In successive rolls of a pair of fair dice, what is the probability of getting 2 sevens before 6 even numbers?
Assume that on the nth roll, the game ends. So this means we roll a seven on the nth round and in the $n-1$ rounds, we have only 1 seven and less than 6 rolls of an even.
So P(1 seven in $n-1$ rounds) = $ { n-1 \choose 1}\frac{1}{6} (1-\frac{1}{6})^{n-1-1}$
P(<6 evens and 1 seven) = P(0 even and 1 seven) + P(1 even and 1 seven) + ...+ P(5 evens and 1 seven) = $ \sum_{n=1,\,0\leq i \leq 5}^{\infty} {n-1 \choose 1}\frac{1}{6} (1-\frac{1}{6})^{n-1-1} {n-2 \choose i} (\frac{1}{2})^i (1-\frac{1}{2})^{n-2-i}$ Can anyone give me some pointers - I know this is incorrect as it is since if the summation starts at n=1, then the first term is undefined