The probability that the first 1 comes in even trials of a six faced fair dice

Question

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?

Answer 1

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 $n>1$})$

$p=\frac56\times\frac16+(\frac56)^2p$

$p=\frac5{11}$