Statistics guessing game This is a game that I wanted to know the answer for.
The Game
There are many players in the game, who each get a guess.
There is a loss distribution, $L$, where:
$L$ ~ Gamma($\alpha$, $\beta$)
i.e. the value of $L$ will be drawn from a Gamma($\alpha$, $\beta$).
The idea is to choose an amount $X$ so that $(X-L)\geq0$.
The player who chooses the smallest $X$, such that $(X-L)\geq0$, will win.
I was wondering what would be the best guess that maximized a players chances of winning?
The obvious constraints for winning are that $X$ must be at least the size of $L$ and also that the minimum guess of all players will win.
--EDIT--
In the event of a tie between players, the winner will be randomly selected from this group (where each member of this group has an equal probability of being randomly selected).
 A: I can think of two approaches. Both require certain subjective judgments in the form of priors or penalties, however. The optimal solution is denoted by $x^*$.
@whuber 's warning: Subjective strategies are a bad idea if you can find an optimal strategy and assume that all players adopt it.
APPROACH 1: Priors
Let there be $n$ players. We need to assume that the responses $X_i, i \in \{1,\ldots,n\}$, are i.i.d samples from a prior distribution on $\mathbb{R}_+$ that represents subjective knowledge about the opponents' guesses.
Denote the cumulative distribution function of the prior by $F_X(x)$, so that $P(X\leq x)=F_X(x)$.
Also, let $G_{\alpha, \beta}(x)$ denote the CDF of the Gamma random variable.
(EDITS: Based on the OP's comment, an additional constraint on $X_i$ was added) 
We are now looking at the probability of winning:
$f(x)=P(\cap_i \color{red}{(X_i \not\in [L,x])} \cap (L \leq x))$
(which, by conditional probability, becomes:)
$=P[\cap_i (X_i \not\in [L,x])|L \leq x] P[L\leq x]$
$=\prod_i P[(X_i \leq L|L\leq x) \cup (X_i \geq x)] G_{\alpha, \beta}(x)$
$\Rightarrow f(x) = (1-\color{red}{c(x)})^n G_{\alpha, \beta}(x)$
Here, $c(x) = P(X_i \in [L,x] |L\leq x)$. You will need to get this as follows:

So,
$c(x)=\int_{v=0}^x \int_{u=0}^{v} \text{jointpdf}_{X,L}(u,v) du dv$
$=\int_{v=0}^x \int_{u=0}^{v} p_X(u) p_L(v) du dv$
$=\int_{v=0}^x p_L(v) \int_{u=0}^{v} p_X(u)\; du dv$
This is assuming that $p_X(x)$ has mass only on $[0,\infty]$.
You're now in a position to proceed with setting the derivative to zero and finding the optimal solution.
APPROACH 2: Penalties
Here, the idea is to maximize your chances of exceeding $L$ subject to a norm constraint on your guess $x$.
First, pick a tradeoff penalty term $\lambda \geq 0$, then maximize:
$$f(x)=P(x\geq L) - \lambda x$$
The solution to this is:
$$f'(x^*)=0 \Rightarrow G_{\alpha, \beta}'(x^*)=\lambda$$
Assuming $G_{\alpha, \beta}'(x)=\dfrac{\beta^\alpha}{\Gamma(\alpha)} x^{\alpha-1}\exp(-\beta x)$, we get:
$$(\alpha-1)\log x^* - \beta x^* + \alpha \log \beta - \log \Gamma (\alpha) - \log \lambda = 0$$
This could be solved numerically.
Note that there is an implicit inequality constraint that $x \geq 0$. So you would have to discard negative values of $x^*$.
Notice that the forms of both solutions are similar. You can take the logarithm of the prior approach's solution to see this. The prior approach is a more principled way of choosing a penalty term.
