What is the probability of obtaining a sum of $10$ in two consecutive throws of two dice in the first $8$ throws? We are playing with two dice until we get $10$ as the sum of the results two times in a row.
(i) What is the probability of having gotten this in $8$ throws?
(ii) What's the probability of having thrown a sum less than $10$ exactly $8$ times before we stopped?
My attempt:
i) The elements of ordered pairs that add up to $10$ are $(4,6), (6,4)$ and $(5,5)$ that appear on the $7$th and $8$th throw and therefore its probability is $3/36$. I don't know how much I correctly tackled the problem. 
The second one a little bit confusing. Any help please. 
 A: (i) What is the probability of having gotten this in 8 throws?
Solution: You didn't consider the all possible events.
In order to calculate the required probability, you have to consider the throws before $7th$ for not getting the sum of $10$ in consecutive throws. 
since you've got the sum in $7th$ & $8th$ throw, therefore in the $6th$ throw you have a constraint of not getting a sum $10$ but all from $1$ to $5$ have a possibility of getting a sum of $10$ (not consecutively).
let $p$ = prob of getting a sum $10$.
$p$ = $3/36$
and  $\bar p$ = prob of $NOT$ geeting a sum $10$.  
$\bar p$ = $33/36$
$p+\bar p= 1$
now there are two possibilities: 
(i) $1,~~~3,~~~5~~$ has an option of getting a sum or not but $2,4$ are constrained of not getting a sum 10.
$\color{red}{1} ~~~2 ~~~\color{red}{3}~~~4~~~\color{red}{5}~~~6~~~7~~~8$
probability = $(p+\bar p)*\bar p*(p+\bar p)*\bar p*(p+\bar p)*\bar p*p*p = p^2 \bar p^3$ 
(ii) $~~1~~~\color{red}{2}~~~3~~~\color{red}{4}~~~5~~~6~~~7~~~8$
probability= $\bar p*(p+\bar p)*\bar p*(p+\bar p)*\bar p*\bar p*p*p = p^2 \bar p^4$
therefore required probability = $p^2 \bar p^3 +p^2 \bar p^4 =  ~p^2 \bar p^3 (1+\bar p) $
PS: long but easy to understand :) .
