With a bit further analysis one can see that the best possible classical strategy can be represented by a table where each cell now contains both Alice and Bob's answers, that may differ. It is possible to make their answers equal in 8 out of 9 cells, while respecting the parity of Alice's rows and Bob's columns. This implies that if the referee asks for a row and column whose intersection is one of the cells where their answers match they win, and otherwise they lose. Under the usual assumption that the referee asks for them uniformly at random, the best classical winning probability is 8/9.