
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
Solution. Method 1. Even throws belong to B. Bons. Therefore, Bons wins only if the total number of throws, including the last successful one, is even. The probability of rolling a six is $\frac{1}{6}$. The probability of the opposite event is $\frac{5}{6}$. Thus, the probability of making an even number of throws in total is

![](https://cdn.mathpix.com/cropped/2024_05_06_542f32b3c267c270a1a7g-2.jpg?height=185&width=1450&top_left_y=958&top_left_x=383)

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

1) At the beginning, 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) The first time, neither Silver nor Bons rolled a 6. After this, the game essentially restarts, and B. Bons wins it with probability $p$. The probability of this development is $\frac{5}{6} \cdot \frac{5}{6} \cdot p=\frac{25}{36} p$.

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

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