# How to mathematically express in a payoff matrix that “not losing” isn't the equivalent of winning

My uncle was watching a documentary on the revolutionary war and one of the historians said, "Washington realized he didn't need to win the war, he only needed to not lose it." Is it possible to express in a pay off matrix that losing is not the antonym to winning?

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A little more background: "[T]he British were much better at those kinds of direct, decisive fights than the colonists were. The British had naval support and well-trained veterans who could be counted on under fire. Washington tried a couple of direct confrontations at the beginning of the war, but only sheer luck saved his army from total annihilation. Eventually, he came to realize that, in the words of historian Joseph Ellis, "the way to win the war was not to lose it."" (Source) –  Glen The Udderboat Jul 11 '13 at 9:08

The easiest way of doing this is saying that there are 2 levels of the game. Let $p_1$ and $p_2$ be two players, let $A$ be a finite set of all possible outcomes and let $f_1,f_2$ be a correspondent payoff matrix. Based on the outcome $A$, we determine new "meta" outcomes: $$S_1 = \{f_1>f_2\},\; S_0 = \{f_1 = f_2\} = \{\text{draw}\}, \;S_2 = \{f_2>f_1\}.$$ Now, the final result of the game is evaluated not based on the first-level-payoffs $f_1,f_2$, but on the second-layer ones $u_1$ and $u_2$ which in your case are given by: $$u_1(S_1) = u_1(S_0) = 1, \;u_1(S_2) = -1.$$ That means, the in the final count the player $p_1$ wants to avoid the loss in the first layer, but both the win and the draw in the first layer he considers as a final win.