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
Best Answer
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 ${\displaystyle{4547\over8192}\doteq{5\over9}}$.