A game of investment - maximizing the chances of winning

Question

This game is based on the concept of finance where we save part of our money and invest the rest hoping to earn a profit. But there is always a risk that the investment might return a loss.

The Game

1. Cards which do not have numbers are not used e.g. ace, king, queen, joker. For the remaining cards, regardless of the color and shape, the value of a card is equal to the number on it. The cards are shuffled and distributed between the players. They get equal number of cards.
2. In every round, both player pick one card each keep it aside as savings without showing the opponent. Then both invest one card each from their unused deck and place it on the table upside down. After this they show their invested cards.
3. The player with the bigger number on the invested card wins the round. The round winner retains his/her invested card (principal) as well as the opponent's invested card (profit). These two cards are added to the round winner's savings.
4. If the number on the invested cards are equal, the round is a draw and both players save their invested cards. Thus at the end of a round, all invested card ends up on somebody's savings
5. The game continues with the rest of unused cards until there is no unused card left. At the end, the player with the greater net worth or total savings wins the game

Strategic dilemma: If we save all the big cards and invest the small cards then the chances of wining rounds with the small cards and making a profit will be slim. If we save all the small cards and invest the bigger cards then the the losses would be heavy e.g. if $A$ invests a $9$ and $B$ invests a $10$, then $B$'s earnings in the round is $9+10 = 19$ points which is a big profit for $B$ and a heavy loss for $A$. Ideally $A$ would hope to invest a small card such as a $2$ when $B$ invests a big card so that $B$ does not make a big profit. Thus in general, a player would have to optimize the balance between investment and saving in each round based on the profits and losses made so far.

Question 1: Assume that the opponent plays randomly without any strategy. Is there a mathematically optimal strategy that maximizes your chances of winning?

Note: This game is an improvement of earlier version of the game mentioned in this post.

Answer 1

Given $a$ to be your saved card, and $b$ to be your played card.
The value gained, on average, is $$\overbrace{a}^{\text{savings}} \cdot \underbrace{1}_{\text{ chance}} \overbrace{(b+c)}^{\text{investment is profitable}}\cdot \underbrace{\frac{b-1}{10}}_{\text{chance}}+ \overbrace{b}^{\text{draw}} \cdot \underbrace{\frac{1}{10}}_{\text{chance}}$$ Where $c$ is the number the opponent put, on average(in the winning case) it is equal to $\frac{b+1}{2}$.(the average of values bigger then $0$ but smaller then $b$) This simplifies the expression to $$a+ \frac{(\frac{3}{2}b + \frac{1}{2})(b-1)}{10}+\frac{b}{10}$$ To judge if it’s better to choose we need to find for when the expression is bigger. $$a+ \frac{(\frac{3}{2}b + \frac{1}{2})(b-1)}{10}+\frac{b}{10} > b+ \frac{(\frac{3}{2}a + \frac{1}{2})(a-1)}{10}+\frac{a}{10}$$

This can be simplified to $$38a-30a^2+9> 20b - 3b^2$$ Which is a test you can run on the cards you got to find the optimal play.