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.
Call the strategies of rock, paper, and scissors A, B, and C: C beats B beats A beats C.
Label the possible positions in this game with $2(n-1)$ dollars in the pot as either: $T_{n-1}$ if the previous result was a tie; $G_{n-1}$ if player I has a winning streak of 1 using strategy A (where A could be any of rock, paper or scissors); or $H_{n-1}$ if player I has a winning streak of 2 using strategy A (and will thus take the pot if he wins the next round using strategy A). We are using $n-1$ for the amount in the pot rather than $n$ for algebraic convenience when presenting the sub-game matrices.
(Clearly, this omits the cases where player II has a winning streak, but the values of those positions $G^{-}_{n-1}$ and $H^{-}_{n-1}$ are just the negatives of the values of the corresponding $G_{n-1}$ and $H_{n-1}$ positions.)
Let the value of any position in this game be the optimum game value minus the result that would happen if the players were to quit and immediately split the pot. Then by symmetry
$V(T_{n-1}) = 0$ for all $n >0$ and the optimal strategy for any $T_{n-1}$ position is the trivial one of choosing A, B, or C each with probability 1/3.
Because the value of $T_{n-1}$ is zero, in the position analyses the values and strategies are unaffected if we were to say we don't care how the $2(n-1$ dollars got into the pot, and that any tie ends the game by splitting the pot (and that the players. So the game overall for some value of $n-1$ is characterized by which of 4 positions you are at.
From $G_{n-1}$ the game may end (by a tie, in which case the players restart the game at $T_n$ but since that has zero value we don't care), transition to $H_{n-1}$ with player I gaining a dollar, or transition to $G^{-}_{n-1}$ with player II gaining a dollar.
From $H_{n-1}$ the game may end by a tie, transition to $G_{n-1}$ with player I gaining a dollar having won using strategy B or C, transition to $G^{-}_{n-1}$ with player II gaining a dollar, or end with player I winning the pot (gaining $n$ dollars altogether) by winning using strategy A.
For $n=1$, with no money in he pot, their is no advantage at all to having a winning streak, the position values are all zero, and the optimal strategies are the trivial equal-probability strategies. All the interesting features are in games where there is someting in the pot.
The two game matrices (for $n>1) are:
$$
G_{n-1} = \left(
\begin{array}{ccc}
0 & -g & h \\
+g & 0 & -g \\
-g & g & 0
\end{array}
\right)
$$
(where for convenience we introduce $g \equiv 1+V(G_{n-1})$ and $h \equiv 1+V(H_{n-1})$) and
$$
H_{n-1} = \left(
\begin{array}{ccc}
0 & -g & n \
+g & 0 & -g \
-g & g & 0
\end{array}
\right)
$$
$$
H_{n-1} = \left(
\begin{array}{cc|c}
0 & -g & n \\
+g & 0 & -g \\
-g & g & 0
\end{array}
\right)
$$
To solve $G_{n-1}$ we do the usual mantra of subtracting columns and taking determinants:
$$
G_{n-1}:
\begin{array}{ccc|cc|c}
0 & -g & h & g & -h & 3g^2 \\
+g & 0 & -g & g & 2g & g^2 + 2hg \\
-g & g & 0 & -26 & -g & 2g^2 + hg \\
\hline
-g & g & h+g & & & \\
g & -2g & h & & & \\
\hline
2g^2 + hg & g^2 + 2hg & 3g^2 & & &
\end{array}
$$
He game value is $\frac{hg-g^2}{3h+g6}$. So
$$
g = 1 + V(G_{n-1}) = 1 + \frac{hg-g^2}{3h+g6}
$$
Similarly, we solve $H_{n-1}$:
$$
H_{n-1}:
\begin{array}{ccc|cc|c}
0 & -g & n & g & -n & 3g^2 \\
+g & 0 & -g & g & 2g & g^2 + 2ng \\
-g & g & 0 & -26 & -g & 2g^2 + ng \\
\hline
-g & g & n+g & & & \\
g & -2g & n & & & \\
\hline
2g^2 + ng & g^2 + 2ng & 3g^2 & & &
\end{array}
$$
The game value is $\frac{hg-g^2}{3h+g6}$. So
$$
h = 1 + \frac{ng-g^2}{3n+g6}
$$
Before working with these two equations, we can look at the situation for $n$ very large. There, $ h = 1 + g/3 + O(1/n)$ and we can substitute to find that
$$
\begin{array}{l}
g = \frac{15+\sqrt{1053}}{46} \approx 1.031521 \\
V(G_{\infty}) \approx 0.031521 \\
h = 1 + g/3 \approx 1.34384 \\
V(H_{\infty}) \approx 0.34384
\end{array}
$$
That is, with a very large pot, if player I has a winning streak of 2 wins playing strategy A, player II will very rarely (probability $\frac{3g}{3N+6g}$) risk using strategy C, which could result in losing the whole pot. Player I will almost always choose strategies B (about 2/3 of the time) or C (about 1/3), and the game is favorable to player I with a value of $+\frac{1}{3}$. And if player 1 has just a 1 game winning streak, then the roughly 1.33 reward in the upper right hand corner (going over to $H_{n-1}$ with another A win) biases the game in player I's favor, but only by about 0.03.
Okay, now look at the case for $n$ not large enought to ignore $1/n$ effects: Substituting the expression for $h$ into the expression for $g$ we find
$$
40g^3 + (23n-21)g^2 - (15n+18)g - 9n = 0
$$
So for example, if there is $2 \times 1$ dollars in the pot ($n = 2$) then the value of the game to a player with a winning streak is $G_1 \approx 0.008118$ and the value of a two game winning streak is $G_1 \approx 0.082991$.
In game $G_1$ the strategy for player I is to choose A (the winning streak choice) to (B which beats A) to C in ratio
$$
3g : g + 2h : 2g + h = 3.024 : 3.172 : 3.099
$$
and in game $H_1$ the ratios are
$$
3g : g + 2n : 2g + n = 3.024 : 5.008 : 4.016
$$
The game values for $n = 11$ (ten dollars in the pot) are $G_{10} = 0.024413$ and
$H_{10} = 0.261048$; the strategies for player 1 in game $H_{10}$ are in ratio
$$
3.073 : 21.049: 1.024 \approx 1 : 7 : 4
$$
and for player II they are in ratio
$$
2n+g : n+2g : 3g = 21:024 : 12.049 : 3.073 \approx 7 : 4 : 1
$$
for A, B and C in that order.
Best Answer
Technically speaking, there's only one option, which is to keep stepping forward. Mathematical games, as far as I understand it, don't include not playing as an option. There's no way to optimize or formulate any strategy, because the other player in the game has perfect play as their only move, and always wins with perfect play.
So I guess the optimal strategy is to keep playing until you lose all your money, since it's the only strategy.
I would look at the definition of a game at this Wikipedia article