Sketch for the solution:
Note: Im assuming here that just player two have some strategy and player one is playing randomly.
You are over-complicating a bit I think, the sample space can be "reduced" drastically just thinking about a generic draw of four cards, that is
$$ \Pr [P_1 \text{ win }]=\Pr [P_1 \text{ win }|P_2 \text{ play it lower card }]\Pr [P_2 \text{ play it lower card }]\\ +\Pr [P_1 \text{ win }|P_2 \text{ play it higher card }]\Pr [P_2 \text{ play it higher card }] $$
As the cards are assumed to be drawn randomly (i.e. each card have the same probability to come up) then the probability $$ \Pr [P_1 \text{ win }|P_2 \text{ play it higher card }] $$ is the same as drawing three cards randomly and the first one is higher than the other two cards, what is easy to handle, and the probability $$ \Pr [P_1 \text{ win }|P_2 \text{ play it lower card }] $$ is equivalent to draw randomly three cards and the second or third card drawn be lower than the first one.
Well, you need to count also (if you want) the rare case where the lower rank beat the higher. But overall it seems that this probability is small and the changes in the probabilities discarding this possibility will be small.
EDIT: if you want to add some strategy to the first player also and $H_1$ and $H_2$ are the hands of player one and two respectively then you can build the model as
$$ \Pr [P_1 \text{ win }]=\Pr [\max H_1>\max H_2]\Pr [\max H_2]\Pr [\max H_1]\\ +\Pr [\min H_1>\max H_2]\Pr [\max H_2]\Pr [\min H_1]\\ +\Pr [\max H_1>\min H_2]\Pr [\max H_1]\Pr [\min H_2]\\ +\Pr [\min H_1>\min H_2]\Pr [\min H_2]\Pr [\min H_1] $$
where, by example, the probability $$ \Pr [\max H_1>\max H_2] $$ is equivalent to the probability that, after we had drawn four cards randomly, the first or the second have higher rank than the third and the fourth.
(Im not assuming again the case where the lowest rank beat the highest.)