# Finding the Shapley value solution for a set of players of size n?

I'm revising Game Theory and have come across this question:

"The Miners’ Game is defined as follows. There are $n (>2)$ miners who discover a large $(>n/2)$ quantity of gold bars. It takes two miners to carry a bar. (Assume that it is not possible to return to the mine for a second round.) Each bar is of equal size and value. Show that the core of this game consists of a single imputation if $n$ is even and is empty if $n$ is odd. Find the Shapley value solution for this game."

I can, with very basic and crude methodology, "prove" the first two facts. However I'm not sure how to find the Shapley value solution. I'm assuming that it consists of $\bar{x} = (0.5, 0.5, ... , 0.5)$ because logic, and I know for subset $S$ of all players we have $v(S∪\{i\})-v(S) = 0$ for $|S|$ even and $= 1$ for $|S|$ odd.

Any advice on where to go from here?

(Also, model proofs would appreciated for the facts relating to the core as mine are lengthy, but this would simply be a bonus.)

• Hi, welcome to math.SE! Here's a reference and tutorial for typesetting math on this site. – joriki Apr 15 '16 at 12:52

Let $v$ be the characteristic function of the game and $N$ the entire set of miners.
If $n$ is even, $v(N)=n/2$. For an imputation to sum to $n/2$, the average over all pairs of the combined allocations to the pairs must be $\frac12$, and for the imputation to be coalitionally rational, the combined allocation must be at least $\frac12$ for each pair. It follows that the allocation is exactly $\frac12$ for each pair.
If $n$ is odd, $v(N)$ is less than $n/2$. The same argument as above shows that the average over all pairs is less than $\frac12$ but the allocation to each pair is at least $\frac12$, a contradiction.
The Shapley value is $v(N)/n=\frac1n\left\lfloor\frac n2\right\rfloor$ for each miner, by efficiency and symmetry.