We roll a fair die repeatedly until we see the number four appear and then we stop.
What is the probability that we needed an even number of die rolls?
For this problem I said that it would be the Sum of (5/36)*(25/36)^j and I got an answer of 5/9? Does that seem correct.
Thanks
Best Answer
Hint: The probability that it takes exactly $k$ rolls is $$ \frac16\left(\frac56\right)^{k-1} $$ Add those up for even $k$. Your setup looks okay, but $$ \begin{align} \sum_{k=1}^\infty\frac16\left(\frac56\right)^{2k-1} &=\overbrace{\frac{5}{36}}^{\text{first term}}\quad\overbrace{\frac1{1-\frac{25}{36}}}^{\Large\frac1{1-\text{common ratio}}}\\ &=\frac5{11} \end{align} $$
Another Approach: Given that we end either on roll $2n-1$ or on roll $2n$, there are $6$ ways to end on roll $2n-1$: $4$-$1$, $4$-$2$, $4$-$3$, $4$-$4$, $4$-$5$, $4$-$6$; and $5$ ways to end on roll $2n$: $1$-$4$, $2$-$4$, $3$-$4$, $5$-$4$, $6$-$4$. Therefore, no matter what $n$ is, the probability of ending on roll $2n$ would be $$ \frac5{6+5}=\frac5{11} $$