A Bayesian Nash Equilibrium dictates an optimal strategy for each type of player. In this game A has only one type, and B has 151 types, each of which corresponds to one possible value for $x$, the monetary value of the company.
Since the game is only played once, these is no reason for B to refuse any offer worth more than $x$ in hopes of receiving a better offer in future. Therefore, the optimal strategy for each type of B is to accept any offer of at least $x+1$ euros, and can accept or refuse offers of value $x$ arbitrarily.
A does not know what type B is, but knows that they will accept any offer of $x+1$ euros. A therefore receives $1.5x - y$ euros if $y \geq x+1$, and recieves $0$ otherwise. A does not know whether B will accept or refuse an offer of $y=x$. If A assumes B will refuse such offers, then only deals where $x \leq y-1$ will be accepted. This means A's expected value is $\sum_{i=0}^{y-1}\frac{1.5i - y}{151}$, where $y-1$ is at most 150. If A assumes that B will accept offers where $x = y$, then the expected value is increased by $\frac{1.5y-y}{151}$, or $\frac{y}{302}$.
The value of the sum is $\frac{n(a_1 + a_2)}{2}$, where $n$ is the number of terms in the series, $a_1$ is the first term in the series, and $a_2$ is the last term in the series. $n = y$, $a_1 = \frac{-y}{151}$, $a_2 = \frac{1.5(y-1) - y}{151} = \frac{0.5y-1.5}{151}$.
The sum is $\frac{y(-y+0.5y-1.5)}{2*151} = \frac{-0.5y^2-1.5y}{302}$. Assuming that B chooses to accept or refuse the $x=y$ offer arbitrarily, half of $\frac{y}{302}$ must be added to the expected value, giving a total of $\frac{-0.5y^2-1.5y}{302} + \frac{0.5y}{302} = \frac{-0.5y^2-y}{302}$.
The maximal value of the quadratic is the point where the derivative is equal to 0.
$v=\frac{-0.5y^2-y}{302}$
$\frac{dv}{dx}=\frac{-y-1}{302}$
$0=\frac{-y-1}{302}$
$0=-y-1$
$y = -1$
The sign of the a coefficient in the quadratic is negative, so the value decreases to the right and left. At $y=1$, the lowest useful offer to make, the expected value is $\frac{-1.5}{302}$, a negative amount. As the curve decreases past $y=-1$, there will never be a more optimal strategy for A than to make an offer of 0 euros. The Nash Equilibrium, therefore, is for A to never offer B any money, and for B to refuse, except in the case where the business is worht 0 euros, in which case it accepts or refuses arbirarily.
Intutively, this makes sense, as A is always going to make a return of between 50% and -100% on their money, with an even distribution of probabilities between those two possibilites.
Erniemist
