Standard games
Load one and the grid fills in:
How each part is computed
Pure-strategy equilibria
A cell is an equilibrium when neither player wants to move alone: the row player’s payoff is the largest in its column, and the column player’s payoff is the largest in its row. That is the whole test, and it is why the calculator marks best responses directly in the grid — a cell with both numbers green is an equilibrium, and you can see it without reading the answer.
Dominated strategies
A strategy is strictly dominated if some other strategy of the same player pays strictly more no matter what the opponent does. It can never be played in equilibrium, so it can be deleted — and deleting it may expose another dominated strategy, which is why the process is iterated. In the Prisoner’s Dilemma both players’ cooperative strategies disappear on the first round, which is the cleanest possible statement of why that game ends where it does.
Mixed-strategy equilibria
These are found by support enumeration. For every pair of candidate supports — the sets of strategies each player might mix over — the calculator solves the indifference conditions: if a player is genuinely mixing between several strategies, all of them must pay the same, otherwise they would drop the worse one. That gives a small linear system, which is solved exactly by Gaussian elimination, then checked: are all probabilities non-negative, and does any strategy left out of the support pay more? Only if both checks pass is it an equilibrium. The equations used are printed under each solution.
The method finds every equilibrium of a non-degenerate game. Degenerate games — where a player happens to be indifferent by accident, typically because payoffs are repeated — can have infinitely many equilibria, and any such continuum is reported by whichever isolated points of it the enumeration lands on. The calculator says so when it detects a singular system rather than quietly dropping it.
Equilibrium exists. That does not mean you get there.
Nash’s theorem guarantees that every finite game has at least one equilibrium in mixed strategies. It says nothing about whether players who learn from experience will end up there, and it is very easy to read the first statement as the second.
Rock–paper–scissors is the sharpest counterexample. Its only equilibrium is 1/3 on each move, forced by symmetry. Running two learners against each other for 200,000 rounds with a learning rate of 0.2, the strategies do not settle at the centre — they spiral outward and end up 3,050 times further from equilibrium than they started, finishing at probability 1.000 on a single move, which is the most exploitable strategy that exists. The time-average over the whole run was (0.3320, 0.3344, 0.3336) — the average converges beautifully while the actual play never does. The full experiment is here.