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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).

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  • $\begingroup$ If each player is behaving optimally, then they will all use the same strategy and therefore all will guess the same $X$. Therefore it is crucial that you explain what the payoff will be in case of a tie. $\endgroup$ – whuber Feb 24 '16 at 13:34
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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:

Region of interest is the complement of the solid blue triangle

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.

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  • $\begingroup$ Hey Salmonstrikes, thanks for the comprehensive response. I just had a quick point. I like the first approach where you assume a prior distribution for your competitors' guesses. Just a further question, how would you alter it so that rather than the probability of winning being equal to the intersection of all $(x<X_{i})$, but instead the intersection of those $X_{i}$ that are above $L$ (because you only need to be less than those $X_{i}$ that are larger than the loss size)? $\endgroup$ – StatsPlease Feb 24 '16 at 7:13
  • $\begingroup$ Good point. The answer has been edited to reflect this additional constraint. $\endgroup$ – Salmonstrikes Feb 24 '16 at 9:32
  • $\begingroup$ Also, you can extend this to tailor each prior differently. This could be useful, for instance, if you believe that some players take greater risks than others. As long as the players guess independently, you will get an objective like $f(x)=G_{\alpha, \beta}(x) \prod_i (1-F_{X_i}(x)+c_i)$ $\endgroup$ – Salmonstrikes Feb 24 '16 at 11:06
  • $\begingroup$ Game theory shows that a choice of priors should not be relevant. You proceed by assuming each player is attempting to optimize their expected loss. Any player who proceeds based on "subjective knowledge" of the other players' choices will be a loser in the long run. $\endgroup$ – whuber Feb 24 '16 at 13:33
  • $\begingroup$ So whuber, are you saying that all players simply set their guess at the expected value of the Gamma? Also, given that this is a single event, is 'long-run' a feasible concept? $\endgroup$ – StatsPlease Feb 24 '16 at 13:46

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