Game Theory: Number Battle

Question

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?

Answer 1

There is no strategy that outperforms a player randomly selecting cards in either version of the game. Here's a way to see it.

Suppose your opponent declares a random strategy by shuffling her cards and laying them out, face down, in a row. From here on out, all choices are yours, including the choice of reshuffling your opponents cards, if you like. The first thing you realize, though, is that there is really no point in ever doing that, since all it would accomplish is to re-randomize an already random arrangement.

For the heck of it (and/or to free your hands so you can eat a sandwich), you lay your own cards out, face up, in a row matching your opponent's cards. You now implement your strategy by deciding which of your cards you want to play first. You play it against the face-down card across from it. After seeing who won the round (by turning your opponent's card over, in the main version of the game, or being told by a neutral third party who examines the face-down card, in the modified version), you continue to implement your strategy by deciding what to play next. You can, if you like, re-randomize your opponent's remaining cards, but again there is no gain in re-randomizing an already random arrangement, so you may as well once again play it against the card it's across from. This continues to the end of the game, at which point you realize that the outcome was essentially determined as soon as the cards were all laid on the table. That is, since there is no point in re-randomizing an already random arrangement, you might as well have simply played the cards as they lay across from each other at the outset, and done so in any order.

Answer 2

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.