# If a mixed strategy is strictly dominated, then there is a strictly dominated pure strategy in its support?

I am looking at the proof of NE survives the iterated removal of strictly dominated strategies (MWG, ex 8.D.2) and in the solution manual, authors say something like if a mixed strategy is strictly dominated (by another mixed strategy say s), then we can find a pure strategy in the support of that strictly dominated mixed strategy that is strictly dominated (by s).

My intuition told me that this statement is (of course) correct, but can someone give me a hint on how to show it formally?

• I apologize for answering your question too hastily and indeed incorrectly. It appears that the following is a counterexample to the claim made in your question. The following matrix shows only the column player's payoffs: \begin{array}{|c|c|c|c|} \hline &L&M&R\\\hline T&1&7&5\\\hline B&7&1&5\\\hline \end{array} Here, the mixed strategy $\frac12L+\frac12M$ is strictly dominated by pure strategy $R$, but neither $L$ nor $M$ is strictly dominated by any strategy. I have deleted my previous answer. – Herr K. Jan 25 '19 at 2:14
• I also couldn't seem to find the statement you mentioned in MWG. Could you give a more precise reference? – Herr K. Jan 25 '19 at 2:16
• Thanks, that's OK! It is confusing and I guess the proof is not that obvious. The proposition is in the solution manual written by Hara, Segal, and Tadelis. When they prove the unique strategy survive the iterated elimination of strictly dominated strategies must be NE, they assume that if there is strategy that strictly dominates NE for player i at some round, that strategy must also strictly dominates some pure strategies in the support of NE strategy. – kokutou_qian Jan 25 '19 at 14:59
• That is an interesting counterexample! Let me double check if I misunderstand the proof. – kokutou_qian Jan 25 '19 at 15:03