Problem: What's the probability that in 100 rolls of a 6-sided die, the side number 1 will come up exactly 10 times?
There are 100C10 (read, "100 choose 10") ways of getting 10 one's out of a total of 6100 different ways to roll the die.
For each way of getting 10 one's, there are 590 different outcomes involving the other rolls (2, 3, 4, 5 and 6) of the die.
That's P(10) = 100C10(590)/6100
≈ (1.73 E13)(8.08 E62)/(6.53 E77)
≈ 2.14 E−2 ≈ 0.0214
or about 1 in every 46.7 trials
In the general case, what's the probability that in n rolls of a m-sided die, the side number 1 will come up exactly k times (k ≤ n)?
Out of a total of mn different ways to roll an m-die n times, there are nCk ways of getting k one's.
For each such combination, there are (m−1)n−k permutations (patterns with order taken into account) involving the other rolls (2, 3, ..., m−1, m) of the die. So we have
P(k) = nCk(m−1)n−k/mn
That is, the number of combinations of n things k at a time, times the number of ways to get any one of the m−1 negative outcomes (outcomes other than one) in the remaining n−k places, divided by the total number of ways to throw the die.
The above formula is also given equivalently as
P(k) = nCk(1/m)k[(m−1)/m]n−k
That is, the number of combinations of n things, k at a time, times the probability of k consecutive positive outcomes, times the probability of n−k consecutive negative outcomes.