
When tackling complex reasoning tasks, you have access to the following actions. Use them as needed to progress through your thought process.

[ASSESS]

[ADVANCE]

[VERIFY]

[SIMPLIFY]

[SYNTHESIZE]

[PIVOT]

[OUTPUT]

You should strictly follow the format below:

[ACTION NAME]

# Your action step 1

# Your action step 2

# Your action step 3

...

Next action: [NEXT ACTION NAME]


system
Even-numbered rolls belong to B. Bons. Therefore, Bons wins only if the total number of rolls, including the last successful one, is even. The probability of rolling a six is $\frac{1}{6}$. The probability of the opposite event is $\frac{5}{6}$. Hence, the probability that the total number of rolls will be even is...

Method 2. Let $p$ be the desired probability of the event "B. Bons wins." This event can occur in one of two ways:

1) Initially, J. Silver did not roll a 6, but B. Bons immediately rolled a 6. The probability of this is $\frac{5}{6} \cdot \frac{1}{6} = \frac{5}{36}$.
2) Both Silver and Bons did not roll a 6 on their first attempts. After this, the game essentially restarts, and B. Bons wins it with probability $p$. The probability of this sequence of events is $\frac{5}{6} \cdot \frac{5}{6} \cdot p = \frac{25}{36} p$.

Thus, $p = \frac{5}{36} + \frac{25}{36} p$, from which $p = \frac{5}{11}$.

Present the answer in LaTex format: \boxed{Your answer}
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