalger Special user St. Louis, MO 996 Posts Posted: Dec 16, 2005 12:30 pm    0 One person: 60.33% Two people: 90.22% stanalger Special user St. Louis, MO 996 Posts Posted: Dec 16, 2005 01:17 pm    0 For n total digits, each randomly selected from the set {2,3,4,5,6,7,8,9}, the probability of their product being a multiple of nine is (8^n - 5^(n-1)*(5+2*n))/8^n. Here are the probabilities for the first dozen values of n, (given as percents rounded to two decimal places): n=1: 12.50% n=2: 29.69% n=3: 46.29% n=4: 60.33% n=5: 71.39% n=6: 79.73% n=7: 85.84% n=8: 90.22% n=9: 93.31% n=10: 95.45% n=11: 96.93% n=12: 97.94% Slim King Eternal Order Orlando 17672 Posts Posted: Dec 16, 2005 07:21 pm    0 Very Cool!!!!!!! Thanks! This is perfect. Dave THE MAN THE SKEPTICS REFUSE TO TEST FOR ONE MILLION DOLLARS.. The Worlds Foremost Authority on Houdini's Life after Death..... The Magic Cafe Forum Index » » Magical equations » » Math Smaries to the rescue! (0 Likes)
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