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Assume the usual continuous nonnegative valuation distribution, so ties occur only on zero-probability events. Put $t=F(v)$. In a monotone symmetric Bayesian Nash equilibrium, a type $v$ is first with probability $t^{n-1}$ and second with probability $(n-1)(1-t)t^{n-2}$. Its rank-order expected prize allocation in $G_1$ is thereforeThe all-pay effort identity gives $b_1(v)={\int }_{\underline{v}}^{v}s{a}_1^{\prime }(s)ds$. It also verifies equilibrium directly: a type $v$ imitating type $z$ has utility $v{a}_1(z)-b_1(z)$, whose derivative is $(v-z)a_1^{\prime }(z)$, so the true type is a best response.

In the first version of $G_2$, the two contests have expected allocationsThere is no common effort budget, and quasilinear utility makes the two effort choices separable. Since $a_{21}+a_{22}=a_1$, adding their all-pay effort identities yieldsThe equality holds type by type for aggregate effort, rather than only after taking expectations. The within-player correlation of the two efforts does not enter these additive expected payoffs.

For the second version of $G_2$, let $V_{[1]}\ge \cdots \ge V_{[n]}$ denote descending order statistics. The expected effort in a rank-order contest with prize vector $(2,1,0,\dots )$ isEquivalently, decompose the allocation into a unit award to the best player and a unit award to each of the best two players, then use revenue equivalence: the corresponding total auction payments are $V_{[2]}$ and $2V_{[3]}$. Two separate first-place contests with prize values one and two instead generateConsequentlyFor a nondegenerate continuous distribution, the inequality is strict. No regularity of virtual valuations is needed for this comparison. With a uniform distribution on $[0,1]$, the two totals are $(3n-5)/(n+1)$ and $3(n-1)/(n+1)$, giving a difference of $4/(n+1)$.

Let $n\ge 2$ players have independent values uniform on $[0,1]$, unit effort costs, and $1\le m<n$ equal prizes of size $w(m/n)>0$. In the symmetric Bayesian Nash equilibrium, total expected effort isThe support $[0,1]$ fixes the scale; a different uniform interval requires adjusting the order-statistic calculation.

One proof uses expected effort in a rank-order contest: only prize layer $m$ is nonzero, so total effort is $mw(m/n)\mathbb{E}{V}_{[m+1]}$. The corresponding descending order statistic of $n$ uniform values has mean $(n-m)/(n+1)$. Equivalently, the all-pay effort identity gives $b(v)=w(m/n){\int }_0^{v}t{q}_m^{\prime }(t)dt$, where the allocation derivative is a Beta distribution density with parameters $n-m,m$.