The winning probability in a card game Suppose that I am playing a card game with my friend - a $1$ vs $1$ card game. All cards in standard card deck (52 cards) are shuffled randomly, then two cards are drawn to each person respectively. (without replacement) Each player is required to play one of these cards. The card is ranked according to its standard value, regardless of the suits, but the absolute weakest card beats the absolute strongest card, i.e. a $2$ wins an A. The winner belongs to the player who shows the larger value on his card. If both cards have the same value, then we have a tie. Cards will be reshuffle again after a match.
The following are the probabilistic assumptions on this game in order to compute the probability to win:

*

*The probability to play any one of these cards are equally likely for me and my opponent.

*There is no other factor that affect the match.

If there is the case, then I have calculated that the result is $P(\text{I win})=P(\text{I lose})=\dfrac{8}{17}$ and $P(\text{Tie})=\dfrac{1}{17}$. This sounds reasonable because the probability to win equals to probability to losing by symmetry argument. However, if I define a new parameter for the tendency of a player to play a larger value card as $p$, then I should get a new function $f(p,q)$ for my probability to win, where $q$ is the tendency of my opponent. Note that $0\leq p,q\leq 1$. (Why define such parameter? Because everyone is not guaranteed to play any card equally likely). This changed the probabilistic assumptions, and I intended to do so. But now I have no idea to calculate $f(p,q)$ because the sample space involved is too large. Say a quick example,
$$\begin{align*}
P(&\text{I win with a }4)\\
&=P(\text{4 being the smaller card and I choose it})P(\text{win }\lvert\text{ 4 being the smaller card and I choose it})\\&\quad +P(\text{4 being the larger card and I choose it})P(\text{win }\lvert\text{ 4 being the larger card and I choose it})
\end{align*}$$
Writing this seems helpless to solve the problem? How do I proceed the next?

With the help of python, the function is
\begin{align}
f(p,q)=\dfrac{564}{1225}pq+\dfrac{5137056}{6497400}p(1-q)+\dfrac{1110432}{6497000}(1-p)q+\dfrac{564}{1225}(1-p)(1-q)
\end{align}
 A: 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.)
A: You haven't been able to solve it because of the large number of cases involved.  Each player has $91$ possible hands.  So, you have to abstract some of the details.  Suppose first that the ranks of all four cards are different, say $r_1<r_2<r_3<r_4$  It's easy to compute the probability of this.  Then there are $6$ equally likely possibilities: player $1$ holds $r_1$ and $r_2$ or $r_1$ and $r_3$, etc.  Now you can compute the probability that he wins in each case.
The foregoing is wrong is the case that one player holds an Ace and the other a $2$, so you have to adjust for that.
Now we have to consider the case where some of the ranks are the same.  Consider first the case that two cards are of the same rank and the other two are of two other ranks.  Again, it's easy to figure out the probability that this occurs.  We have to consider the cases where the repeated rank is the lowest, middle, and highest rank separately.  Now there are $4$ possibilities for the hand that player $1$ holds, and they're not equally likely this time.  Again, we have to adjust the calculations to account for the case when one player has an Ace and another a $2$.
We still have to consider the cases where $3$ cards have the same rank, or where we have $2$ pairs of cards with the same rank, or where all cards have the same rank.
If you decide to do this, I suggest you do it first without the extra rule that $2$ beats Ace, writing it up carefully, and then adjust the calculations for the extra rule.
This will be an arduous, tedious task, and in my view, not a very instructive one.  I think you'd do better to look for a more interesting problem.  If you do decide to go ahead, though, I hope the foregoing will help.  If you need more details, please ask.
