Last lesson you learned to name a best response: the move that scores highest for you, given one specific thing the other player does. That’s a conditional answer — “if they play Left, I play Top” — and it’s always available, even when no single move dominates. This lesson takes that one building block and clicks two of them together. When both players are best-responding to each other at the same time, the reasoning stops spinning: nobody wants to move, so the game has found its resting point. That resting point is the Nash equilibrium, and it’s the single most important prediction in all of game theory. We’re going to define it with full precision, and then learn the mechanical recipe that finds every one of them in a 2×2 game — whether there’s one, several, or none.
The definition, precisely
A Nash equilibrium is a combination of strategies — exactly one for each player — in which every player is simultaneously playing a best response to the others. The plain-language test that matters most:
A Nash equilibrium is an outcome where no player can do better by changing only their own move, given what everyone else is doing.
Read that clause slowly, because every word is load-bearing. No player — the test must pass for everyone at once, not just one lucky person. Do better — strictly higher payoff, a genuine improvement. Changing only their own move — they alter their strategy and nobody else’s. Given what everyone else is doing — the others’ choices are frozen while we run the test. Put it together and a Nash equilibrium is a point of no unilateral regret: stand in the equilibrium cell, look only at your own options, ask “could I switch to my other move and score higher, holding their choice fixed?” — and the honest answer is no. And it’s no for every player simultaneously.
Here’s the clean way to hold it in your head, and it’s a direct sequel to last lesson. A best response is one player’s arrow pointing at their favourite reply. A Nash equilibrium is the cell where all the arrows point inward at once — a knot of mutual best responses. Each player is there not because they’re forced or being kind, but because, given what everyone else is doing, staying put is the best they can do. That mutual stability is exactly why it’s the natural forecast for where rational play settles: it’s the only kind of outcome no one has a private reason to abandon.
'Unilaterally' is the whole trick
The Nash test lets each player change their own move, one at a time, while everyone else stays frozen. It does not ask “could we all switch together and all do better?” That’s a completely different question — and the gap between the two is exactly what makes the next lesson’s dilemma hurt. A cell can be rock-solid against any lone deviation and still be an outcome every player would escape in a heartbeat if only they could move together. Hold onto “unilaterally”; it is the hinge the whole course turns on.
Finding it: the best-response method
Enough theory — here’s the mechanical recipe, the one worth doing by hand at least once so the definition stops being words and becomes a procedure. For any 2×2 game:
- Mark the row player’s best responses. Go column by column. Fix the column (that’s the assumption “the rival plays this”), then compare the row player’s two payoffs in that column and mark the bigger one. Repeat for the other column. You’ve now marked the row player’s best reply to each of the opponent’s moves.
- Mark the column player’s best responses. Go row by row. Fix the row, compare the column player’s two payoffs in that row, mark the bigger one. Repeat for the other row.
- Read off the equilibria. Any cell where both payoffs are marked is a Nash equilibrium — both players are best-responding at once. A cell with only one mark (or none) is not: whoever isn’t marked is itching to deviate.
The interactive matrix does exactly this, live. It rings each player’s best-response payoff and outlines any cell where both rings coincide with an NE badge. Below is the standards game — two startups each picking a connector format, where the payoff is highest when they match and worthless when they clash. Try to find the equilibria by eye using the three steps above, then check yourself against the rings. (It’s editable — nudge the numbers and watch the equilibria move.)
Standards / coordination game
Two equilibria: match on A, or match on B
Go column by column to mark the row player's best response, then row by row for the column player. 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)
Let’s walk the method on this game, out loud.
Row player, column by column. In the Format A column, your payoffs are 3 (top) versus 0 (bottom) — 3 wins, so mark the top. In the Format B column, your payoffs are 1 (top) versus 2 (bottom) — 2 wins, so mark the bottom. Column player, row by row. In the Format A row, the rival’s payoffs are 3 (left) versus 1 (right) — 3 wins, mark the left. In the Format B row, the rival’s payoffs are 0 (left) versus 2 (right) — 2 wins, mark the right. Now read it off: the top-left cell (A, A) has both marks, and the bottom-right cell (B, B) has both marks. Two Nash equilibria.
Now the worked check on a good cell versus a bad one, which is the same test from a second angle.
- Check (A, A) = (3, 3). Could you improve by switching to Format B while they hold on A? That drops you from 3 to 0 — no. Could they improve by switching to B while you hold on A? That drops them from 3 to 1 — no. Neither wants to move, from either side. It’s an equilibrium, and (B, B) = (2, 2) passes the identical test.
- Check the mismatch (A, B) = (1, 1). Here you’d rather drop to Format B to match them, lifting your 1 to 2 — you want to deviate. And they’d rather jump to Format A to match you, lifting their 1 to 3 — they also want to deviate. Two players itching to bolt means the cell fails. In any mismatch, at least one player always has a profitable lone switch, so no mismatch can ever be an equilibrium.
In the standards game, the players are stranded at the both-B equilibrium (2, 2), even though both-A (3, 3) is better for everyone. Why doesn't the Nash concept simply move them to the better cell?
Nash versus dominant strategy
You already have one tool for solving games — the dominant strategy, a move that beats its alternative no matter what the opponent does. How do the two ideas relate? Cleanly, and in one direction:
A dominant-strategy outcome is always a Nash equilibrium. But most Nash equilibria involve no dominant strategy at all.
