1
A payoff as a function of the others
Fix a herder and count how many of the other herders restrain. Call the count k. Then the herder's payoff from restraining is C(k) and from adding is D(k), and the whole game is those two lists.
Lesson · Games and the Social Contract · Many players
The Tragedy of the Commons
Ten herders share a pasture, and each can keep a single animal or add a second. An added animal pays its owner in full and costs every herder a share of the grass, so adding is best whatever the others do. The lab finds the one equilibrium, the best total for the group, and the gap between them. No herder needs to be greedy for this to happen.
my animal's gain is mine, its cost is shared
adding dominates, so every herder adds
the equilibrium falls short of the best
make the adder pay and the gap closes
A gain kept and a cost shared
Core ideas
What this lesson establishes
Writing a herder's payoff as a function of the others is the new skill; the three ideas below set it up.
2
The equilibrium and the optimum are different questions
The equilibrium asks where no herder gains by changing alone. The optimum asks where the herders' payoffs sum to most. In a dilemma these are different answers, and the gap is the tragedy.
3
A cost on others is not a wish for more
Greed would mean wanting more animals than the pasture can carry. The tragedy needs only that the herder who adds does not pay the whole cost of adding, which is a fact about who bears the grass, not about the herder.
The material
Ten herders and one pasture
A payoff for each choice, the equilibrium it leads to, and the optimum it misses.
Ten herders graze one pasture. Each keeps one animal and may add a second. Every animal on the pasture yields 22 less the number of animals grazing there, so each animal added makes every animal, the herder's own included, worth one less. A herder owns either one animal or two.
Count the others who restrain, k, from 0 to 9. If I restrain, the pasture carries ten animals plus the second animals of the nine other herders who add, which is 2·n − 1 − k in all, with n = 10 herders, and my one animal is worth 22 − (2·n − 1 − k). If I add, the pasture carries 2·n − k animals, and my two are each worth 22 − (2·n − k).
C(k) = 22 − (2·n − 1 − k)
D(k) = 2·(22 − (2·n − k))
With ten herders these are C(k) = 3 + k and D(k) = 4 + 2·k. The difference is D(k) − C(k) = 1 + k, which is positive for every k. Adding pays more than restraining whatever the others do, so adding dominates. The lab's dominance tile says so, and it reports one equilibrium, with no herder restraining. A lone restrainer among nine adders would get 3 where adding gets 4, so no herder can improve alone.
Compare the equilibrium with the best the group could do. Each herder gets 4 there, 40 in all. If all ten restrain, each gets 12 and the total is 120. The lab searches every number of restrainers for the largest total, and finds it at nine restraining and one adding, 121, because the first extra animal does little harm to a pasture that is otherwise lightly stocked. The point survives: the equilibrium is far below the optimum.
What one herder gains and what everyone loses
A herder who adds while the other nine restrain gets D(9) = 22 against C(9) = 12, a gain of 10. If every herder does the same thing, each gets 4 where all restraining would have given 12, a loss of 8 each. The lab prints both in the tile for what defecting gains. The gain is private and the loss is shared, and that asymmetry is the whole of the tragedy.
Make the adder pay
Put a fee of 11 on every animal added. The most a herder ever gains by adding is 10, when the other nine restrain, so 11 is more than anyone gains in any case. Restraining now pays more at every k, and the only equilibrium is all ten restraining. A fee of exactly 10 is the border, and the lab's fourth preset sits on it: a herder facing nine restrainers is then indifferent, so the equilibria tile lists nine restrainers as well as ten, while the dominance tile already says restraining dominates, because this lab counts dominance the way “The Decision Matrix and Dominance” did, at least as good everywhere and better somewhere, where the two-player lab of “The Prisoner's Dilemma” counted only strict dominance. The herders' wishes are the same as before: the cost of the extra animal has been moved onto the herder who adds it. Giving each herder his own pasture would do the same.
Three herders on the same pasture do not reach a tragedy. Adding is still best for each, but three herders do not crowd the grass enough for the equilibrium to fall below the optimum. The lab finds the equilibrium total, 96, to be the best total as well, and it shows that each herder would even get 13 more if all three added than if all three restrained. The payoffs are written with the number of herders in them, so the slider shows where the gap opens: at five herders the equilibrium is still the best total, and at six it is not. It is a fact about how large the cost to others is compared with the gain, and a lightly used commons can be left alone.
Two limits. The herders here cannot talk, and real commons are often managed by their users with rules that change the payoffs; Elinor Ostrom's work on such arrangements is the reason this lesson claims no more than a conditional. And the payoffs are chosen: a pasture that regrows differently is a different table.
Interactive
The many-player dilemma
payoffs as exact expressions in k, and what follows from them
Method
Writing a commons as a game
Four steps, and the last is the comparison the lesson exists for.
1
Fix one herder and count the others who restrain
Call the count k. Everything the herder needs to know about the others is that number.
2
Write what restraining and adding pay at each k
Count the animals on the pasture in each case and value each at the pasture's yield per animal. These are C(k) and D(k).
3
Compare the two lists
If adding pays at least as much at every k and more at some, adding dominates, and the only equilibrium is that nobody restrains.
4
Add up the herders' payoffs for each number of restrainers
The largest sum is the optimum. If it exceeds the sum at the equilibrium, there is a gap, and that gap is what a fee or an owner is for.
Gain, cost and gap
Score 0 / 0Choose the answer you can justify.
Ten herders at 22 less the animals
The pasture yields 22 less the number of animals grazing, with ten herders.
k others restrain: restrain pays 3 + k, add pays 4 + 2k
Add minus restrain = 1 + k, positive for every k
So every herder adds: 10 x 4 = 40 in all
A lone adder among nine who restrain: 22 against 12
All ten restrain: 10 x 12 = 120
Best total, nine restrain and one adds: 121
No herder, looking at the nine others adding, would restrain: restraining pays 3 and adding pays 4. The total of 40 is a third of what restraint would give, and each step that led there was individually sensible.
Common mistakes
1
Thinking the tragedy requires greedNo herder in the table wants more than one extra animal, and the herders of the regulated preset want exactly the same. The equilibrium is bad because the cost of the extra animal falls on the nine others: add the fee that moves the cost to the adder and all ten restrain. Greed would be a change in wishes; the tragedy needs only a change in who pays.
2
Mistaking the equilibrium for the best the herders can doNo herder gains by changing alone at four each, and the total of 40 is still below the 120 of all restraining and the 121 of the best split. Stability is not merit, as in “The Prisoner's Dilemma”.
3
Thinking every shared resource is a tragedyThree herders on the same pasture have no gap between the equilibrium and the optimum, since three animals added do not crowd it. The lab's verdict depends on the size of the cost to others in comparison with the gain, and different payoffs can remove it.
Check your understanding
✓
Finish when you can find the gap and say what closes it.
Write each herder's payoff as a function of how many others restrain, find the dominance verdict and the equilibrium, find the best total, state the difference, and choose a fee that makes restraint dominate.
Note: The yield rule here, a fixed number less the animals grazing, is the simplest that makes each animal worse for all the others. Drag the number of herders from three up to ten and read where the equilibrium total stops matching the best total. The last tile, the share of cooperators after some generations, is the replicator step of “The Evolution of Cooperation”; with the generations at zero it repeats the starting share.