# Verification: modelling a game theory problem

Each of two sellers has an indivisible unit of a good. Seller 1 posts the price $p_1$ and seller 2 posts the price $p_2$. Each of two buyers would like one unit of the good, and no more. if both buyers approach the same seller they have 50% chance of trading. The buyer who misses out will not have the chance to trade with the other seller.

Each buyer’s preferences are represented by the expected value of a function that assigns the payoﬀ 0 to not trading and the payoﬀ $1−p$ to purchasing one unit of the good at the price $p$. So I have created a matrix model where the payoff for buyer 1 is the left side and the payoff for buyer 2 is the right side.

$$\begin{array}{c|lcr} & \text{seller 1} & \text{seller 2} & \\ \hline seller \quad 1 & \frac{1-p_1}{2},\frac{1-p_1}{2}& 1-p_1,1-p_2 \\ seller \quad2 & 1-p_2,1-p_1 & \frac{1-p_2}{2},\frac{1-p_2}{2} \\ \end{array}$$

Now I want to find the Nash equilibrium for any pair of prices where $0 \le p_1,p_2 \le 1$ in pure and mixed strategies.

Now there are only 3 cases, $p_1=p_2$, $p_1 \lt p_2$ and $p_1 \gt p_2$.

So all I would need is to look at each of the cases and compare the diagonals to see which value is greater. For example in the case where $p_1=p_2$ the pure nash equilibria would be for each buyer to approach different sellers, as any other movement would give a lower payoff. Is this approach correct?

Your reasoning is generally correct, and applies to all cases.

Some points you may want to pay attention to: 1) the case $p_1>p_2$ is a permutation for $p_1<p_2$ so you can study only two cases; 2) for $p_1=p_2$ you have found the two equilibria in pure strategies; there is a third one in mixed strategies.

• See that is where I got confused, suppose i assign probability 'a' to buyer 1 picking seller 1, and probability (1-a) to buyer 1 picking seller 2. Then how would I generate mixed equilbria? Aren't there too many unknowns? Commented Sep 2, 2016 at 13:30
• The probabilities in the randomised strategies should depend on the parameters $p_1,p_2$.
– mlc
Commented Sep 2, 2016 at 18:26
• Never mind I worked it out! Commented Sep 3, 2016 at 20:33