Here’s a game-theory game I don’t think I’ve ever seen explicitly described before: Vicious Stag Hunt, a two-player non-zero-sum game elaborating on both Stag Hunt and Prisoner’s Dilemma. (Or maybe Chicken? It depends on the obvious dials to turn. This is frankly probably a whole family of possible games.)
The two players can pick from among 3 moves: Stag, Hare, and Attack.
Hunting stag is great, if you can coordinate on it. Playing Stag costs you 5 coins, but if the other player also played Stag, you make your 5 coins back plus another 10.
Hunting hare is fine, as a fallback. Playing Hare costs you 1 coin, and assuming no interference, makes you that 1 coin back plus another 1.
But to a certain point of view, the richest targets are your fellow hunters. Preparing to Attack costs you 2 coins. If the other player played Hare, they escape you, barely recouping their investment (0 payoff), and you get nothing for your boldness. If they played Stag, though, you can backstab them right after securing their aid, taking their 10 coins of surplus destructively, costing them 10 coins on net. Finally, if you both played Attack, you both starve for a while waiting for the attack, you heartless fools. Your payoffs are symmetric, though this is one of the most important dials to turn: if you stand to lose less in such a standoff than you would by getting suckered, then Attack dominates Stag. My scratchpad notes have payoffs at (-5, −5), for instance.
To resummarize the payoffs:
(H, H) = (1, 1)
(H, S) = (1, −5)
(S, S) = (10, 10)
(H, A) = (0, −2)
(S, A) = (-10(*), 20)
(A, A) = (-n, -n); n <(=) 10(*) → A >(=) S
So what happens? Disaster! Stag is dominated, so no one plays it, and everyone converges to Hare forever.
And what of the case where n > 10? While initially I’d expected a mixed equilibrium, I should have expected the actual outcome: the sole Nash equilibrium is still the pure all-Hare strategy—after all, we’ve made Attacking strictly worse than in the previous case! (As given by https://cgi.csc.liv.ac.uk/~rahul/bimatrix_solver/ ; I tested n = 12.)
Here’s a game-theory game I don’t think I’ve ever seen explicitly described before: Vicious Stag Hunt, a two-player non-zero-sum game elaborating on both Stag Hunt and Prisoner’s Dilemma. (Or maybe Chicken? It depends on the obvious dials to turn. This is frankly probably a whole family of possible games.)
The two players can pick from among 3 moves: Stag, Hare, and Attack.
Hunting stag is great, if you can coordinate on it. Playing Stag costs you 5 coins, but if the other player also played Stag, you make your 5 coins back plus another 10.
Hunting hare is fine, as a fallback. Playing Hare costs you 1 coin, and assuming no interference, makes you that 1 coin back plus another 1.
But to a certain point of view, the richest targets are your fellow hunters. Preparing to Attack costs you 2 coins. If the other player played Hare, they escape you, barely recouping their investment (0 payoff), and you get nothing for your boldness. If they played Stag, though, you can backstab them right after securing their aid, taking their 10 coins of surplus destructively, costing them 10 coins on net. Finally, if you both played Attack, you both starve for a while waiting for the attack, you heartless fools. Your payoffs are symmetric, though this is one of the most important dials to turn: if you stand to lose less in such a standoff than you would by getting suckered, then Attack dominates Stag. My scratchpad notes have payoffs at (-5, −5), for instance.
To resummarize the payoffs:
(H, H) = (1, 1)
(H, S) = (1, −5)
(S, S) = (10, 10)
(H, A) = (0, −2)
(S, A) = (-10(*), 20)
(A, A) = (-n, -n); n <(=) 10(*) → A >(=) S
So what happens? Disaster! Stag is dominated, so no one plays it, and everyone converges to Hare forever.
And what of the case where n > 10? While initially I’d expected a mixed equilibrium, I should have expected the actual outcome: the sole Nash equilibrium is still the pure all-Hare strategy—after all, we’ve made Attacking strictly worse than in the previous case! (As given by https://cgi.csc.liv.ac.uk/~rahul/bimatrix_solver/ ; I tested n = 12.)