To determine the probability that A wins the game, we need to calculate the probability of A rolling a 6 before B does.

We can start by analyzing the possible outcomes of A's rolls. A has 6 possible outcomes when rolling the die: 1, 2, 3, 4, 5, or 6.

Let's consider the probability of A winning on his first roll. The probability of rolling a 6 on the first roll is 1/6.

If A doesn't roll a 6 on his first roll, the game continues to the next round, and now B has a chance to roll the die. In this scenario, the probability of A winning on his second roll depends on B not rolling a 6 on his first roll. The probability of B not rolling a 6 is 5/6, as there are 5 outcomes (1, 2, 3, 4, or 5) that are not a 6.

If B doesn't roll a 6 on his first roll, the game continues to the next round, and the probability of A winning on his third roll depends on B not rolling a 6 on his second roll. Again, the probability of B not rolling a 6 is 5/6.

We can continue this pattern for each subsequent round, where the probability of A winning on his nth roll is (5/6)^(n-1), as B needs to not roll a 6 in each previous round.

Now, to calculate the overall probability of A winning, we need to take into account the probabilities of A winning on each round. We sum up these probabilities for each round until infinity, as there is no limit on the number of rounds. This can be represented mathematically as follows:

P(A wins) = (1/6) + (5/6)(1/6) + (5/6)^2(1/6) + ...

This is a geometric series with a common ratio of 5/6. We can use the formula for the sum of an infinite geometric series to calculate the probability:

P(A wins) = (1/6) / (1 - 5/6) = (1/6) / (1/6) = 1/6

Therefore, the probability that A wins the game is 1/6, which corresponds to option 'A'.