Hmm, if you were talking about "rock-paper-scissors" as an example of games without a pure Nash equilibrium, then some games have it and some don't. Intuitively, the more complex the game (the larger the number of pure strategies) the less likely that there is a pure Nash equilibrium, but that's not a strict rule.
However, under weak assumptions, there is always at least one, generally mixed, Nash equilibrium.

If by "winning strategy" you meant a mixed strategy that guarantees a greater than zero payoff, then in a two-player symmetric zero-sum game it can't exist:
if it existed both player would use it at the Nash equilibrium, and the sum of their expected payoffs would be greater than zero.