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A finite game has finitely many players, each with finitely many pure strategies.

Use net monetary gain as payoff and order each player's pure strategies as $1,2,4$. The payoff matrix for the first player isThe second player's matrix is $-A$. Equal choices return both stakes, so the diagonal is zero; when the first player wins, their net gain is the other's stake, and when they lose it is minus their own stake. This is a matrix game, hence a zero-sum game and a normal-form game.

The game is degenerate in the nondegeneracy of a bimatrix game sense. Against the first player's pure strategy $4$, the second player's choices $1$ and $2$ are both best responses, with payoff four. A mixed strategy of support size one therefore has two pure best responses. The usual Lemke-Howson algorithm path needs additional tie handling or perturbation in such a game; the zero-sum game structure gives a simpler direct linear program.

Add five to every entry of the first player's matrix, obtainingThis does not change either player's best responses or Nash equilibria; it raises the game value by five. Since all entries are positive, its value $v_B$ is positive. If $p$ is a row mixed strategy guaranteeing $v_B$, put $x=p/v_B$. Then $B^{T}x\ge 1$ and $1^{T}x=1/v_B$.

Conversely any feasible $x$ has $s=1^{T}x>0$, and $p=x/s$ guarantees payoff $1/s$ in the shifted game. Maximizing that guaranteed payoff is therefore equivalent to minimizing $s$ under $B^{T}x\ge 1$, $x\ge 0$. These are precisely the displayed constraints. The dual program maximizes $1^{T}y$ subject to $By\le 1$, $y\ge 0$; normalizing an optimal $y$ gives the column strategy. This is positive-payoff linear programming for a matrix game.

Write $S=R+\delta$ and let $C_i=\delta +{\sum }_{j\ne i}{b}_j$. Player $i$'s payoff in the proportional contest with outside effort is $u_i(b)=v_{i}b/(b+C_i)-b$. When $C_i>0$, this is strictly concave andThe best response is zero when $v_i\le C_i$, and otherwise is $\sqrt{v_i{C}_i}-C_i$. Thus the nonnegative-effort Karush-Kuhn-Tucker conditions at a pure strategy Nash equilibrium giveFor an active player the first-order equation is $v_i(S-b_i)=S^2$. For an inactive player the derivative at zero is $v_i/S-1$. Strict concavity makes these conditions sufficient as well as necessary.

If $\delta =0$, an equilibrium cannot have just one active player: that player would win with certainty and could lower its positive effort. An all-zero profile also cannot be an equilibrium under the usual completion of proportional allocation at zero: at least one player can gain by investing an arbitrarily small amount. Consequently there are at least two active players, so every $C_i$ is positive. If $\delta >0$ this positivity is automatic. The allocation rule's otherwise undefined all-zero value at $\delta =0$ is therefore immaterial to the equilibrium calculation.

Let $k=\hat{n}>0$, $H_k={\sum }_{i=1}^{k}1/v_i$, and use the harmonic mean ${\bar{v}}_k=k/H_k$. Summing the active efforts givesThe positive root supplies the total-effort formula for a proportional contest with outside effort:At $\delta =0$ this reduces to $R=(k-1){\bar{v}}_k/k$. There is also a genuine zero-active case: if $\delta \ge v_1$, then $k=0$ and $R=0$; no harmonic mean of an empty family is needed.

For a fully explicit active-set rule, setThe equilibrium equation is $G(S)=0$. For $\delta >0$, $G$ is strictly decreasing from $+\infty$ to $-1$. For $\delta =0$ its limit at zero is $n-1>0$, and it is strictly decreasing wherever a zero could occur. Hence the positive root is unique. Moreover $S\ge \delta$, since its equation gives $1-\delta /S\ge 0$, and $b_i=S(1-S/v_i{)}_{+}$ supplies an equilibrium with total effort $S-\delta$.

