A quick reminder of the definition of Schmidt's game:

Let ${X}$ be a metric space and ${S\subset X}$ be a subset. Let ${0<\alpha,\beta<1}$ be constants. Bob chooses any open ball ${B_0\subset X}$ with radius ${\rho_0}$. Then Alice chooses a ball ${B_1\subset B_0}$ with radius ${\rho_1=\alpha\rho_0}$. Then Bob chooses a ball ${B_2\subset B_1}$ with radius ${\rho_2=\beta\rho_1}$, then Alice chooses a ball ${B_3\subset B_2}$ with radius ${\rho_3=\alpha\rho_2}$ and so on. Let ${x}$ be the (single) point in the intersection of all balls ${B_n}$. If ${x\in S}$ then Alice wins the game. Otherwise Bob wins. If Alice can force a victory, then the set ${S}$ is called

${(\alpha,\beta)}$-winning. $S$ called$\alpha$-winningif it's $(\alpha,\beta)$-winning for all $0 < \beta < 1$. One can also define $windim(S)$ to be the least upper bound on all $\alpha$ such that $S$ is $\alpha$-winning.

It's easy to see that $S$ need to be dense to be $(\alpha,\beta)$-winning. It's suprising though, that some sets of lebesgue measure $0$ are $\alpha$-winning (badly approximated numbers for $\alpha < \frac{1}{2}$).

Does some criterion exist for the inverse claim: sets of full measure (in a sense that $\mu(S^c)=0$) that are not $\alpha$-winning for some $\alpha$'s? It seems like the complement of the ternary Cantor set might be an example of that, but I couldn't find a good reasoning.