Backward-induction threshold in an elimination all-pay contest ($C_r=(1-\delta ){\sum }_{j=2}^r{\delta }^{j-2}{v}_j+{\delta }^{r-1}{v}_{r+1}$)

With ordered values $v_1>\cdots >v_k$ and $r<k$ prizes remaining, define $C_r=(1-\delta ){\sum }_{j=2}^r{\delta }^{j-2}{v}_j+{\delta }^{r-1}{v}_{r+1}$. In the recursively constructed discounted subgame perfect equilibrium, it is the second active player's effective prize and the effort-support upper endpoint. The highest player's utility is $v_1-C_r$. The coefficients form a convex combination, so $C_r\ge v_{r+1}$, allowing lower-player deviations to be bounded.

As $\delta \uparrow 1$, both active players' effective prizes in every nonfinal subgame tend to the marginal valuation $v_{r+1}$. The two-player complete-information all-pay equilibrium then gives each a winning probability $1/2$. In the final stage, actual valuations determine the asymmetric winning probabilities. Taking this limit from discounted equilibria specifies the continuation selection instead of independently choosing an undiscounted game equilibrium.

For $m<n$ prizes and ordered distinct valuations, the discounted-equilibrium limit gives $x_1=1-2^{-m}{v}_{m+1}/v_1$ and $x_i=1-2^{-(m-i+2)}{v}_{m+1}/v_i$ for $2\le i\le m$. A player must lose its successive fair nonfinal contests and then the asymmetric final contest to receive no prize. The marginal player's probability is $m-{\sum }_{i=1}^m{x}_i$, and lower players have probability zero.