Game Theory: Number Battle There's a little game in which two players take part. The game consists of ten rounds. In each round, the two players simultaneously choose a number between $1$ and $10$ inclusive. The player with the larger number receives two points. If their numbers are equal, then each player gets one point. Once a player has used a number, they cannot pick it again. 
In reality, the game is played with poker cards so each player receives 1..10 from one color. Players place the cards face down and then turn them over once both players have picked a card and then the continue with the remaining cards. 
There's a modified version where a third person looks at the played cards so players do not learn what cards the opponent has played the third person will only announce who had the higher card or draw/tie.
The game ends after all cards have been played. The winner is the one with the highest score.
Is there a strategy that beats a player randomly selecting cards and does this change in the modified version?


Is there a strategy that beats a player randomly selecting cards?

No, it seems there isn't. We can see this by looking at all the possible game continuations and in each configuration deciding whether or not we can steer the game in our favor by making the correct choices. 
To illustrate this, let's look at a variant of the game where each player has a low number of $N$ cards from $1$ to $N$, say $N=2$ or $N=3$ (there are $(N!)^2$ possible paths through a game with $N$ cards, so anything more than $N=3$ becomes really tedious).
For $N=2$, we can have the following paths:
$$\begin{matrix} 1 & 2 \\ 1 & 2 \end{matrix}\rightarrow
\cases{
\underline{\begin{matrix} X & 2 \\ X & 2 \end{matrix} \rightarrow 2-2} \\ 
\underline{\begin{matrix} X & 2 \\ 1 & X \end{matrix} \rightarrow 2-2} \\
\underline{\begin{matrix} 1 & X \\ X & 2 \end{matrix} \rightarrow 2-2}\\
\begin{matrix} 1 & X \\ 1 & X \end{matrix} \rightarrow 2-2
}$$
Here, each row represent the cards of a player. The $X$'s represent what has been chosen. The final score is shown to the very right.
So we see that all game ends in a draw. Well, this was a boring case, one might object, since decision only comes into play after the first card has been removed (where neither player has any information to utilize), and the outcome of the game is fixed by then. Okay, let's do the same analysis in the case of $N=3$ then. I won't write out all the paths, but only show the possible outcomes of each of the $3^2$ possible positions after the first card has been removed:
$$\begin{matrix} 1 & 2 & 3 \\ 1 & 2 & 3 \end{matrix}\rightarrow
\cases{3-3 \\ 3-3 \;\vee\;2-4\;\vee\;2-4\;\vee\;3-3 \\ 3-3\;\vee\;4-2\;\vee\;4-2\;\vee\;3-3 \\ 3-3 \\ 2-4\;\vee\;3-3\;\vee\;3-3\;\vee\;2-4 \\ 4-2\;\vee\;3-3\;\vee\;3-3\;\vee\;4-2 \\ 3-3
}
$$
We see that if all outcomes are equally likely, each player will be expected to score the same on average.
In the case that there are four possible outcomes after the first card has been removed, can one not influence the game continuation in one's favor? No, since if you choose any card, it is equally likely you land in either one of the two different outcomes.
By induction (and just by noticing that we didn't gain any advantage even though we had more information about the state of the game), this can also be shown to be valid for any $N$. 

Does this change in the modified version?

No, since it doesn't matter what we do in the first place.
