XXVII OM - III - Task 5

A fishing vessel is fishing in the territorial waters of a foreign state without permission. Each cast of the net results in a catch of the same value. During each subsequent cast, the probability of the vessel being intercepted by the border guard is $ \frac{1}{k} $, where $ k $ is a fixed natural number. We assume that the event of the vessel being intercepted or not intercepted during each subsequent cast is independent of the previous course of the fishing. If intercepted by the border guard, the entire catch is confiscated, and further fishing is impossible. The captain plans to return after the $ n $-th cast. Considering the risk of interception, the profit from fishing is a random variable. Find the number $ n $ for which the expected value of the profit is maximized.
The probability of an event where a ship is not caught with a certain cast of the net is $1 - \frac{1}{k}$. Since events involving catching or not catching a ship with each subsequent cast of the net are independent, the probability of the event where the ship is not caught after $m$-times casting the net is $\left(1 - \frac{1}{k}\right)^n$. The expected value of the profit from $n$-times casting the net is therefore equal to

where $w$ is the value of the profit from one cast of the net.
The task is to investigate for which natural number $n$ the function $f$ attains its maximum value.
From (1), it follows that

Since the inequality $\displaystyle 1 + \frac{(k-1)-n}{kn} \geq 1$ is equivalent to the inequality $(k - 1) - n \geq 0$, which is $n \leq k - 1$, we have

Therefore, the function $f$ attains its maximum value twice at $n = k - 1$ and $n = k$.