There is no simplified description of the Nash equilibrium of this game.
You can compute the best strategy starting from positions where both players are about to win and going backwards from there.
Let $p(Y,O,P)$ the probability that you win if you are at the situation $(Y,O,P)$ and if you make the best choices. The difficulty is that to compute the strategy and probability to win at some situation $(Y,O,P)$, you make your choice depending on the probability $p(O,Y,0)$. So you have a (piecewise affine and contracting) decreasing function $F_{(Y,O,P)}$ such that $p(Y,O,P) = F_{(Y,O,P)}(p(O,Y,0))$, and in particular, you need to find the fixpoint of the composition $F_{(Y,O,0)} \circ F_{(O,Y,0)}$ in order to find the real $p(O,Y,0)$, and deduce everything from there.
After computing this for a 100 points game and some inspecting, there is no function $g(Y,O)$ such that the strategy simplifies to "stop if you accumulated $g(Y,O)$ points or more". For example, at $Y=61,O=62$,you should stop when you have exactly $20$ or $21$ points, and continue otherwise.
If you let $g(Y,O)$ be the smallest number of points $P$ such that you should stop at $(Y,O,P)$, then $g$ does not look very nice at all. It is not monotonous and does strange things, except in the region where you should just keep playing until you lose or win in $1$ move.
You can directly compute the optimal strategy and its expected value with a dynamic program (backward induction).
Consider the possible states of the game, which can be completely described by the (number of -1 cards remaining, number of +1 cards remaining), with 16 possibilities.
Arrange them in a square grid as follows (it might be better on paper if you make it a diamond with (3,3) on the left and (0,0) on the right)
$$\begin{array}{ccccccc}
\stackrel{(0)}{(3,3)} & \rightarrow & \stackrel{(1)}{(3,2)} & \rightarrow & \stackrel{(2)}{(3,1)} & \rightarrow & \stackrel{(3)}{(3,0)}\\
\downarrow && \downarrow && \downarrow && \downarrow\\
\stackrel{(-1)}{(2,3)} & \rightarrow & \stackrel{(0)}{(2,2)} & \rightarrow & \stackrel{(1)}{(2,1)} & \rightarrow & \stackrel{(2)}{(2,0)}\\
\downarrow && \downarrow && \downarrow && \downarrow\\
\stackrel{(-2)}{(1,3)} & \rightarrow & \stackrel{(-1)}{(1,2)} & \rightarrow & \stackrel{(0)}{(1,1)} & \rightarrow & \stackrel{(1)}{(1,0)}\\
\downarrow && \downarrow && \downarrow && \downarrow\\
\stackrel{(-3)}{(0,3)} & \rightarrow & \stackrel{(-2)}{(0,2)} & \rightarrow & \stackrel{(-1)}{(0,1)} & \rightarrow & \stackrel{(0)}{(0,0)}\\
\end{array}
$$
The entries are on top (score if you stop here), bottom (#-1, #+1) remaining in the deck.
The trick is to work backwards from the (0,0) corner and decide at each state whether you want to continue or not. Examples:
- There is no decision at (0,0), it is worth 0.
- At (0,1) the choice is between taking -1, or drawing a card which gets 0. Since drawing is better, we now know (0,1) is also worth 0.
- At (1,0) we take 1 instead of drawing which gets 0.
- At (1,1) a real decision comes up. Stopping is worth 0. Drawing gets you 1/2 chance to move to (1,0) [worth 1] and 1/2 chance to move to (0,1) [worth 0]. So drawing is worth 1/2 on average, and it is optimal to do so.
You can continue filling in all the states to find the optimal strategy. Note that unequal card counts matter: at say (1,2), drawing gives you 1/3 chance to move to (0,2) and 2/3 chance to move to (1,1).
The filled in square looks like:
$$\begin{array}{ccccccc}
\stackrel{17/20}{(3,3)} & \rightarrow & \stackrel{6/5}{(3,2)} & \rightarrow & \stackrel{\mathbf{2}}{(3,1)} & & \stackrel{\mathbf{3}}{(3,0)}\\
\downarrow && \downarrow && &&\\
\stackrel{1/2}{(2,3)} & \rightarrow & \stackrel{2/3}{(2,2)} & \rightarrow & \stackrel{\mathbf{1}}{(2,1)} & \rightarrow^{?} & \stackrel{\mathbf{2}}{(2,0)}\\
\downarrow && \downarrow && \downarrow^{?} &&\\
\stackrel{1/4}{(1,3)} & \rightarrow & \stackrel{1/3}{(1,2)} & \rightarrow & \stackrel{1/2}{(1,1)} & \rightarrow & \stackrel{\mathbf{1}}{(1,0)}\\
\downarrow && \downarrow && \downarrow &&\\
\stackrel{0}{(0,3)} & \rightarrow & \stackrel{0}{(0,2)} & \rightarrow & \stackrel{0}{(0,1)} & \rightarrow & \stackrel{\mathbf{0}}{(0,0)}\\
\end{array}$$
States where you stop have their value bolded. At (2,1) it doesn't matter if you draw or stop.
Since you have made value-maximizing choices at every step including the effects of later choices, Strategy 2 is proven optimal, with value exactly 17/20.
Best Answer
The expected value has to be at least $\frac 12$ because that is the value of the following strategy. Draw the first card, if P quit (winning $1$), if N draw all the others (breaking even). You can do better. If you have drawn an equal number of cards of each type you are at $0$ and the same logic applies, so you should draw. If you are ahead the expected value of the next draw is negative, so you should quit. The correct strategy is to draw any time you are even or behind and to quit if you ever hit $+1$ or run out of cards.
The number of sequences that get you $0$ are counted by the Catalan number $C_{10}=\frac 16{10 \choose 5}=42$ The number of sequences is ${10\choose 5}=252$, so the chance you get $0$ is $\frac 16$ and the value of the game is $\frac 56$