I am interested in finding the probabilities of hands being dealt after a given number players have folded. For each player, I randomly select a hand. If it's within his folding range, I remove the two cards from the deck. If not, repeat. Repeat this procedure until n players have folded, leaving me with a deck with 52-2*n cards remaining. With the stub deck, I deal out all remaining two-card combinations and increment the appropriate counter depending on hand type (ie 55, A9o, T2s, etc). Then, begin new trial. Note that each single trial after (for example) 8 players fold generates 630 possible hands for the 9th player.

After n trials, I can find the expected probability of each hand type by dividing its counter by the sum of all counters.

How many trials do I need to run to be reasonably confident (95%?) that the resulting probability is a "good" estimate of the population probability? Say, accurate to 6 decimal places? Does the number of trials needed depend on n the number of players folding?

There are 3 basic hand types – pairs, non-pair suited, non-pair off-suit with probabilities of 6/1326, 4/1326 and 12/1326 respectively.

Say you want your estimate to be within E of the true value with 95% confidence. For example, E might be 0.001. Then an approximate formula for sample size is

N= 4*(PQ)/E^2, where Q= 1-P.

Now which P you use will depend on which one is most important. To be conservative choose the largest P, which will be 12/1326. The real determinant is what value you use for E. For example, if you chose E=0.001, you need about 36,000 samples which should be quite doable. For E=0.0001, you need 3.6million; that will take a little longer!!!

This formula assumes that the hand probabilities are binomially distributed – not true because of the dependencies but close enough I think for this purpose.