XXVI - I - Problem 11

A ship is moving at a constant speed along a straight line with an east-west direction. Every $ T $ minutes, the direction of movement is randomly chosen: with probability $ p $, the ship moves in the eastern direction for the next $ T $ minutes, and with probability $ q= 1-p $, it moves in the western direction. At a point outside the line, there is a submarine whose task is to torpedo the ship. The travel time of the torpedo from the point of firing to any point on the ship's track is $ 2T $. The captain of the submarine knows the value of $ p $ and aims to maximize the probability of hitting the ship. How should $ p $ be chosen to minimize the probability of the ship being torpedoed?
Let's assume that a submarine fires a torpedo at the moment when a lottery is taking place on the ship. In the case of firing the torpedo at a different time, the reasoning proceeds similarly. We can also assume, without loss of generality, that $ p \geq q $, i.e., $ \displaystyle p \geq \frac{1}{2} $. After $ T $ minutes from the moment the torpedo is fired, the next lottery takes place. The results of these two lotteries can be as follows: east-east, east-west, west-east, west-west with probabilities $ p^2 $, $ pq $, $ qp $, $ q^2 $, respectively. In the first case, after $ 2T $ minutes from the moment the torpedo is fired, the ship will be located at point $ A $ to the east of the initial point $ O $, in the second and third cases - at point $ O $, and in the last case - at point $ B $ to the west of $ O $. The probability of the event that the ship will be at point $ A $, $ O $, $ B $ at that time is therefore $ p^2 $, $ 2pq $, $ q^2 $, respectively.
Since $ p \geq q $, then $ p^2 \geq q^2 $. The inequality $ p^2 > 2pq $ holds if and only if $ p > 2q $. Therefore, if $ p > 2q $, i.e., $ \displaystyle p > \frac{2}{3} $, then the largest of these three numbers is $ p^2 $ and thus the torpedo should be fired in the direction of point $ A $. If, however, $ 2q > p \geq q $, i.e., $ \frac{2}{3} > p \geq \frac{1}{2} $, then the number $ 2pq $ is the largest and the torpedo should be fired in the direction of point $ O $. For $ \displaystyle p = \frac{2}{3} $, we have $ p^2 = 2pq > q^2 $ and thus the torpedo can be fired in the direction of point $ A $ or $ O $ with the same probability of hitting. The probability of hitting $ f(p) $ is therefore

The function $ f(p) $ is increasing in the interval $ \displaystyle \langle \frac{2}{3};\ 1 \rangle $ and decreasing in the interval $ \displaystyle \langle \frac{1}{2};\ \frac{2}{3} \rangle $. Therefore, the smallest probability of hitting is at $ \displaystyle p = \frac{2}{3} $. It is $ \displaystyle \frac{4}{9} $.