Outcome probabilities for set number of dice rolls with conditional extra rolls If we are allowed $n$ rolls of a dice, where each roll of 1 gives us an extra roll, what is the probability of rolling m 1s in the sequence of available rolls, and likewise what is the probability of rolling m 2s (say)?
(Title and question edited to remove trivial recurrence relation)
 A: Without extra rolls, the probability of getting $m$ of any given result out of $n$ rolls is given by the Binomial distribution (http://en.wikipedia.org/wiki/Binomial_distribution):  in this case, it is $\binom{n}{m}p^m(1-p)^{n-m}$, where $p=\frac{1}{6}$ is the probability of rolling a particular number on the die.
Now suppose you begin with $n$ rolls, and whenever you roll a 1, you get an extra roll.  With probability $1$ you will eventually stop rolling your dice, and there will be $n$ total rolls which are not 1s. Any of the other values is equally likely for each such roll, so the probability that the number of 2s (3s, etc) is $m$ is given by $\binom{n}{m}q^m(1-q)^{n-m}$, where $q = \frac{1}{5}$.
What about the number of 1s rolled?  This is given by the Negative Binomial distribution (http://en.wikipedia.org/wiki/Negative_binomial_distribution), where a success is rolling a 1, a failure is any other roll, and we want the number of successes before $n$ failures.  Hence the probability that $m$ 1s are rolled is $\binom{m+n-1}{m}(1-p)^np^m$, where $p = \frac{1}{6}$ is the probability of rolling a 1 on any given roll.
