Geometric conditional Probability In order to start a game, each player takes turns throwing a fair six-sided dice until a $6$ is obtained. Let $X$ be the number of turns a player takes to start the game. Given that $X=3$, find the probability that the total score on all three of the dice is less than $10$.
So initially I thought this was straight forward. Total number of outcomes that three dice can sum up to less than $10$ is equal to $60$. Total amount of outcomes is $6^3=216$.
Therefore probability of the total summing to less than $10$ is $$P(\text{sum} < 10)=\frac{60}{6^3}=\frac{5}{18}$$.
However, this did not match the provided solution of $1/12$ and so I thought it had something to do with conditional probability. i.e. $P(\text{sum} < 10 | X = 3)$.
So naturally, $$P(\text{sum} < 10 | X = 3)=\frac{P(X = 3 | \text{sum} < 10) P(\text{sum} < 10)}{P(X=3)}$$
Out of the $60$ outcomes that sum less than $10$, only $6$ of those outcomes contain sixes. So that when we have to calculate the probability of taking three turns to get one $6$ we have to fail twice and succeed once and hence$$P(X = 3 | \text{sum} < 10) = \left(\frac{9}{10}\right)^2\left(\frac{1}{10}\right)=\frac{81}{1000}$$
$P(X=3)$ is simply $\left(\frac{5}{6}\right)^2\left(\frac{1}{6}\right)=\frac{25}{216}$ and so finally
$$P(\text{sum} < 10 | X = 3)=\frac{P(X = 3 | \text{sum} < 10) P(\text{sum} < 10)}{P(X=3)}=\frac{\frac{81}{1000} \frac{5}{18}}{\frac{25}{216}}=\frac{243}{1250}$$
Where did I go wrong? Alternatively, maybe the textbook is incorrect?

When we look at the sum of dice thrown, there's no importance if we got $(1,2,3),(1,4,1),(3,1,2)$ or any other permutation that gives us 6. This is called unordered sampling with replacement. "Unordered" because the order of results doesn't matter; "with replacement" because a result can occur more than once.
The number of possible outcomes for which the sum of $n$ positive variables is $k$ is given by $\binom{n+k-1}{k}$. We have some more limitations, so a more comlex formula could be found here.
In any case, there's 1 possible rolls for sum of 3, 3 possible rolls for sum of 4, 6 possible rolls for sum of 5 and so on, totalling 81 possible rolls with sum up to 9. That probability is $\frac{81}{216}=\frac{3}{8}=0.375$, which differs from your result.
