$A$ and $B$ alternately throw a pair of dice. $A$ wins if he throws a sum of 5 before $B$ throws a sum of 8, and $B$ wins if he throws a sum of 8 before $A$ throws a sum of 5. The probability, that $A$ wins if $A$ makes the first throw, is

Step 1: Calculate probabilities of sums

First, we need to determine the probability of rolling a sum of 5 and a sum of 8 with a pair of dice. The total number of outcomes when rolling two dice is $6 \times 6 = 36$. The combinations that sum to 5 are (1,4), (2,3), (3,2), (4,1), so there are 4 ways. The combinations that sum to 8 are (2,6), (3,5), (4,4), (5,3), (6,2), so there are 5 ways.

Step 2: Define probabilities of success and failure

Let $P(A)$ be the probability that A throws a sum of 5, and $P(B)$ be the probability that B throws a sum of 8. We also need the probabilities that A does not throw a 5, $P(\text{not } A)$, and B does not throw an 8, $P(\text{not } B)$.

Step 3: Set up the equation for A winning

A wins if he throws a 5 on his first turn, OR if he doesn't throw a 5, B doesn't throw an 8, and then A throws a 5 on his second turn, and so on. This forms an infinite geometric series.

Step 4: Simplify the equation

Let $P_W$ be the probability that A wins. We can express this as $P_W = P(A) + P(\text{not } A)P(\text{not } B)P_W$. This is because if A doesn't win on the first turn and B doesn't win on their first turn, the game effectively restarts from A's turn with the same probability of A winning.

Step 5: Solve for P(A wins)

Now, we substitute the calculated probabilities into the simplified equation and solve for $P_W$. We perform the arithmetic operations to find the final probability.

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