<s>[INST] Conditional Probability with dice 
Two dice are rolled and their number is added. If the sum equal 7, the game is lost. If the sum equal 9, the game is won. If the sum is any other number, the dice are rolled again, until the sum equals 7 or 9. What is the probability that someone wins 1 out of 5 games?

My approach:
Sum of 9: $2 \times (6,3) + 2 \times (5,4) \implies p(9) = \frac{1}{9}$
Sum of 7: $2 \times (6,1) + 2 \times (5,2) + 2 \times (4,3) \implies p(7) = \frac{1}{6}$
$$P(x=1) = {5 \choose 1} \bigg(\frac{1}{9}\bigg)^1 \bigg(\frac{1}{6}\bigg)^{4} = 4.29 \times 10^{-4}$$
 [/INST] Let us consider $(a, b)$ a pair of values for the two dice, with $a$ the value for die 1 and $b$ the value for die 2. There are six equiprobable cases in which you lose:
$$(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)$$
Similarly, there are four equiprobable cases in which you win:
$$(3, 6), (4, 5), (5, 4), (6, 3)$$
All other combinations result in a new dice roll, and thus need not be considered. As such, the probability of winning a game equals $\frac{4}{10} = 0.4$. Based on the binomial distribution, the probability of winning exactly 1 out of 5 games now equals:
$${5 \choose 1} \bigg(\frac{4}{10}\bigg)^1 \bigg(\frac{6}{10}\bigg)^4 = 0.2592$$
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