

Slim King Eternal Order Orlando 17670 Posts 
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 
One person: 60.33%
Two people: 90.22% 
stanalger Special user St. Louis, MO 996 Posts 
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^(n1)*(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 17670 Posts 
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.....

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