A fair die is rolled repeatedly. Let X be the number of rolls needed to obtain a 5 and Y is the number of rolls needed to obtain a $6$. Calculate the Expected value of X given Y = 2.

I have read few posts on this. However, I still am not able to understand the methods used in those posts to get to the final answer. This is how I approached the problem after getting some ideas from some discussions about the topic.

> 
> Let P(X) be the conditional distribution of X given Y=2  
> 
>  P(1) = 5 on first roll; 6 on second =(($\frac 16$^2))  
> 
>  P(2) = 5 on second roll(impossible since Y=2) =0  
> 
>  P(3) = no 5 or 6 on the first roll; 6 on the second roll; 5 on third roll
>  =($\frac 23$ )($\frac 16$ ^2)  
> 
>  P(4) = no 5 or 6 on first roll; 6 on second roll; no 5 on third roll; 5 on the fourth roll  
> 
>  =$\frac 23$ ($\frac 16$ ^2)($\frac 56$ )  
> 
>   
> 
>  E[X given Y=2]  
>  
>  =1(($\frac 16$ ^2)) + 2(0) +3[($\frac 23$) ($\frac 16$ ^2)] + 4[($\frac 23$ )(($\frac 16$ )^2)($\frac 56$ )] +5[($\frac 23$ )(($\frac 16$ )^2)(($\frac 56$ ^2)] + ........../P(Y=2)  
> 
>   
> 
>  =(($\frac 16$ )^2)[1 + 3(($\frac 23$ ) +4(2/3)(5/6) + 5(($\frac 23$ )((5/6)^2) + ........../P(Y=2)]  
> 
>  =(($\frac 16$)^2)[1+ (($\frac 23$ )[3+4(($\frac 56$ ) +5((($\frac 56$ )^2) +............../P(Y=2)]  
> 
>   
> 
> 
> 
> 

Okay. So, at this point, I have an infinite geometric series where a=3 and r=5/6. The sum of this infinite series would be a/(1-r) which is 18. However, the correct sum should be 48 and the rest of the solution goes like this:

> 
> =(($\frac 16$)^2[1+ (($\frac 23$ )[48]] / (($\frac 5/36)  
> 
>  =(33/36)/(5/36) = 33/5 = 6.6
> 
> 
> 

Can someone explain to me the final steps? And if there is a better alternative to this solution, could you please share it to me? Thanks. Help would be highly appreciated!