Determine probability if player wins game Two players play a game with a infinite deck of cards. This deck consisting of these types of cards:


*

*Queen of Spades

*Jack of Spades

*King of Hearts

*Queen of Hearts

*Jack of Hearts

*Two of Clubs

*Three of Clubs

*Four of Clubs

*Five of Clubs

*Six of Clubs


(There are 10 different card types)
Player 1 starts and the two players take turns. When it is $P_1$'s turn they cheat and pick 2 cards uniformly at random, and when it is $P_2$'s turn they pick 1 card uniformly at random.
If $P_1$ picks a card that is of the suit Spades $P_1$ won. If $P_2$ picks a card that is of the suit Spades or Hearts $P_2$ wins. What is the probability of $P_1$ winning what is the probability of $P_2$ winning.
I am trying to come up with formulas for these events and the sample sapce but I am stuck
I was thinking that I can do something like this in the event (A) that $P_1$ wins: 
A = $P_1$ wins -> {$Not$ $Spade^n$, $Spade^m$; $n\ge0$, $m>0$}
B = $P_2$ wins -> {$Club^n$, $Spade$ $or$ $Heart^m$; $n|ge0$, $m>0$}
Any ideas if I am on the right track?
 A: For brevity, write A if Player A gets at least one spade when it is her turn, and X if she doesn't. Write B if Player B gets a spade or heart when it is her turn, and Y if she doesn't.  Then here is a beginning of the infinite list of ways Player A can win the game:
$$\text{A; XYA; XYXYA; XYXYXYA; XYXYXYXYA}$$
The probability A does not get a spade when it is her turn is $(0.8)^2$. So the probability she wins when it is her turn is $1-(0.8)^2$, which is $0.36$.
Thus $\Pr(\text{X})=0.64$. It is easy to see that $\Pr(\text{Y})=\frac{1}{2}$, so the probability of XY is $0.32$.  Thus the probability A ultimately wins  is
$$0.36+(0.32)(0.36)+(0,32)^2(0.36)+(0.32)^3(0.36)+\cdots.$$
This is an infinite geometric series, first term $0.36$, common ratio $0.32$. By a standard formula it has sum $\frac{0.36}{1-0.32}$.
Another way: We condition on the result of the first pick. Let $p$ be the probability A ultimately wins. She can win in two ways, (i) immediately or (ii) later. The probability of an immediate win is $0.36$.
Winning later happens if A and B strike out on their first turns. If that happens, A has probanility $p$ of ultimately winning. Thus
$$p=0.36+(0.32)p.$$
Solve for $p$.
Remark: The "words" for which A wins can be described in standard shorthand as $(\text{XY})^n\text{A}$, $n\ge 0$.
