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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?

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I would never say yes (but I am not sure I'm in Nash equilibrium ;-)) Ok, I if have one orange in my bag I might say yes but I don't expect to get more than an orange back. – Fabian Apr 14 '12 at 21:05
I'll say no with ten in my bag, and so will the other. So if I have nine ... – Mark Bennet Apr 14 '12 at 21:24
Do the players like oranges? – joriki Apr 14 '12 at 22:02
@Matt: It's obvious that you say "no" for $10$ only because you have nothing to gain from saying "yes". But that's not enough to say something about all Nash equilibria; you might still say "yes" for $10$ if you have nothing to lose from doing so. – joriki Apr 14 '12 at 22:27
How are payoffs calculated? Expected payoffs are not really defined without a prior and you need them for the definition of equilibrium. – Michael Greinecker Apr 15 '12 at 7:03

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.

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The probability of saying yes by other person can be proportional to $(10-number\ of\ oranges\ in\ his\ bag)/10$ so in this way we can associate probability and with each player but i'm not able to get payoff matrix. – user997704 Apr 15 '12 at 8:16
@user997704: I don't understand that. What do you mean by "can be"? The question was specifically about what it can be in a Nash equilibrium, and I showed that it can't be non-zero for any number other than $1$ in a Nash equilibrium. – joriki Apr 15 '12 at 9:02
But you need analysis based on payoff matrix right?Your argument is correct but still it is considering arbitrary probability for user,i'm saying probability can be assigned as mentioned in first comment – user997704 Apr 15 '12 at 9:06
@user997704: I'm sorry, I'm unable to make any sense of what you're saying. There's no "user" mentioned in the question. As I mentioned in a comment under the question, I believe that the only assumption I'm making about the payoffs is that each of the numbers in the range actually has a non-zero probability of occurring. If you disagree with that, please point out specifically where in my answer you believe I'm assuming more than that. – joriki Apr 15 '12 at 9:31

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