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With values $v_i>0$ and effort costs $b_i^2$, the unique pure Nash equilibrium has the efforts displayed above. Interior first-order conditions give $v_i{b}_j/(b_1+b_2{)}^2=2b_i$, hence $b_1/b_2=\sqrt{v_1/v_2}$ and $(b_1+b_2{)}^2=\sqrt{v_1{v}_2}/2$. Against a positive rival effort, each payoff is a strictly concave function, so these conditions identify global best responses. On the axes, lowering a positive uncontested effort or adding a sufficiently small effort at the zero tie rules out equilibrium.

A player maximizes the worst payoff it can receive against the other player's actions. Allowing the opponent to mix does not lower the worst payoff beyond a pure action, since the payoff is linear in that mixture. A security level payoff is an individual guarantee, not generally a Nash equilibrium payoff.

A mechanism or social choice function is strategyproof if truthful reporting is a weakly dominant strategy: for every true type and every collection of other reports, no misreport yields strictly higher utility or a strictly preferred outcome. This is stronger than merely requiring truthful reports to form a Nash equilibrium.

The support of a finite mixed strategy consists of the pure strategies assigned positive probability. At a Nash equilibrium, every action in the support is a best response. In a game satisfying nondegeneracy of a bimatrix game, the two equilibrium supports have equal size.

A strategy is strictly dominated when another strategy gives a strictly larger payoff against every opposing strategy. A strictly dominated strategy cannot receive positive probability in a Nash equilibrium, because moving that probability to its dominator increases expected payoff.

A strategy profile that is a Nash equilibrium in every subgame, including those outside the realized play path. In finite sequential games, backward induction constructs such profiles by solving continuation games before earlier decisions. When a stage is an all-pay contest, its solution can involve mixed strategies.

Enumerate pairs of strategy supports, solve the equations making each supported action indifferent, then check strict positivity of supported probabilities and all omitted-action payoff inequalities. These conditions suffice for a Nash equilibrium. Under nondegeneracy of a bimatrix game, only pairs of equal support size need be tested.

A square bimatrix game is symmetric when exchanging players exchanges their payoffs: its payoff matrices are $P,Q$ with $Q=P^T$. If $(x,y)$ is a Nash equilibrium, then $(y,x)$ is also a Nash equilibrium. A symmetric bimatrix game can still have asymmetric equilibria, so the two strategy vectors must be kept separate in the Lemke-Howson algorithm.

In a nondegenerate symmetric bimatrix game, player interchange pairs every nonsymmetric Nash equilibrium with a distinct swapped equilibrium. The equilibrium oddness theorem gives an odd total; the remaining fixed points, which are the symmetric equilibria, therefore also number an odd amount.

With $k>0$ active players of harmonic mean valuation ${\bar{v}}_k$, unit-cost Nash equilibrium effort satisfies $R+\delta =\frac{{\bar{v}}_k}{2k}[(k-1)+\sqrt{(k-1{)}^2+4k\delta /{\bar{v}}_k}]$. Sum the active-player first-order equations $b_i=(R+\delta )(1-(R+\delta )/v_i)$ to obtain the quadratic. If $\delta$ is at least the largest valuation, no one is active and $R=0$. With no outside effort, at least two players must be active.

For effective prizes $A_1\ge A_2>0$ and unit effort costs, equilibrium effort CDFs on $[0,A_2]$ are $G_1(b)=b/A_2$ and $G_2(b)=1-A_2/A_1+b/A_1$. The weaker player has an atom at zero. Incremental utilities are $A_1-A_2$ and zero, and winning probabilities are $1-A_2/(2A_1)$ and $A_2/(2A_1)$. Direct payoff indifference and exclusion of larger bids verify the Nash equilibrium. A further player with effective prize at most $A_2$ cannot profit by entering against these distributions.

In the zero-sum game with row payoff matrix $\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right)$, the second row is weakly dominated by the first. Nevertheless second row and first column form a Nash equilibrium: both rows tie against that column, and both columns tie against that row. Weak dominance therefore cannot be used to discard all possible equilibrium strategies without further qualifications.