I have the following problem. It is about how to analyze the coin game and finally find its Nash equilibrium.
The coin game: 2 players each get $n$ coins and will play $k$ rounds. Each round each of the players places an arbitrary number of their remaining coins (placing $0$ coins is also allowed). Simultaneously they show each other their coins. The player who placed more wins the rounds and get one point. The placed coins will be removed. The game continues that way till the last round ($k$-th round) is played. At the end of the game all coins must be placed. The player who won more rounds is the winner of the game.
This game is easy to display as an extensive form game with a game tree. Since the placing of the coins is simultaneous, it is a game with (sometimes) incomplete information. (The second player knows how much coins the first player had before the round, but does not know how much he puts down this round).
For finding the Nash equilibrium, I think I first have to translate the game into normal form. Is that right or is there an other way?
But how to do that? I tried just to note down all combinations of how a player can place $n$ coins in $k$ rounds and then generated a matrix. The entires represented the payoffs for each player. BUT this representation ignores that after each round the players are allowed to adapt their strategy since they then know how the other player played and know the number of remaining coins.
How would you try to solve this problem?
I did lot of research but couldn't find a case which was comparable to mine.
Edit: Thanks Henry. I tried out your suggestion, but did not get any further insights. I fixed $k$ = 3 and then focused only on what can happen in the second and third round starting with different numbers of coins for both players (other cases are trivial). After each round both players show the amound of their placed coins. And all coins have to be placed at the end of the game, so the last two rounds serve complete information and are easy to display in a normal form game (matrix). Each round itself is a simultaneously decision and so representable in a matrix. But how to display the complete game (e.g. $k$ = 3) in such a matrix is still a mystery to me. (But different literatures confirm that extensive and normal form games are equivalent, so it has to be possible.)
You do not want to represent the entire game in normal form; there are far too many possible strategies, even for moderate $n$ and $k$. Instead, you should represent the various subgames in normal form, and use that to find a subgame perfect Nash equilibrium. You can solve the entire game by solving each of these subgames, starting with the subgames closest to the end of the game and moving towards the earlier subgames.
Suppose we are at a state of the game where player $1$ has won $k_1$ rounds and has $n_1$ coins remaining, while player $2$ has won $k_2$ rounds and has $n_2$ coins remaining. Necessarily, $0\le k_1,0\le k_2,k_1+k_2\le k$, $n_1,n_2\in \{0,\dots,n\}$. We now want to find both player's optimal decisions in this subgame, which we will do by writing as a normal form game. Let $f(n_1,n_2,k_1,k_2)$ denote the expected score for player $1$ when both players play optimally. This is well-defined because this is a zero-sum game.
For the next round, player $1$ can use between $0$ and $n_1$ coins, while player $2$ has can use between $0$ and $n_2$ coins, so the situation can be written in normal form as a matrix with $(n_1+1)$ rows and $(n_2+1)$ columns. If player $1$ chooses to place $c_1$ coins, and player $2$ places $c_2$ coins, then they proceed to a different subgame where the expected score for player $1$ is $$ \begin{cases} f(n_1-c_1,n_2-c_2,k_1+1,k_2) & \text{if $c_1>c_2$}, \\ f(n_1-c_1,n_2-c_2,k_1,k_2+1) & \text{if $c_1<c_2$}, \\ f(n_1-c_1,n_2-c_2,k_1,k_2) & \text{if $c_1=c_2$}. \end{cases}\tag1 $$ The exception is the case where $k_1+k_2=k$. In this case, the game is over, and the output of $f$ is $1$ when $k_1>k_2$, the output is $-1$ when $k_1<k_2$, and the output is zero when $k_1=k_2$.
Therefore, the players' decision at this subgame is a normal-form game, where the payoffs are the solutions to simpler subgames. Provided you have already solved the subgames for each pair of options, you can use standard methods to solve this normal form game.
Let us look at a concrete example, say $n=10$ and $k=5$. You start by computing $f(n_1,n_2,0,4)$, $f(n_1,n_2,1,3)$, $f(n_1,n_2,2,2)$, $f(n_1,n_2,3,1)$, and $f(n_1,n_2,4,0)$. These all represent the final round of the game, where the unique optimal strategy for both players is to use all of their remaining coins, so they are easy to solve. Regardless of $n_1$ and $n_2$, we find $$ f(n_1,n_2,0,4)=-1,\quad f(n_1,n_2,1,3)=-1,\\ f(n_1,n_2,3,1)=+1,\quad f(n_1,n_2,4,0)=+1. $$ The last case, $f(n_1,n_2,2,2)$ depends on which of $n_1$ or $n_2$ is greater, or if they are tied.
Next, you would solve the simpler cases $f(n_1,n_2,0,3)$, $f(n_1,n_2,1,2)$, $f(n_1,n_2,2,1)$, and $f(n_1,n_2,3,0)$. Two of these have shortcuts; we know $f(n_1,n_2,0,3)=-1$, since player $2$ has an insurmountable lead, and similarly $f(n_1,n_2,3,0)=+1$. Let us look at the matrix for $f(3,5,2,1)$. $$ \begin{array}{c|cccccc} \begin{array}{cc}&\text{P2}\\\text{P1}&\end{array} &0&1&2&3&4&5 \\\hline 0&0&-1&0&1&1&1 \\ 1 & 1&0&-1&0&1&1 \\ 2 & 1 &1 &0&-1&0&1 \\ 3 & 1 & 1 & 1 & 0 & -1 & 0 \end{array} $$ This matrix was filled out using $(1)$, but I can offer some strategic commentary. Player $2$ is losing by one round, so the only hope of player $2$ winning is to beat player $1$ by exactly one coin this round, so they have enough leftover to win the next round. If player $2$ ties this round, they win the next round and tie the whole game. If player $2$ plays $2$ more coins than player $1$ plays, then player $2$ wins this round, but they tie the next round and therefore tie the whole game. In all other case, player $1$ wins. The strategies for player $2$ of placing $0$ coins or $5$ coins are both dominated, so we can reduce this to a $4\times 4$ matrix, which then must be solved via the usual linear programming methods.