What is the probability that the first 1 comes in even trials of a six faced fair dice? A six faced fair dice is thrown until 1 comes, 
then what is the probability that 1 comes in even no. of trials?
I tried my best and my efforts include:
I find the
probability of getting a 1 on 2nd throw $p(2) = (5/6) (1/6)$
probability of getting a 1 on 4th throw $p(4) = (5/6)^3 (1/6)$
probability of getting a 1 on 6th throw $p(6) = (5/6)^5 (1/6)$
then I combine them like $p =  p(2)+p(4)+p(6)+ \ldots$ 
now, which formula should I apply to solve this problem?
 A: Let $p$ be the probability that $1$ comes in even number of trials.
$p=P(\textrm{first }1 \textrm{ comes in the second trial})+P(\textrm{first }1 \textrm{ comes in the $2n$th trial, where $n>1$})$
$p=\frac56\times\frac16+(\frac56)^2p$
$p=\frac5{11}$
A: It is an infinite sum of the following:


*

*
*

*is not-1, but the second throw is 1


*1-3. are not-1, but the fourth is 1

*1-5. are not-1, but the sixth is 1

*1-7 are not-1, but the eighth is 1

*...and so on.


So, the probability is
$$\sum_{k=0}^{\infty}(\frac{5}{6})^{2k+1}\frac{1}{6}$$
In closed form, using the identity $\frac{1}{1-x}=1+x+x^2+x^3+...$:
$$\sum_{k=0}^{\infty}(\frac{25}{36})^{k}\frac{5}{6}\frac{1}{6}=
=\frac{5}{36}\frac{1}{1-\frac{25}{36}}=
=\frac{5}{36}\frac{36}{11}=
\underline{\underline{\frac{5}{11}}}$$
A: Observe that if we define $\Pr(2k)$ to be the probability that the first 1 appears on the $2k$th throw, then 
$$\Pr(2k) = \left(\frac{5}{6}\right)^{2k - 1}\left(\frac{1}{6}\right)$$
Hence, the probability that the first one appears on an even-numbered throw is
$$\sum_{k = 1}^{\infty} \left(\frac{5}{6}\right)^{2k - 1}\left(\frac{1}{6}\right)$$
By extracting a factor of $\left(\frac{1}{6}\right)\left(\frac{5}{6}\right)$, we can obtain a convergent geometric series.
\begin{align*}
\sum_{k = 1}^{\infty} \left(\frac{5}{6}\right)^{2k - 1}\left(\frac{1}{6}\right) & = \frac{1}{6} \sum_{k = 1}^{\infty} \left(\frac{5}{6}\right)^{2k - 1}\\
& = \left(\frac{1}{6}\right)\left(\frac{5}{6}\right)\sum_{k = 1}^{\infty} \left(\frac{5}{6}\right)^{2k - 2}\\
& = \frac{5}{36} \sum_{k = 1}^{\infty} \left[\left(\frac{5}{6}\right)^2\right]^{k -1}\\
& = \frac{5}{36} \sum_{k = 1}^{\infty} \left(\frac{25}{36}\right)^{k - 1}\\
& = \frac{5}{36} \cdot \frac{1}{1 - \frac{25}{36}} & \text{geometric series with $r = \frac{25}{36}$}\\
& = \frac{5}{36} \cdot \frac{1}{\frac{11}{36}}\\
& = \frac{5}{36} \cdot \frac{36}{11}\\
& = \frac{5}{11}
\end{align*}
A: What's the difference between rolling two times and getting a 1 in neither, and rolling no times?
There isn't one! So getting a result like this is the same as doing nothing.  So we don't even count these scenarios as a thing that happens at all.
So: out of $36$ original scenarios (odd die roll, then even die roll, so $6\times6$), only $11$ count: the $6$ where the odd die roll is $1$, and the $5$ where the even die roll is $1$ and the odd die roll isn't.  So, the probability that the game ends on an odd roll is $\frac{6}{11}$ and the probability that the game ends on an even roll is $\frac{5}{11}$.
