Suppose a game is played between $A$ and $B$, in which there exists a winning strategy for $A$.
Suppose $A$ and $B$ play their moves at random, do we have $\mathbb{P}(A\;\text{wins})=1$? Also, is the converse true (does $\mathbb{P}(A\;\text{wins})=1$ imply the existence of a winning strategy?)
If the winning strategy implies that $A$ always wins, and $A$ uses it, then yes, $P(A \text{ wins}) = 1$.
If there are non-winning strategies, it is probable that $A$ does not play the winning strategy. Then $\mathbb{P}(A \mbox{ wins})<1$ if the non-winning strategies are noticeable compared to the whole set of strategies (have measure greater than zero).
The converse may not be true either. If we know that some event happens with probability one, in this case $\mathbb{P}(A \mbox{ wins})=1$, that does not certainly imply that other events may not happen. Zero probability events can also happen. See, for instrance, here, and here.
