Solution

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To find the probability that A wins, we can approach the problem by considering the probability of each player winning on their turn.

Let \( p \) be the probability that A wins

There are two cases :

• If B rolls a 6 on his first turn (which happens with probability \( \frac{1}{6} \) ), A doesn't get a chance to roll and thus can't win. So, the probability of A winning in this case is 0.
  


• If B doesn't roll a 6 on his first turn (which happens with probability \( \frac{5}{6} \)), the game continues. Now it's A's turn. At this point, the situation is essentially the same as the original situation, but with roles reversed. Therefore, the probability of A winning from this point on is 1 - \( p \). (because if A wins, the game ends; if not, it's B's turn, and the game continues).

We can write the equation for \( p \) as:

\( p = \frac{1}{6} \times 0 + \frac{5}{6} \times (1 - p) \)

Solving above equation for \( p \)
\( p = \frac{5}{6}(1 - p) \)

\( p = \frac{5}{6} - \frac{5}{6}p \)

\( p + \frac{5}{6}p = \frac{5}{6} \)

\( \frac{11}{6}p = \frac{5}{6} \)

\( p = \frac{5}{11} \)

So, the probability that A wins is \( \frac{5}{11} \).