# Nash-equilibrium for two-person zero-sum game

How one can find each player’s optimal strategy and the value of the two-person zero-sum game in the following Table:

\begin{array}{c|c|c|c|c|} & L & CL & CR & R \\ \hline T& 4& 5 & 1 & 4 \\ \hline M & 2& 1&6 &3\\ \hline B & 1& 0 &0 &2\\ \hline \end{array}

I know this game has no saddle point. I want to use mixed strategy. But, how can I reduce this matrix using iterated elimination of dominated strategy first? Moreover, what would be the Nash-equilibrium of this game?

Anyone help?

• Delete the third row because it is strictly dominated for Player 1 by either of the two top row. In the next round, delete the right-most column because it is strictly dominated for Player 2 by a convex combination of the two central columns (can you see why?) – mlc Mar 18 '17 at 23:08
• Thanks. I understand for the last row why it is deleted. I haven't realized for the right most column. I thought the first column should have been deleted because the right most column dominates the first column. – UserAb Mar 18 '17 at 23:47
• The Column player wants to minimize so the rightmost column is weakly dominated by the leftmost column. (Rightmost should be deleted, not leftmost.) – mlc Mar 19 '17 at 8:59

For $0 \le p \le 1$, suppose that Player 1 adopts the mixed strategy that plays $T$ with probability $p$ and $B$ with probability $1-p$. Then, if 2 plays $L$, the expected utility of Player 1 is $u_1(p,L) = 4p+2(1-p)=2+ 2p$; if 2 plays $CL$, it is $u_1 (p, CL) = 1 + 4p$; and, if 2 plays $CR$, it is $u_1 (p,CR) = 6-5p$. Compare these three functions on a graph and check that $$\min \{ 1 + 4p, 2+ 2p, 6-5p\} = \left\{ \begin{array}{cc} 1 + 4p & \mbox{if } 0 \le p \le 1/2 \\ 2 + 2p & \mbox{if } 1/2 < p \le 4/7 \\ 6 - 5p & \mbox{if } 4/7 < p \le 1 \\ \end{array} \right.$$ (The graph of this function is the lower envelope of the three component functions.) Recall that Player~1 wants to play maximin, so he chooses $p$ in order to $$\max_p \min \{ 1 + 4p, 2+ 2p, 6-5p\} = \max$$ which is attained at $p^*=4/7$ for a value of $22/7$. Given $p^*$, 2 is indifferent between L and CR and thus she randomises between these two strategies with probability $q^*=5/7$ and $1-q^*$, respectively, to make 1 indifferent. The value of the game is $22/7$.