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

One person: 60.33%
Two people: 90.22%

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%

Very Cool!!!!!!!
Thanks!
This is perfect.
Dave