Two players take turns rolling a fair die. Player A wins if he rolls a 6, Player B wins if he rolls a 5 or 6. If A starts, find the probability that A wins.

Step 1: Define probabilities for each player

First, we determine the probability of each player winning on their turn. Player A wins if they roll a 6, so the probability is $1/6$. Player B wins if they roll a 5 or a 6, so the probability is $2/6 = 1/3$.

Step 2: Define probabilities of not winning

Next, we calculate the probability that each player does not win on their turn. For Player A, this is $1 - 1/6 = 5/6$. For Player B, this is $1 - 1/3 = 2/3$. These probabilities are crucial for considering turns where neither player wins.

Step 3: Formulate the probability of A winning

Player A can win on their first turn, or on their second turn (if both A and B fail on their first turns), or on their third turn (if both A and B fail on their first two turns), and so on. This forms an infinite geometric series.

Step 4: Write the geometric series

Substituting the probabilities, the series becomes $1/6 + (5/6)(2/3)(1/6) + (5/6)(2/3)(5/6)(2/3)(1/6) + \dots$. This is a geometric series with first term $a = 1/6$ and common ratio $r = (5/6)(2/3)$.

Step 5: Calculate the common ratio

The common ratio $r$ is the probability that neither player wins in a round (A's turn then B's turn). This is calculated as $P(A_{\text{not win}}) \times P(B_{\text{not win}}) = (5/6) \times (2/3) = 10/18 = 5/9$.

Step 6: Apply the sum of an infinite geometric series formula

For an infinite geometric series with $|r| < 1$, the sum is given by $S = a / (1-r)$. Here, $a = 1/6$ and $r = 5/9$.

Step 7: Calculate the final probability

Substitute the values into the formula: $P(A_{\text{wins}}) = (1/6) / (1 - 5/9) = (1/6) / (4/9)$. This simplifies to $(1/6) \times (9/4) = 9/24 = 3/8$.

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