A game is played where a standard six sided dice is rolled until a $6$ is rolled, and the sum of all of the rolls up to and including the $6$ is taken. What is the probability that this sum is even?
I know that this is a geometric distribution and the expected number of rolls is $\frac1{1/6} = 6$ rolls until a $6$ occurs, along with the expected value being $21$ ($6$ rolls times expected value of $3.5$ per roll), but I'm not certain how to proceed from there. Would the expected range (if it is even relevant) be from $11 = 1·5+6$ to $31 = 5·5+6$? The answer is supposedly $\frac47$. I'm also curious about how this question would change if the stopping number was anything else, say a $3$ stopping the sequence rather than a $6$. Thank you in advance!
Best Answer
Let $p$ be the desired probability, and consider the first roll. It is either a $6$, in which case we're done and the sum is even, a $2$ or $4$, in which case we want the sum of the rest of the terms to be even, or a $1,3$, or $5$, in which case we want the sum of the rest to be odd.
Thus $$p = \frac{1}{6}+ \frac{1}{3}p+\frac{1}{2}(1-p)$$ which simplifies to $p=\frac{4}{7}$.