Does a payoff matrix like
$$\begin{array}{c|cc} &B& B&\\ \hline A& (0,1)& (1,1)\\ A& (1,1)& (0,1) \end{array}$$
has infinite number of mixed-strategy Nash Equilibriums?
Assume $B$ chooses column and $A$ chooses row. The probability that $A$ chooses the first row is $p$, so second is $1-p$. So the expected payoff of $B$ is $1-p+p= 1-p+p$ (first column = second column). So I think p can be randomly choosed ranges from $1$ to $0$. But assume the probability that $B$ chooses the first column is $q$ and second column is $1-q$. Then the expected payoff of $A$ is $1-q = q$ (first row = second row). So $q = 0.5$. But it seems that $p$ cannot be randomly choosed...
Let A choose row, either Up or Down. B chooses column, Left or Right. Given any mixed strategy by A, choosing U with p and D with (1-p), then B's payoff is q(payoff to L)+(1-q)(Payoff to R) or q[p+1-p]+(1-q)[p+1-p]=1. Hence, given any mixed strategy of A's, any mixed strategy of B's will be a best reply. Now, what is the payoff to A, and what is the best reply to B? It is to choose p to maximize p(payoff to U)+(1-p)(payoff to D), remembering that B is choosing between L and R with probability q and (1-q).