TITLE: Winning sets of full measure (Schmidt's game)
QUESTION [6 upvotes]: 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.

REPLY [3 votes]: 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.