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
>  **$\alpha$-winning** if 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.



Winning sets of full measure (Schmidt's game)
I'm not sure if you're more interested in general conditions for this or an example with a simple proof. I can give a simple example and explanation. Let $S$ be the set of numbers normal in base $b$. $S$ is a set of full measure (follows by SLLN, Birkhoff ergodic theorem, etc.) W. Schmidt proved that $S^c$ is $1/2$-winning. So $S^c$ is $\alpha$-winning for all $\alpha \in (0,1/2]$. But the property of being $\alpha$-winning is preserved under countable intersections. So if $S$ were $\alpha$-winning, then $S \cap S^c=\emptyset$ would also be $\alpha$-winning. Does that answer your question? There are many other examples along these lines in some of the more recent literature on Schmidt games that I can give if you're interested.


