Slim King
Eternal Order
Orlando
18091 Posts
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Posted: Dec 16, 2005 01:45 pm
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.....
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stanalger
Special user
St. Louis, MO
998 Posts
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Posted: Dec 16, 2005 07:17 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%
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Slim King
Eternal Order
Orlando
18091 Posts
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Posted: Dec 17, 2005 01:21 am
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.....
|
Good to here.