Consider a 3 by 3 board and suppose that Player I hides a destroyer(length 2 squares) vertically or horizontally on this board. Then Player II shoots by calling out squares of the board, one at a time. After each shot, Player I says whether the shot was a hit or a miss. Player II continues until both squares of the destroyer have been hit.The payoff to Player I is the number of shots that Player II has made. Let us label the squares from 1 to 9 as follows.
Solve the game.
Note: this problem is invariant under rotations and reflections, hence can be forged into a 2 by n game where n is the number of invariant strategies of player II. However, I don't understand the relationship between payoff of I and strategy of II here. What is the payoff for I? The cost II pays each time when he shoots a number from 1 to 9 or something else? And how to deduce this? Via probability of each calling and its consequence?
The strategy of II influences the payoff to I because the expected number of shots changes. Say II uses the (very bad) strategy of always guessing $1,3,5,7,9$ for the first five guesses. The destroyer cannot be sunk before guess $6$. If I has placed the destroyer randomly, the payoff will be uniform over $[6,9]$ with average $7.5$. If II uses a better strategy, the average payoff to I will be lower.
You are expected to come up with a list of strategies for each player and find the average payoff under each. For I, there are only two different types of position: in a corner and touching the center. Having chosen one of these two, clearly I should randomly pick which of the positions to use. I just has to choose the mix of the two types of position. II has a wider range of choices. The first move has three types of choices: center, edge, corner. The strategy starts with a ratio between these, then says what to do if you miss. Clearly if you hit you just randomly try the possible other squares.