When no one has a dominant strategy, the hall of mirrors returns: your best move depends on theirs, theirs depends on yours, and the chain of “but then I’d switch, so then they’d switch…” seems to spin forever. In 1950, a 21-year-old graduate student named John Nash published a two-page paper that stopped the spin. His idea — the Nash equilibrium — is the concept the rest of game theory is built around, the one that won him a Nobel Prize and a film, and the single most useful thing in this entire course. It answers the question “where does the reasoning land?”
First, a smaller idea: the best response
Before the equilibrium, one building block. Your best response to a particular choice by the other player is simply the move that gives you the highest payoff given that they do that. It’s a conditional thing: “if they play Left, my best response is Top.” You compute it by fixing their column and picking your highest-payoff row in it.
Notice you can always name a best response, even in games with no dominant strategy — because now you’re answering an easier question. Not “what’s best, period?” (often unanswerable) but “what’s best if they do this specific thing?” (always answerable). The equilibrium is built entirely out of these.
The Nash equilibrium, defined
A Nash equilibrium is a combination of strategies — one for each player — in which every player is simultaneously playing a best response to the others. The famous, plain-language test:
A Nash equilibrium is an outcome where no player can do better by changing only their own move, given what everyone else is doing.
It’s a point of no unilateral regret. Stand in a Nash cell, look at your own option, and ask “could I switch to my other move and score higher, holding their choice fixed?” If the answer is no — and it’s no for every player at once — you’re in a Nash equilibrium. Nobody is held there by force or kindness; each is there because, given what the other is doing, it’s the best they can do. That mutual stability is why it’s the natural prediction for where rational play settles.
'Unilaterally' is the whole trick
The Nash test only lets each player change their own move, one at a time, while everyone else stays put. It does not ask “could we both switch and both do better?” That’s a different (and crucial) question — and the gap between the two is exactly what makes the prisoner’s dilemma so painful. Hold onto this distinction; it’s the hinge of the next lesson.
Finding it: the best-response method
Here’s the mechanical way to find every Nash equilibrium in a 2×2, and it’s worth doing by hand once. For the row player, go column by column and mark their best-response payoff in each. For the column player, go row by row and mark their best-response payoff in each. Any cell where both payoffs are marked is a Nash equilibrium — both players are best-responding at once.
The interactive matrix does exactly this for you: it rings each player’s best-response payoff and outlines any cell where both rings coincide as an NE. Here’s the standards game from earlier — two startups picking a connector format. Find the equilibria by eye first, then check against the rings.
Payoff matrix
Two equilibria: both-A and both-B
A ringed number is that player's best response to the rival's choice. A cell where both numbers are ringed is a Nash equilibrium — neither can improve by switching alone.
| Rival startup | ||
|---|---|---|
| Format A | Format B | |
| Your startupFormat A | NE33 | 11 |
| Format B | 00 | NE22 |
A ringed payoff is that player’s best response to the rival’s choice. A cell where both are ringed is a Nash equilibrium.
What the matrix says
Your startup — dominant strategy: none
Rival startup — dominant strategy: none
Nash equilibrium (pure): (Format A, Format A) · (Format B, Format B)
Check the top-left, (3, 3). Could you do better by switching to Format B while
they stay on A? That would drop you from 3 to 0 — no. Could they do better by
switching to B while you stay on A? From 3 to 1 — no. Neither wants to move, so
both-A is a Nash equilibrium. Run the same check on (2, 2) and you’ll find it’s an
equilibrium too. The two mismatch cells fail the test: in the top-right (1, 1),
the row player would rather drop to Format B (lifting their 1 to 2) and the column
player would rather jump to Format A (lifting their 1 to 3) — both are itching to
deviate, so it can’t be an equilibrium. In a mismatch, at least one player always
wants to bolt.
In the standards game, the players are stuck at the both-B equilibrium (2, 2), even though both-A (3, 3) would be better for everyone. Why doesn't the Nash concept just move them to the better cell?
Equilibria can be multiple, or absent
The standards game shows that a game can have more than one Nash equilibrium — here, two. When that happens, the equilibrium concept tells you the candidates for where play settles but not which one you’ll land on; that becomes a problem of coordination, focal points, and expectations (the subject of a later lesson).
Stranger still, some games have no Nash equilibrium in pure strategies at all — no single cell where both are best-responding.
Picture matching pennies. You and a friend each secretly flip a coin to Heads or Tails. If the two faces match, you win their penny; if they differ, they win yours. It’s pure conflict — what’s good for you is exactly bad for them.
Try to find a stable cell. Suppose you’re both on Heads (a match): you win, so they instantly want to switch to Tails. But once they’re on Tails (a mismatch), you want to switch to Tails too, to match again. Then they want Heads… The chase never settles into a cell — every outcome has someone itching to deviate. There is no pure-strategy Nash equilibrium.
The resolution, which Nash also proved, is a mixed strategy: each player randomises — here, flip genuinely 50/50. That is the equilibrium: when you’re truly unpredictable, your opponent can’t exploit you, and vice versa. Mixed strategies are why bluffing in poker, varying your serve in tennis, and randomised audits all make strategic sense. Nash’s great theorem: allow mixed strategies and every finite game has at least one equilibrium.
Nash equilibrium is a prediction, not a blessing
The biggest misreadings of “equilibrium” treat it as good, fair, or agreed. It is none of those. It’s the outcome that’s stable against lone defections — which can be collectively awful (next lesson), can be one of several (the standards game), or can require randomising (matching pennies). Read it as “where rational play gets stuck,” not “where everyone ought to want to be.”
Match each term to its precise meaning.
Pick a term, then click its definition.
Is the described outcome a Nash equilibrium?
Place each item in the right group.
- Both shops accept cards; neither earns more by switching to cash-only alone
- You play Format A while the rival plays B; you would gladly switch to B to match
- Everyone drives on the right; any lone driver switching to the left crashes
- In matching pennies, both pick Heads; the loser instantly wants to flip
- Both firms set the low price; either raising alone just loses customers
State the definition in your own words.
Pick the right option for each blank, then check.
A Nash equilibrium is an outcome in which no player can get a higher payoff by changing move, assuming the others keep theirs fixed — so it is against lone deviations, though not necessarily the best outcome for the group.
When to use it
Reach for the Nash equilibrium whenever there’s no dominant strategy and you need to predict where a strategic situation will actually settle — a negotiation, a competitive market, a standards war, a standoff. Find every cell that survives the “can anyone improve alone?” test. If there’s one, that’s your prediction. If there are several, the real question becomes which one — coordination. If there are none in pure moves, expect randomisation. And always remember the warning above: the equilibrium tells you where play gets stuck, not where anyone deserves to be. The most important case of all — a stable equilibrium that leaves everyone worse off — is so important it gets its own lesson. That’s the prisoner’s dilemma, next.