This is based on a case from a question asking for the optimal strategy for the original $11$ card version of the game.
In short, I want to formulate this part of a case as a question on its own:
Imagine a game like rock-paper-scissors, but you have a Swordsman, a Rouge, and a Wizard. For simplicity lets label them with numbers $3$-$2$-$1$.
Instead of each beating the other, they work in the following way in a close up combat in broad daylight:
- $3$ always wins against the $2$ - rouge can't hide in the shadows during the day.
- $2$ always wins against the $1$ - rouge is quicker than the wizard in close up combat.
- The battle between $1$ and $3$ has a $\frac{1}{3}$ chance of resulting in win, lose or draw for either.
- outcome depends on the success of the spell the wizard attempts to use.- Two equal numbers result in a draw. ($3,3$) ($2,2$) ($1,1$)
Both players can send out any one of the three troop types, and the troop that wins gives a point to its master, both troops safely return home.
Now consider a set of $n$ games is being played. The overall winner is the one with the most points. Lets assume that a draw must be resolved by playing additional games until the winner is found.
If your opponent plays a troop at random, and you play a specific troop, you have the following chances to $\color{blue}{Win}$ / $\color{gray}{Draw}$ / $\color{red}{Lose}$ :
$$(3)\to \color{blue}{4}/\color{gray}{4}/\color{red}{1}$$
$$(2)\to \color{blue}{1}/\color{gray}{1}/\color{red}{1}$$
$$(1)\to \color{blue}{1}/\color{gray}{4}/\color{red}{4}$$
The optimal strategy would be to always play a $3$ - Swordsman, because you will either win, draw or have and equal chance to win, lose or draw.
But if both players follow the same "best method", the game will always be a draw forcing you into a endless loop of draw games. But we want the best method that will result in more wins for us than the opponent.
So to avoid this, we can add to our "best method" to play a $1$ if your opponent plays a $3$ for $k$-times in a row (Just after we are sure he won't play a $2$ to counter our $1$), and this pattern will result in $\frac{1}{3}$ chance of either winning,losing or drawing at the end.
But a better method would be to try to predict when the $1$ will be played and counter it with a $2$ to get a better than $1/3$ win rate (Not to be confused with a rate of scoring a "win" as a individual point). But if you fail of course, you give the advantage to the opponent.
So my question is, what would be the best method and a proof for it? The best method should have the best win rate against any other method including the exact copy of itself. If my method is the "best method", there are still some details to determine. Like how large should $k$ be and how often should it change to avoid collision between $1$ vs $1$, and a possible random play against a $2$?