Consider the following strategic game played between two players:
Each player can demand from the other as much as desired between zero and one, but there is a penalty for being too greedy. The player who demands more than his opponent must pay a fine of $b$ to the other, where $b$ is a fixed number, $0\leq b \leq 1/2$. Thus we have the game $(X,Y,A)$ where $X$ is the set of strategies for player I, $Y$ is the set of strategies for player II and where $A(x,y)$ is the payoff function (player II pays $A(x,y)$ to player I). We have $X = Y = [0,1]$ and
$$A(x,y) = x - y + \begin{cases} +b \text{ if }x <y\\ 0 \text{ if }x = y\\ -b \text{ if }x > y \end{cases} $$
Solving this game apparently starts by recognising that for player I choosing $x \in [0, 1- 2b]$ is dominated by choosing $x = 1$, and similarly for player II choosing $y\in[0,1 - 2b]$ is dominated by choosing $y =1$, so that the game reduces to a game on $[1-2b,1]^2$.
I can see that choosing $x\in[0,1-2b]$ is dominated by $x = 1$. Suppose that we choose $x\in[0,1-2b]$. There are two cases: $x<y$ and $x>y$.
If $x<y$ we have $$ A(x,y) = x - y + b \leq 1 - 2b - y + b = 1 - b- y $$ and if $x > y$ we have $$ A(x,y) = x - y - b \leq 1 - 2b - y -b = 1 - 3b - y $$ Since the payoff for $x = 1$ is equal to $A(x,y) = 1 - b - y$, $x = 1$ dominates choosing $x\in[0,1 - 2b]$.
Question: How do you see that $x = 1$ dominates choosing $x\in[0,1-2b]$ beforehand? I can check that it is true, but I have no idea how to arrive at this conclusion myself. Do you just try a lot of values or is there a smart way to find out which interval of $X$ is dominated so that you can remove it from the game?
Intuitively, the maximally greedy strategy is to abuse the game! We demand just under $y$ to take bonus $b$ (assuming $b>0$), OR demand $1$, depending on which is better. If $b=0$, the game is obvious and everyone demands $1$.
Demanding just under $y$ assuming $b>0$ strictly dominates matching, since you essentially get $y+b>0$. If we're going to choose $x<y$ and take the bonus, we might as maximize $x$ as much as possible. And if we're going to choose $x>y$ and pay the penalty, we might as well choose $x=1$. We compare the two options and pick the one with the higher payoff.
You can check that this is the best strategy (for both players as the game is symmetric). There is no NE as the game cycles at choosing $0.5-b$ to choosing $1$.
For example, with $b=0.3$ and $y=0.3$, you choose $x\lessapprox0.3$ and receive $x+0.3\lessapprox0.6$. Your opponent does the same and this repeats until someone, say you, chooses $0.5-0.3=0.2$ for a payoff of $0.2+0.3=0.5$. Now picking $y\lessapprox0.2$ has payoff $y+0.3\lessapprox0.5$ while picking $y=1$ has payoff $1-0.2-0.3=0.5$, so $y=1$ will be chosen. It is now your turn and you of course choose just under $1$...