# How to calculate the payoff in Battleship Game Theory

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?

• What is the criterion for judging strategies? Expected number of shots? Maximum number of shots? Something else? – Henry Mar 23 '16 at 14:37
• "The payoff to Player I is the number of shots that Player II has made." It seems to me if Player II misses three times before sinking the destroyer, hence five shots altogether, the payoff to Player I is $5$. – David K Mar 23 '16 at 14:42
• "Solve the game" must refer to some solution concept. Presumably you're meant to find one or all Nash equilibria? – joriki Mar 23 '16 at 23:23

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.