Payoffs question 
Suppose players can pay 1 unit of payoff to pick what kind of person they meet (ie. football fan or opera fan). How do I calculate the proportion of football fans in the long run?
 A: It seems a bit unrealistic that any number of people can pay to decide whom to meet no matter how many of those they want to meet are still around, but never mind...
Let the proportions of football fans and opera fans be $o$ and $f$, respectively. Then the four possible strategies yield the following payoffs:
opera without paying: $2o+6f$
football without paying: $0o+4f$
opera with paying: $5$
football with paying: $3$
So opera without paying dominates football without paying and opera with paying dominates football with paying, and thus eventually everyone will go to the opera after picking phantom football fans they can feel culturally superior to.
A: The obvious tactic is to pay 1 to meet a football fan and then go to the opera alone for a net payoff of +6-1 = +5.  So all those with a choice will become snobish solitary opera watchers.
I suspect that your question would be slightly more interesting if there was a free market and people had the choice between


*

*Pay $x_{00}$ and go to the opera with the other person (net $2-x_{00}$)

*Pay $x_{01}$ and go to the opera, sending the other to football (net $6-x_{01}$) 

*Pay $x_{10}$ and go to football, sending the other to the opera (net $-x_{10}$)

*Pay $x_{11}$ and go to football with the other person (net $4-x_{11}$)

*Accept $x_{00}$ and go to the opera with the other person (net $2+x_{00}$)

*Accept $x_{01}$ and go to football leaving the other to go to the opera (net $+x_{01}$) 

*Accept $x_{10}$ and go to the opera leaving the other to go to football (net $6+x_{10}$)

*Accept $x_{11}$ and go to football with the other person (net $4+x_{00}$)


and prices adjusting until the numbers match in pairs, i.e. $n_1 =n_5$, $n_2 = n_6$, $n_3 =n_7$, and $n_4 =n_8$.  
If all the net payoffs have to match where the numbers are strictly positive, you will find that $x_{11}=0$ and everybody goes to football (i.e. $n_4 = n_8 = n_{\text{total}}/2$).      
A: Note that this is an instance of the Prisoner's Dilemma.  The stable strategy is to always choose Opera, since you'll never have fewer "points" than the other person from doing so.  However, this ultimately leads both players to choose a less-than-optimal outcome.
Iterated prisoner's dilemmas are more interesting because they allow more strategy.  You can begin to incorporate an element of trust into your decision based on what your partner chose before.  Here, the maximum number of points can be gained if both players consistently choose Football, but each player must resist the chance to betray the other for short-term gain.
Circumstances like this appear everywhere, like evolutionary biology or politics.
Unfortunately, I don't think your new situation would lead to different behavior.  If players could choose what kind of person they could meet, then everyone would like to meet a Football spectator, then choose Opera for themselves.  Pretty quickly you would have no Football spectators, just like in the case where people have to decide what to pick without any knowledge of the other person.  Football playing just isn't a stable strategy, since no Football spectator personally benefits from their choice, only others.
I highly recommend The Selfish Gene by Richard Dawkins.  There is an excellent section there about modeling Prisoners' Dilemmas in evolutionary biology.
