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
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Step-by-Step Solution
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 6×6=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(not A), and B does not throw an 8, P(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 PW be the probability that A wins. We can express this as PW=P(A)+P(not A)P(not B)PW. 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 PW. We perform the arithmetic operations to find the final probability.