Final answer:
The probability of Player A winning the die rolling game is 6/17. This is calculated by assessing the chance of A winning on the first try and then incorporating the probabilities for subsequent rounds through an infinite series which converges to the calculated value.
Step-by-step explanation:
The student is asking about the probability of Player A winning a sequential game involving a fair six-sided die. To solve this, we'll analyze the different scenarios where Player A can win.
First, Player A's win condition involves rolling a 1 and then rerolling a number that is less than 4 (1, 2, or 3). Since there is a 1/6 chance of rolling a 1 and, upon rerolling, a 1/2 chance of rolling a number less than 4, the probability of A winning on the first try is (1/6) * (1/2) = 1/12.
Player B wins if they roll a 6. If B gets a turn, there is a 1/6 chance B will win. The chance of B rolling is dependent on A not winning, so it is 1 - 1/12 = 11/12. Thus, the probability that B rolls and wins is (11/12) * (1/6) = 11/72.
To account for multiple rounds, we must consider that the game can go on indefinitely. Let's denote 'P(A)' as the probability that A wins overall. Then, after both players fail to win in the first round, the game essentially restarts. So, 'P(A)' also represents the probability of A winning in any subsequent round. The probability of the game continuing after the first round is the probability of A not winning times the probability of B not winning, which is (11/12 * 5/6) = 55/72. Therefore, we have the following equation:
P(A) = Probability A wins on the first try + Probability the game continues and A wins in a subsequent round
P(A) = (1/12) + (55/72) * P(A)
Solving for P(A) yields:
P(A) = (1/12) / (1 - 55/72) = (1/12) / (17/72) = 6/17
So, the probability Player A wins is 6/17.