
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
Sam makes odd-numbered shots. Therefore, he wins only when the total number of shots, including the last successful one, is odd. The probability of hitting the target with one shot is \(\frac{2}{5}\). The probability of the opposite event is \(\frac{3}{5}\). Thus, the probability that a total odd number of throws will be made is:

\[
\frac{2}{5} + \frac{3}{5} \cdot \frac{3}{5} \cdot \frac{2}{5} + \frac{3}{5} \cdot \frac{3}{5} \cdot \frac{3}{5} \cdot \frac{3}{5} \cdot \frac{2}{5} + \ldots
\]

Method 2. Let \( p \) denote the desired probability of the event "Sam hits the target." This event can occur in one of two ways:

1) Sam hits the target immediately. The probability of this is \(\frac{2}{5}\).
2) Both Sam and Billy miss once. After this, the competition essentially restarts, and Sam wins with a probability of \( p \). The probability of this sequence of events is:

\[
\frac{3}{5} \cdot \frac{3}{5} \cdot p = \frac{9}{25} p
\]

Thus, \( p = \frac{2}{5} + \frac{9}{25} p \), from which \( p = \frac{5}{8} \).

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