Optimal strategy for dominoes game

Question

Here is the basic principle of the game I'm trying to find an optimal strategy for:

Two players (say P and Q) are given a 2x3 grid and a domino. Then P chooses one way of positioning the domino on the grid (there are 7, according to me) and Q chooses a square of the grid (one out of 6). If Q chooses one of the covered squares, he wins and P looses (say 1 point). Otherwise, P is the winner and Q has lost. I'm looking for the optimal strategy for both winners and trying to find out who has more chances of winning in the end. And then I'm wondering whether or not I could generalize some of the ideas to larger grids (using symmetries and other particularities of the problem)...**

** Numbering the squares in the grid : first row 1,2,3, second row 4,5,6; the different ways of placing the domino are covering 1 and 2, 2 and 3, 4 and 5, 5 and 6 (horizontal positions) as well as 1 and 4, 2 and 5, 3 and 6 (vertical positions).

What I've tried : - Payoff matrices but it doesn't give anything concluent for an unknown reason - Thinking that P might want to cover each square with the same probability so he might want to cover (1 and 2) or (4 and 5) or (3 and 6) (and no other possibility) with probability 1/3. - Q's strategy might be to play square 2 or 5 all the time (since they're most likely to be covered)

After that, I'm stuck and turning around the bush... Do you have an idea? Can you help me please?

Thank you so much!

Answer 1

By symmetry, you really have only three strategies for P and two strategies for Q. Q can choose a middle or a corner. P can do a horizontal, an end vertical, or the middle vertical. You can now make a payoff matrix. If P does horizontal and Q chooses corner, Q wins with probability $\frac 14$ You can fill in the other five locations the same way and use your usual game theory techniques. It will work the same for larger boards, but there will be many more possibilities.

Answer 2

For all pairs of neighbouring squares, Q must have the same probability of choosing a square in that pair. This is because the sum over all pairs is twice the sum over all squares, which is fixed at $1$, so if the probabilities aren't the same for all pairs, there will be one below average, and by covering that pair P has a strategy that is worse for Q than the equiprobable one.

It follows that Q must have the same probability for choosing all white squares and the same probability for choosing all black squares; since the total is fixed at $1$, there is one free parameter in his strategy.

P's situation is more complicated. On an $m\times n$ board she has $m(n-1)+n(m-1)=2mn-m-n$ different moves. She must achieve the same probability of covering each square, since otherwise Q could choose a below-average square and P would be worse off. That yields $mn$ linear constraints on the $2nm-m-n$ probabilities, leaving $nm-m-n$ free parameters. In your case, this also happens to be $1$, and we can calculate the corresponding set of strategies. P can make a horizontal move with probability $p$, an end vertical move with probability $q$ and the middle vertical move with probability $1-4p-2q$. Then the probability of covering one of the corners is $p+q$, and this must be $\frac13$. Thus the probabilities are $p$ for a horizontal move, $\frac13-p$ for an end vertical move and $\frac13-2p$ for the middle vertical move, with any $p\in[0,\frac16]$.

Any pair of these strategies for Q and P forms a mixed Nash equilibrium, and these are the only Nash equilibria.