It follows from the fact that, for any strategy $\sigma$, $$\sum_CP_{\sigma}(C)=\frac{1}{r}.$$ This is true for deterministic strategies because if you partition configurations into sets of $r$ that differ only in the content of the box that the mathematician chooses, then the strategy works on precisely one configuration from each set of $r$. (In fact, this is pretty much equivalent to Bjorn's answer.) It follows for probabilistic strategies since they are weighted averages of deterministic strategies.

It follows from the fact that, for any strategy $\sigma$, then the average over configurations of a correct guess is precisely $1/r$: $$\frac{1}{r^k}\sum_CP_{\sigma}(C)=\frac{1}{r}.$$ This is true for deterministic strategies because if you partition configurations into sets of $r$ that differ only in the content of the box that the mathematician chooses, then the strategy works on precisely one configuration from each set of $r$. (In fact, this is pretty much equivalent to Bjorn's answer.) It follows for probabilistic strategies since they are weighted averages of deterministic strategies.