Fair die: Probability of rolling $2$ before rolling $3$ or $5$ Independent trials consisting of rolling a fair die are performed, what is the probability that $2$ appears before $3$ or $5?$
There are $36$ cases if we take two trials like $11 12 13 14 15 16 
..21 22 23 24 25 26..31 32 33 34 35 36$ like this . But two has occurred before , so total $6$ cases , favourable just two$(23 25)$ so ${2\over 6} ={1\over 3}$
what is wrong in this approach , answer given is $3\over 8$ .

Hint: The process is a renewal process since it starts over after each roll (or terminates). So, let $X_1$ denote the result of the first draw and $p$ the required probability. Then you have that $$p=P(X_1=2)+P(X_1=1,4 \text{ or }6)p+P(X_1=3 \text{ or }5)\cdot0$$ 
(can you see why?). 
