Let’s define the situation in terms of probabilities.

• A wins if he throws a 6 before B throws a 1.
• B wins if he throws a 1 before A throws a 6.

Step-by-Step Analysis:

1. Possible Outcomes on a Die: Each die has 6 faces, so when A or B throws a die, there are 6 possible outcomes. The possible favorable outcomes for A are when a 6 is thrown, and for B, the favorable outcome is when a 1 is thrown.

2. Defining Probabilities:

   • The probability of A throwing a 6 on any throw is $P(A_6) = \frac{1}{6}$.
   • The probability of B throwing a 1 on any throw is $P(B_1) = \frac{1}{6}$.
   • The probability of A not throwing a 6 (i.e., throwing any other number except 6) is $P(A_{\text{not 6}}) = \frac{5}{6}$.
   • The probability of B not throwing a 1 (i.e., throwing any other number except 1) is $P(B_{\text{not 1}}) = \frac{5}{6}$.

3. Game Sequence:

   • A and B alternate throws. A goes first.
   • On A’s first turn, he can either throw a 6 and win, or throw a number other than 6 (which leads to B’s turn).
   • If B’s turn comes up, he can either throw a 1 and win, or throw a number other than 1 (which leads back to A’s turn).

   This alternation will continue until one of them wins.

Define Probability $p$ for A’s Win:

Let’s define the probability of A winning as $p$. On A's first turn, two things can happen:

1. A wins immediately: This happens if A throws a 6. The probability of this is $\frac{1}{6}$.
2. The game continues: If A does not throw a 6 (which happens with probability $\frac{5}{6}$), it becomes B's turn to throw. If B throws a 1 (probability $\frac{1}{6}$), B wins. Otherwise, the game returns to A, with the same situation as before.

Thus, we can set up the following equation for $p$:

$p = \left( \frac{1}{6} \right) + \left( \frac{5}{6} \times \left( \frac{5}{6} \times p \right) \right)$

Here:

• $\frac{1}{6}$ represents the probability that A wins on the first throw by throwing a 6.
• $\frac{5}{6} \times \frac{5}{6}$ is the probability that both A and B fail to win on their first throws, after which the situation resets.

Simplifying the Equation:

$p = \frac{1}{6} + \frac{25}{36}p$

Now, solve for $p$:

$p - \frac{25}{36}p = \frac{1}{6}$

$\frac{11}{36}p = \frac{1}{6}$

$p = \frac{1}{6} \times \frac{36}{11} = \frac{6}{11}$

Thus, the probability that A wins the game is $\frac{6}{11}$.

---

Would you like further clarification on the steps or any additional details about this probability model?

Here are 5 related questions for you:

1. How does the probability change if A goes second instead of first?
2. What would the probabilities look like if the dice were biased, say A has a 50% chance to throw a 6?
3. How can we extend this model to include more players?
4. What would happen if B’s favorable outcome changed to a 2 instead of a 1?
5. How does the sequence of independent trials in probability theory apply here?

Tip: When dealing with alternating events in probability problems, setting up recursive equations (like the one we used above) is a powerful method for finding the solution.