They weren’t taking from me, they weren’t free riding. They just weren’t coordinating around my stag.
I’d like to note that with respect to my formal definition, defection always exists in stag hunts (if the social contract cares about everyone’s utility equally); see Theorem 6:
A 2×2 symmetric game is a Stag Hunt when R>T≥P>S. In Stag Hunts, due to uncertainty about whether the other player will hunt stag, players defect and fail to coordinate on the unique Pareto optimum (Stag1, Stag2). In (b), player 2 will defect (play Hare2) when P(Stag1)<12. In Stag Hunts, formal defection can always occur against mixed strategy profiles, which lines up with defection in this game being due to uncertainty.
What’s happening when P(Stag1)<12 is: Player 2 thinks it’s less than 50% probable that P1 hunts stag; if P2 hunted stag, expected total payoff would go up, but expected P2-payoff would go down (since some of the time, P1 is hunting hares while P2 waits alone near the stags); therefore, P2 is tempted to hunt hare, which would be classed as a defection.
If that’s a stag hunt then I don’t know what a stag hunt is. I would expect a stag hunt to have (2,0) in the bottom left corner and (0,2) in the top right, precisely showing that player two gets no advantage from hunting hare if player one hunts stag (and vice versa).
T >= P covers both the case where you’re indifferent as to whether or not they hunt hare when you do (the =) and the case where you’re better off as the only hare hunter (the >); so long as R > T, both cases have the important feature that you want to hunt stag if they will hunt stag, and you want to hunt hare if they won’t hunt stag.
The two cases (T>P and T=P) end up being the same because if you succeed at tricking them into hunting stag while you hurt hare (because T>P, say), then you would have done even better by actually collaborating with them on hunting stag (because R>T).
I’d like to note that with respect to my formal definition, defection always exists in stag hunts (if the social contract cares about everyone’s utility equally); see Theorem 6:
What’s happening when P(Stag1)<12 is: Player 2 thinks it’s less than 50% probable that P1 hunts stag; if P2 hunted stag, expected total payoff would go up, but expected P2-payoff would go down (since some of the time, P1 is hunting hares while P2 waits alone near the stags); therefore, P2 is tempted to hunt hare, which would be classed as a defection.
If that’s a stag hunt then I don’t know what a stag hunt is. I would expect a stag hunt to have (2,0) in the bottom left corner and (0,2) in the top right, precisely showing that player two gets no advantage from hunting hare if player one hunts stag (and vice versa).
Also note that you can indeed make that entry (2,0) by subtracting 1 from every payoff in the game. The same arguments still hold.
T >= P covers both the case where you’re indifferent as to whether or not they hunt hare when you do (the =) and the case where you’re better off as the only hare hunter (the >); so long as R > T, both cases have the important feature that you want to hunt stag if they will hunt stag, and you want to hunt hare if they won’t hunt stag.
The two cases (T>P and T=P) end up being the same because if you succeed at tricking them into hunting stag while you hurt hare (because T>P, say), then you would have done even better by actually collaborating with them on hunting stag (because R>T).
I see, thx.