I have a difficult probability question to resolve.
Say you have 2 chances to roll a dice. If you roll a 6, you're awarded 2 additional rolls. You can receive infinite number of additional 2 rolls every time you roll a 6.
Let X be the average number of rolls. The chance of rolling a 6 is 1/6 = 16.67%. Then to calculate X: (my calculation, it may be wrong)
X = 2*(1-16.67%)^2 + 4*(1-16.67%)^3*(16.67%) + 6*(1-16.67%)^4*(16.67%)^2 + 8*(1-16.67%)^5*(16.67%)^3 + ...
How do you determine the exact value of X?
Many thanks in advance for the answer!
If $n$ is the number of rolls left before a given roll, after the roll, there are either $n-1$ or $n+1$ rolls left, with respective probabilities $p=\frac56$ and $1-p=\frac16$. One asks for the mean number of rolls $X_2$ starting from $2$ rolls left.
For every $n$, the mean number of rolls $X_n$ starting from $n$ rolls left is $n$ times the mean number of rolls $X_1$ starting from $1$ roll left. Furthermore, the usual one-step analysis starting from $1$ roll left yields $$X_1=1+(1-p)X_2+pX_0,$$ that is, $X_1=1/(2p-1)$. Finally, $X_2=2X_1=2/(2p-1)=3$.