Geometric conditional Probability

Question

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?

Answer 1

It is a bit difficult to point out where the mistake is because your solution is "too informal". Is the random variable $sum$ the sum of the first three dice or the sum of all dices? Depending on the definition, your probabilities change. Moreover, if $X=1$, then $sum$ might not even be defined if you consider the first definition (the sum of the three dices).

To solve the problem, lets be a bit more formal, but not too much so we do not get lost. Let $(D_i)_{i=1}^\infty$ be an infinite sequence of dice rolls, where $D_i$ is the dice result in the $i$-th roll. We have to define these variables for all possible natural numbers because your game can begin at any natural number. Let $X$ be the index $i$ of the first dice such that $D_i = 6$. That is, $X = i$ if, and only if, $D_i = 6$ but $D_j \neq 6$ for all $j < i$. Therefore, $X = 3$ if, and only if, $D_1 \neq 6, D_2 \neq 6, D_3 = 6$.

Now we define the sum as $S = \sum_{j=1}^X D_j$. That is, we sum all values up to $D_X$. The exercise asks us to compute

$$P(S < 10 | X = 3) \quad.$$

But when $X=3$, we have $S = D_1 + D_2 + 6$. Hence

\begin{align} P(S < 10 | X = 3) &= P(D_1 + D_2 < 4 | D_1 \neq 6, D_2 \neq 6, D_3 = 6)\\ &= P(D_1 + D_2 < 4 | D_1 \neq 6, D_2 \neq 6) = 0.12\quad. \end{align}

I leave to you as an exercise to show that the last probability is $3/25 = 0.12$. Intuitively, think of the probability of the sum of two five-sided dices being smaller or equal to $3$.

Answer 2

For a more brute force approach, consider the following. You have three numbers, you know the last one is 6. There are $6 \cdot 6=36$ possible combinations for the first two numbers (assuming they can be any number from 1 to 6, which your problem seems to state that they cannot). Asking that the sum of these three numbers is $<10$ is equivalent to asking that the sum of the first two numbers is $<4$. Counting it all out you find there are 3 such combinations. Therefore $3/36=1/12$.

Answer 3

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.