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The equilibrium oddness theorem gives a finite odd number of Nash equilibria in a finite nondegenerate bimatrix game. One way to see the parity is through the endpoint argument underlying the Lemke-Howson algorithm. Add sufficiently large constants to each payoff matrix so all entries are positive; this does not change best responses. Form the bounded convex polytopesLabel an $s_i=0$ facet by row label $i$ and a tight $(Q^{T}s{)}_j=1$ facet by column label $n+j$; in $Y$, label a tight $(Pt{)}_i=1$ facet by $i$ and $t_j=0$ by $n+j$. A completely labelled vertex of a polytope pair has either both vectors zero, or both nonzero and normalizes to a Nash equilibrium. In the latter case the labels express exactly the equilibrium best-response and zero-probability conditions. Nondegeneracy makes each vertex of a polytope have exactly $n$ distinct incident labels and each equilibrium correspond to one such pair.

Fix a label to drop and retain all pairs carrying every other label. Include the product-convex polytope edges that retain those labels. At a completely labelled pair there is one possible outgoing edge, obtained by dropping the fixed label. At any other retained vertex of a polytope, that label is missing and one other label is duplicated; dropping either copy gives the two incident edges. Thus this finite graph consists of paths and cycles, with its endpoints precisely the completely labelled pairs. A finite graph has an even number of degree-one vertices. Since one endpoint is the artificial zero pair, the number of genuine equilibrium endpoints is odd. This proves the required equilibrium oddness theorem and finiteness, without claiming that every equilibrium is reached from the zero pair on the same path.

For a symmetric bimatrix game, swapping players maps $(s,t)$ to $(t,s)$. Every nonsymmetric equilibrium lies in a distinct two-element pair; the fixed points of this involution are exactly the symmetric equilibria. Removing even-sized pairs from an odd total leaves an odd number of symmetric equilibria. In the present game, two equilibria form the swapped pair and the remaining one is $(x,x)$ with $x=(1/2,0,1/2)$. There is exactly one symmetric equilibrium, as required.