# Mixed strategy Nash equilibria for an all-pay auction between two players?

I am actually unsure if this game would be considered a minimum effort game or an all-pay auction, so please forgive me if I'm misusing these names for the title.

In a game where 2 players each choose a level of effort and the player with a higher level of effort wins, how would I go about finding the mixed strategy Nash equilibria? Would the form of the answer look like finding which levels of effort would result in payoffs of 0 for both players?

I would like to understand the most general case for this, but for more context on the specific question I am working on, each player at level $$s_{i}$$ also incurs a cost $$\beta_{i}s_{i}$$, with $$\beta_{1}$$ < $$\beta_{2}$$. This cost is incurred regardless of whether a player wins. The values of winning and losing are normalized to 1 and 0 respectively, and if the players choose the same level of effort, the probability that either one wins is $$\frac{1}{2}$$.

I think that there are no pure strategy Nash equilibria.

Whether or not there is a pure Nash Equilibrium depends on whether or not depends on the exact structure of the levels. Suppose that at level $$s_1$$ the cost $$B_1s_1=0.3$$ but for any $$s_i$$ such that $$i>1$$, $$B_is_i > 1$$, then it is obvious that both agents will use level $$s_1$$ as even a guaranteed win with any other $$s_i$$ will result in a loss for that agent and $$s_i$$ is still a gain in utility for both.
Now I really took this question to be that any cost $$B_is_i$$ can happen with completely uniformity- that is it can take on any value. In such a case, no, there is no pure strategy Nash equilibria as you have correctly pointed out.
Suppose that you and I are the agents in this game (we are both rational and take all other standard game theory question assumptions). If you were to always give some level of effort $$s_i$$ then if I were to mirror you we both get $$\frac{1}{2}$$ for a payout, but if I just 'one up' you then then my new payout is double, so as long as $$B_{i+1}s_{i+1} > 1$$ holds, I will one up you, and likewise you will one up me as well.
Given that this is an equal information game, we will both have the same strategy and thus both win with probability $$\frac{1}{2}$$ and thus receive $$\frac{1}{2}$$ payout in expectation. Average cost of an agent's choice of $$B_is_i$$ must therefore also be $$\frac{1}{2}$$. Something contrary to this would either imply cooperation or suboptimal play by an agent(s) as this is a zero sum game. With linear uniformity of the cost of all $$B_is_i$$ uniformly picking an $$s_i$$ from $$0$$ to $$k$$ where $$B_0s_0=0$$ and $$B_ks_k=1$$.
Regardless of the distribution of the cost of all $$B_is_i$$, the mixed Nash equilibrium will always have it such that an agent's average cost is $$\frac{1}{2}$$.