Discussion

Answer: B

The correct statement is Nash equilibrium is reached when each of them purely randomises their behavior.

Check best responses. Against Left, \(X\) prefers Bottom (2 over 1); against Right, Top (4 over 3). Against Top, \(Y\) prefers Left (2 over 1); against Bottom, Right (4 over 3). In every cell one player wants to switch, so there is no pure-strategy equilibrium. Let \(X\) play Top with probability \(p\) and \(Y\) play Left with probability \(q\):

\[ \begin{aligned} U_{Y}(L) &= 2p + 3(1 - p) = 3 - p \\ U_{Y}(R) &= p + 4(1 - p) = 4 - 3p \\ 3 - p &= 4 - 3p \Rightarrow p = \tfrac{1}{2} \end{aligned} \]

\[ \begin{aligned} U_{X}(T) &= q + 4(1 - q) = 4 - 3q \\ U_{X}(B) &= 2q + 3(1 - q) = 3 - q \\ 4 - 3q &= 3 - q \Rightarrow q = \tfrac{1}{2} \end{aligned} \]

The only equilibrium is for each player to mix 50:50, a purely random choice. There are no pure-strategy equilibria, hidden or otherwise. Battle of the Sexes has two pure-strategy equilibria, so it is unlike this game. The game shares Matching Pennies' cycle of best responses, but Matching Pennies is zero-sum, while here the payoff totals differ across cells (3, 5, 5 and 7); the key accepts only the statement about randomisation.