# Prove in a $4 \times 4$ battleship game the expected payoff of player one is $\frac{3}{4}$

A battleship game is played on a $$4 \times 4$$ matrix, player one can place their domino(that takes up $$2$$ adjacent spaces) in one of $$24$$ places(There are $$3$$ places in each row and each column it can be placed so $$3\cdot 4+3 \cdot 4=24$$). Player $$1$$ secret places the domino and player $$2$$ secret guesses, which one of the $$16$$ indexes it's in. If player $$2$$ guess right they win $$1$$ dollar and player $$1$$ loses $$1$$ dollar. If player $$2$$ guesses wrong then player $$1$$ wins $$1$$ dollar and player $$2$$ loses $$1$$ dollar. The expected payoff for player $$1$$ is $$\frac{3}{4}$$. Prove this is true.

Reducing this to a 3x4 payoff matrix

$$\begin{bmatrix}2&2&2\\2&2&2\\2&2&2\\2&2&-2\end{bmatrix}$$

The above is an example of if player 1 being the row picked a corner then p2 picked the same corner(not entirely sure this is done correctly)

• What have you tried? – Arthur Nov 22 at 20:24
• I actually drew out the matrix and drew a pay off matrix, but I think for expected value you normally need the Stochastic vectors representing their mixed strategy cause the expected value is normally $\sum \sum a_{ij}x_iy_j$. – user8714896 Nov 22 at 20:58
• @Arthur but I guess I don't need the stochastic vectors, because it's supposed to be expected value regardless of what player 2 does, but they never taught us how to calculate that. – user8714896 Nov 22 at 21:04

No need to count the number of arrangements. Two out of $$16$$ squares are covered, so the probability to hit a covered square is $$\frac2{16}=\frac18$$. The expected payoff follows directly.
• It's not that simple. The domino isn't placed randomly so this is actually evaluating a zero-sum game with mixed strategies. Player $2$ can select a corner, a side, or a central square, yielding various probabilities of winning depending on which strategy Player $1$ selected. (For example, if Player $1$ selects a corner and Player 2 chooses a side square, then Player $2$ has a probability of $\frac 18$ of winning.) Player $1$ has $4$ distinct strategies (corner, along side, jutting out from side, central). I can set up the matrix but I don't remember how to find the Nash equilibrium. – Robert Shore Nov 22 at 21:01
• @user8714896 Use my comment above to reduce it to a $3 \times 4$ matrix. – Robert Shore Nov 22 at 21:06
• I don't see why it's not that simple. Clearly in a Nash equilibrium the first player must ensure that all 16 squares are equally likely to be covered; otherwise the second player's strategy would be to pick one of the more likely squares and thus get a higher payoff (which, since it's a zero-sum game, means a lower payoff for the first player). So we don't need matrices or anything; the first player's expectation is simply $p\cdot(+1)+(1-p)\cdot(-1)=2p-1=\frac34$, where $p=\frac78$ is the probability for the first player to win a dollar if her strategy uniformly covers the squares. – joriki Nov 22 at 22:44
• @joriki It follows from your analysis that $50$% Corner, $50$% Dangle should be an optimal strategy for Player 1. – Robert Shore Nov 24 at 5:01