bability of sums:

Sum of 6: The outcomes that sum to 6 are (1,5), (5,1), (2,4), (4,2), and (3,3). Therefore, there are 5 outcomes.

Probability of rolling a 6: $ P(\text{Sum } 6) = \frac{5}{36} $

Sum of 7: The outcomes that sum to 7 are (1,6), (6,1), (2,5), (5,2), (3,4), and (4,3). This gives us 6 outcomes.

Probability of rolling a 7: $ P(\text{Sum } 7) = \frac{6}{36} = \frac{1}{6} $

Calculate the probability of $A$ winning:

Since $A$ rolls first, for $A$ to win, the following sequence can occur:

$A$ rolls a 6 on the first turn.

If $A$ does not roll a 6, with probability $\frac{31}{36}$ (the complementary probability of not rolling 6), and $B$ does not roll a 7, with probability $\frac{5}{6}$, then the game is in the same state again after one complete cycle (both $A$ and $B$ have thrown once each), and we recalculate.

This forms a geometric series where the success probability is $\frac{5}{36}$, and the continuation state probability is $\frac{31}{36} \times \frac{5}{6}$.

Use the formula for geometric series to find the probability $A$ wins:

$ P(A \text{ wins}) = \frac{\frac{5}{36}}{1 - \frac{31}{36} \times \frac{5}{6}} $

Solving this evaluates to:

$ P(A \text{ wins}) = \frac{30}{61} $

Therefore, the probability that $A$ wins is $\frac{30}{61}$.

Probability

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