Probability of 2 sevens before 6 evens 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
 A: Distinguish $12$ states $S_{i,k}$ $\ (0\leq i\leq 1, \ 0\leq k\leq 5)$, plus two terminal states $S_{2\, \cdot}\ $, $S_{\cdot\, 6}\ $.  
$S_{i,k}$ denotes the state where so far $i$ sevens and $k$ evens have been thrown. Denote by $P_{i,k}$ the probability that you win when the game is in state $S_{i,k}$. Then
 you have the $12$ equations
$$P_{ik}={1\over6} P_{i+1,k}+{1\over2} P_{i,k+1}+{1\over3} P_{ik}\ ,$$ 
where  $P_{2,k}=1$ and $P_{i,6}=0$. Solving this linear system gives you in particular $P_{0,0}$, the a priori probability that you win the game.
In the end this probability comes out to ${\displaystyle{4547\over8192}\doteq{5\over9}}$.
A: There are three possibilities: a 7 is rolled with probability 6/36, an even is rolled with probability 18/36, and a non-7 odd is rolled with probability 12/36. The third can be ignored, giving rescaled probabilities of 6/24 = 1/4 for a 7 and 18/24 = 3/4 for an even.
You 'win' if you roll a 7 on your n-th roll and your previous n-1 rolls include a seven and at most 5 even numbers.
A: Hint:  You can ignore all odd throws except $7$ and consider that $P(7)=\frac 14, P(even)=\frac 34$.  Then think of having seven throws-you will have either 6 evens or 2 sevens, but not both.  So you win if there are 2 or more sevens among the first seven throws.  It is probably easier to calculate the chance you lose-you need 6 or 7 evens among 7 throws.
