The first one is beatable if -- and only if, I believe -- a square with two three-in-a-rows in it counts as both an X and an O. If this is the case, then O can

The first one is beatable if -- and only if, I believe -- a square with two three-in-a-rows in it counts as both an X and an O. If this is the case, then O can win one round before X has a chance to it. But otherwise -- as you have it configured, where the first to get a three-in-a-row sets the larger square to either an X or an O -- it's unbeatable.