Wise decision using game theory?

Question

There are 2 persons and two bags of oranges present in system.A bag is assigned to person.Each bag contains some oranges in range 1-10.After opening each person has been asked if they want to trade or not if they both say yes then trading will happen,if they contradict they will get whatever present in their respective bags(no exchange).What is the maximum number of oranges for which either player says yes in a Nash equilibrium?

Answer 1

There is a two-dimensional continuum of Nash equilibria in which both players have an arbitrary probability to say "yes" for $1$ and zero probability for all other numbers.

Now assume that at least one player has a non-zero probability of saying "yes" for at least one number other than $1$. If the other player never says "yes", she could improve her strategy by always saying "yes" for $1$. Thus both players sometimes say "yes". Let $n_i$ be the highest number for which the probability of player $i$ saying "yes" is non-zero. If $n_1\gt n_2$, player $1$ could improve her strategy by never saying "yes" for $n_1$. Thus $n_1=n_2=n$. If either player also had a non-zero probability of saying "yes" for some number less than $n$, the other player could improve her strategy by never saying "yes" for $n$. Thus both players have a non-zero probability of saying "yes" only for $n$. If $n\ne1$, then either player could improve her strategy by always saying "yes" for some number less than $n$. Thus $n=1$, in contradiction to the assumption.

The Nash equilibria with zero probability for all numbers other than $1$ are therefore the only Nash equilibria.