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It is useful to prove the general core of the miners game calculation when $r$ miners carry one unit: $v(S)=\lfloor |S|/r\rfloor$, with $r\ge 2$. Individual rationality gives $x_i\ge 0$, and efficiency gives ${\sum }_i{x}_i=q:=\lfloor n/r\rfloor$.

If $n<r$, the grand coalition value is zero and the only core allocation is zero. If $n\ge r$, every $r$-person coalition must receive at least one. Sum these inequalities over all $\left(\genfrac{}{}{0ex}{}{n}{r}\right)$ such coalitions. Each player appears $\left(\genfrac{}{}{0ex}{}{n-1}{r-1}\right)$ times, soIf $r$ does not divide $n$, then $rq<n$, a contradiction. Thus the core of a cooperative game is empty in these cases.

If $n=r$, every proper coalition has value zero. The core is exactly the simplex of nonnegative allocations with total one. If $n$ is a multiple of $r$ with $n\ge 2r$, the averaged inequality is equality. Since each $r$-person coalition sum is at least one, every such sum equals one. For any two players $i,j$, choose $r-1$ other players and compare the two resulting $r$-person coalition sums. They imply $x_i=x_j$. Hence $x_i=1/r$ for all players. This vector belongs to the core because $|S|/r\ge \lfloor |S|/r\rfloor$ for every coalition.

For two miners per lump, therefore,In particular the even case has a small-group exception at $n=2$.

For three miners per lump, the same proof givesThus divisibility controls the existence of a stable allocation once enough miners are present, while exactly one possible lump leaves freedom to divide its value.

The core is the singleton $\{(1/2,1/2,1/2,1/2)\}$. Here the characteristic function of a coalitional game is $v(C)=\lfloor |C|/2\rfloor$, the two-person instance of the core of the miners game calculation.

Let $x$ be in the core. Efficiency requires $x_1+x_2+x_3+x_4=2$, and every pair coalition requires $x_i+x_j\ge 1$. Summing the six pair inequalities counts each coordinate three times, giving $3{\sum }_i{x}_i\ge 6$. Efficiency makes this an equality; since every individual slack is nonnegative, all six pair inequalities must be equalities. In particular $x_1+x_2=x_1+x_3=x_1+x_4=1$ forces $x_2=x_3=x_4$, and comparison with $x_2+x_3=1$ gives every $x_i=1/2$.

Conversely, for the uniform allocation every coalition satisfiesand its total is $2=v(N)$. It is therefore a core member, completing both existence and uniqueness. In particular, a three-player coalition receives $3/2\ge 1$; the pair constraints are the tight ones.