The seven matrices
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The seven matrices
Every variant names two integer K x K matrices, rowPay and colPay. For
every symmetric variant colPay = transpose(rowPay); only
bach-or-stravinsky differs. Each cell below is row payoff / column payoff.
running-with-scissors (default) — K = 3, rock, paper, scissors
| rock | paper | scissors | |
|---|---|---|---|
| rock | 0 / 0 | -3 / 3 | 3 / -3 |
| paper | 3 / -3 | 0 / 0 | -3 / 3 |
| scissors | -3 / 3 | 3 / -3 | 0 / 0 |
Cyclic and zero-sum, scaled x3 so one clean win pays the same 3 as mutual cooperation in the prisoners' dilemma. No fixed policy survives: whatever you commit to is visible in your inventory before the beam lands, so feints and token denial are the game.
prisoners-dilemma — K = 2, cooperate, defect
| cooperate | defect | |
|---|---|---|
| cooperate | 3 / 3 | 0 / 5 |
| defect | 5 / 0 | 1 / 1 |
Conditional cooperation with strangers. Defect strictly dominates per encounter; a room that manages to hold a cooperative convention out-earns a room that does not.
chicken — K = 2, dove, hawk
| dove | hawk | |
|---|---|---|
| dove | 3 / 3 | 1 / 4 |
| hawk | 4 / 1 | 0 / 0 |
Anti-coordination: T = 4 > R = 3 > S = 1 > P = 0, and mutual hawk is the crash, 0 / 0. Who yields — with no words, because there is no inter-seat channel.
stag-hunt — K = 2, stag, hare
| stag | hare | |
|---|---|---|
| stag | 4 / 4 | 0 / 2 |
| hare | 2 / 0 | 2 / 2 |
Assurance. Stag/stag is payoff-dominant, hare is risk-dominant, and the whole question is whether the room can trust the other cog to have bought stag.
bach-or-stravinsky — K = 2, bach, stravinsky, ASYMMETRIC
| bach | stravinsky | |
|---|---|---|
| bach | 3 / 2 | 0 / 0 |
| stravinsky | 0 / 0 | 2 / 3 |
The row (blue) camp — slots 0-3 — is paid 3 for Bach; the column (orange) camp — slots 4-7 — is paid 3 for Stravinsky. Interactions only resolve BETWEEN camps: a beam at your own camp is a no-contest.
pure-coordination — K = 3, red, green, blue
The identity matrix: all three matches pay 1 / 1, everything else pays 0 / 0. Common interest with no conflict at all — the only problem is agreeing.
rationalizable-coordination — K = 3, bronze, silver, gold
The diagonal pays 1 / 1, 2 / 2 and 3 / 3; everything else pays 0 / 0. Same coordination problem, but now one convention is strictly better than another, so a room that settles on bronze has left money on the table.
What is derived
Two best-response tables are computed once per variant and shipped in every seat's observation, so a seat never has to work the counter out itself:
bestResponseRow[j] = argmax_i rowPay[i][j]
bestResponseCol[i] = argmax_j colPay[i][j]
Ties go to the lowest index.