Given a voting game where $v(1),v(2),v(3) = 0, v(1,2)= \frac{1} {3}, v(2,3) = \frac{5} {6}$, $v(1,3)= \frac{1} {6}$ and $v(1,2,3) = 1$
I know the Shapely coefficients for a 3 player game, for $|s|=1, |s|=3$, the Shapely coefficient is $\frac{1} {3}$, for $|s|=2, \frac{1} {6}$. Now I calculate the Shapely values thus:
$SH_1 = \frac{1} {3} [0] + \frac{1} {6} [\frac{1}{3}-0]+\frac{1} {6} [\frac{1}{6}-0]+\frac{1} {3} [1 - \frac{5}{6}] = \frac{5}{36}$
$SH_2 = \frac{1} {3} [0] + \frac{1} {6} [\frac{1}{3}-0]+\frac{1} {6} [\frac{5}{6}-0]+\frac{1} {3} [1 - \frac{1}{6}] = \frac{17}{36}$
$SH_3 = \frac{1} {3} [0] + \frac{1} {6} [\frac{1}{6}-0]+\frac{1} {6} [\frac{5}{6}-0]+\frac{1} {3} [1 - \frac{1}{3}] = \frac{14}{36}$
And I note that $SH_1 + SH_2 + SH_3 = 1$.
Now, what are the conditions for it to be in the core? $SH_1 > 0, SH_2 > 0, SH_3 > 0$ is one condition, I think. Any help would be appreciated, thanks!
To my mind, the most useful definition of the core is that an allocation is in the core if it is efficient and coalitionally rational (see Wikipedia). You've already noted that the allocation according to the Shapley values is efficient. It is clearly also coalitionally rational with respect to the empty coalition, all single-player coalitions and the grand coalition. So you just have to check that it's coalitionally rational with respect to the three two-player coalitions. This is indeed the case, since $v(1,2)\lt1-\textsf{SH}_3$, $v(2,3)\lt1-\textsf{SH}_1$ and $v(3,1)\lt1-\textsf{SH}_2$. Thus the allocation according to the Shapley values is in the core.