Player $j$ is active exactly when $v_j>S$, equivalently $G(v_j)<0$. Ordering the valuations, including ties, givesTherefore the active-set threshold for a proportional contest with outside effort isThe qualifying indices form a prefix because $G$ is decreasing. Equality excludes the marginal player, as required by strict positivity of effort. Equivalently, the positive root $S_k$ of the displayed quadratic must satisfy $v_k>S_k\ge v_{k+1}$, with $v_{n+1}=0$. These formulas include one active player when $\delta >0$ and exclude that case when $\delta =0$.

Let $u,v$ be the row and column mixed strategies, so their entries are nonnegative and sum to one. The payoffs are $u^{\mathsf{T}}Pv$ and $u^{\mathsf{T}}{P}^{\mathsf{T}}v$. A Nash equilibrium is a pair in which each strategy is a best response to the other: neither player can improve its expected payoff by a unilateral change of mixed strategy. Equivalently, every positive-probability pure strategy attains that player's maximum pure-strategy payoff against the opposing mixture.

For the complementarity construction of a symmetric Nash equilibrium put $s={\sum }_i{x}_i>0$ and $u=x/s$. The equation $Px+z=1$ gives $Px\le 1$. Nonnegativity and $x^{\mathsf{T}}z=0$ imply $x_i>0\Rightarrow z_i=0\Rightarrow (Px{)}_i=1$. Therefore every used pure strategy has payoff $1/s$ against $u$, while all other pure strategies have payoff at most $1/s$. The same vector of pure-strategy payoffs applies to the column player in this symmetric bimatrix game. Thus a symmetric Nash equilibrium is

For a genuine Lemke-Howson algorithm path, use two separate vectors, $\xi$ for the row player and $\eta$ for the column player, with slacksGive labels $i=1,2,3$ to ${\xi }_i=0$ or $r_i=0$, and labels $3+j$ to ${\eta }_j=0$ or $s_j=0$. A nonzero completely labeled pair has ${\xi }_i{r}_i=0$ and ${\eta }_j{s}_j=0$, and normalization yields mutual best responses. At the artificial pair $(\xi ,\eta )=(0,0)$ all six labels are present.

Drop label $3$ by increasing ${\xi }_3$. The two limiting rows of $P\xi \le 1$ are $3{\xi }_3\le 1$ and $2{\xi }_3\le 1$, so the minimum-ratio pivot stops at ${\xi }_3=1/3$, where $s_2=0$. Label $5$ is now duplicated, since ${\eta }_2=0$ also carries it. Increase ${\eta }_2$ to remove that duplicate. The limiting inequalities are $3{\eta }_2\le 1$ and $4{\eta }_2\le 1$, so the next pivot stops at ${\eta }_2=1/4$, where $r_3=0$ restores the dropped label $3$. The endpoint isIt has all six labels and satisfies both complementarity conditions. After normalization, the Lemke-Howson algorithm returnsAgainst column strategy $2$, row strategy $3$ pays $4$, exceeding $3$ and $0$. Against row strategy $3$, column strategy $2$ pays $3$, exceeding $0$ and $2$. This verifies the endpoint directly.

To find every other Nash equilibrium, observe that row strategy $3$ strictly dominates row strategy $1$: the payoff differences against the three columns are $(2,1,2)$, all positive. By symmetry, column strategy $1$ is also strictly dominated. Neither can appear in an equilibrium. The reduced strategies $2,3$ have row payoff matrixThe two pure Nash equilibria are $(e_2,e_3)$ and $(e_3,e_2)$. If the opponent uses strategy $2$ with probability $q$, the payoffs of strategies $2,3$ are $3(1-q)$ and $2+2q$. Indifference requires $q=1/5$. The same computation applies to the other player. A player mixing both strategies forces this exact opposing mixture; a player playing a pure strategy has a unique opposing best response, producing one of the two pure equilibria. Hence there are exactly three Nash equilibria:For the symmetric mixed equilibrium, the original complementarity construction can use $x=(0,1/12,1/3)$ and $z=(3/4,0,0)$; it gives payoff $12/5$ to each player after normalization.