The first half is a short logical step. If a strategy is a best response to everything the opponent might do, then in particular it is a best response to whatever they’re actually doing in this cell. So a player with a dominant strategy is automatically best-responding — the Nash test can’t catch them out. When both players have a dominant strategy, both are best-responding by definition, and the cell where their dominant moves meet is a Nash equilibrium, guaranteed. Dominance is the special, easy case; it hands you the equilibrium for free.
The second half is where beginners get stranded. The standards game above has no dominant strategy for anyone — your best move flips depending on the rival’s choice — and yet it has two equilibria. If you’d walked in hunting only for a dominant move, you’d have found nothing and concluded, wrongly, that the game has no solution. It has two. The Nash equilibrium is the general tool; the dominant strategy is a lucky shortcut that only some games hand you.
Never assume a dominant strategy exists
Dominance is a bonus, not a guarantee. Plenty of important games — every coordination game, every game of pure conflict — have no dominant strategy for either player, so “find the dominant move” comes up empty and leaves you stuck. The best-response method never comes up empty in that way: run it and you’ll always surface every pure equilibrium the game has, dominant strategies or not. Reach for dominance when it’s there; fall back on mutual best response — the real workhorse — when it isn’t.
One, several, or none
How many Nash equilibria does a game have? The honest answer — and it surprises people — is that it depends on the game, and all three counts really happen.
Exactly one. The prisoner’s dilemma has a single equilibrium, and it lands where both dominant strategies meet. Here each player is strictly better off defecting no matter what the other does (5 beats 3 if they cooperate, 1 beats 0 if they defect), so Defect is dominant for both. Their dominant moves collide at (Defect, Defect), and — since a dominant-strategy outcome is always a Nash equilibrium — that lone cell is the unique equilibrium. Notice the sting we’ll unpack next lesson: it’s the only stable outcome, yet both players would rather be at mutual cooperation (3, 3).
Dominant-strategy game
Exactly one: the prisoner's dilemma
Both players' best responses point to Defect in every column and every row. The single cell where both rings meet is the unique Nash equilibrium.
| Prisoner B | ||
|---|---|---|
| Cooperate | Defect | |
| Prisoner ACooperate | 33 | 05 |
| Defect | 50 | NE11 |
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
Prisoner A — dominant strategy: Defect
Prisoner B — dominant strategy: Defect
Nash equilibrium (pure): (Defect, Defect)
Several. The standards game, which you just solved, has two pure equilibria — (A, A) and (B, B). When a game has more than one, the equilibrium concept tells you the candidates for where play lands but not which one you’ll actually reach; that becomes a question of coordination, focal points, and shared expectations (a later lesson). Coordination games almost always come in this multi-equilibrium flavour.
None — in pure strategies. Some games have no cell where both players are best-responding. The cleanest example is matching pennies: each player secretly picks Heads or Tails; if the faces match, the row player wins; if they differ, the column player wins. It’s pure conflict — exactly what’s good for one is bad for the other — and the chase never settles.
Pure-conflict game
None in pure strategies: matching pennies
Follow the rings around the grid: no single cell ever has both. Every outcome leaves one player wanting to switch, so there is no pure-strategy Nash equilibrium.
| Mismatcher | ||
|---|---|---|
| Heads | Tails | |
| MatcherHeads | 1-1 | -11 |
| Tails | -11 | 1-1 |
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
Matcher — dominant strategy: none
Mismatcher — dominant strategy: none
Nash equilibrium (pure): none in pure strategies
Trace the rings and you’ll see them circle the grid forever without ever landing on the same cell. Suppose you’re both on Heads — a match, so the Matcher wins, and the Mismatcher instantly wants to switch to Tails. Once they’re on Tails (a mismatch), the Matcher wants Tails too, to match again. Then the Mismatcher wants Heads. The whole thing spins; every outcome has someone itching to deviate. There is no pure-strategy equilibrium at all.
That “none” case looks like a hole in the theory — and it was, until Nash filled it. Here is his great theorem, in words:
If you let players randomise — choose their moves according to probabilities rather than committing to one pure move — then every finite game has at least one Nash equilibrium.
A mixed strategy (randomising, e.g. flipping genuinely 50/50 in matching pennies) always restores an equilibrium: when you’re truly unpredictable, your opponent can’t exploit you, and vice versa. That’s why bluffing in poker, varying your tennis serve, and randomised tax audits are all strategically sound. The full treatment — how to compute the right randomisation — is lesson 05. For now, keep the headline: a game can have one pure equilibrium, several, or none, but allow mixing and there is always at least one.
Match each term to its precise meaning.
Pick a term, then click its definition.
Is the described outcome a Nash equilibrium? Run the test: from this cell, can any single player do better by changing only their own move?
Place each item in the right group.
- You ship Format A while the rival ships Format B; you would gladly switch to B to match them
- Everyone drives on the right; any lone driver switching to the left crashes
- Both startups ship Format A; either switching alone to B would clash and pay less
- Both prisoners defect; each is already best-responding, since defect beats cooperate whatever the other does
- In matching pennies, both pick Heads; the mismatcher instantly wants to flip to Tails
Why must a dominant-strategy outcome always be a Nash equilibrium?
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.
A 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 deviations — which can be collectively awful (the prisoner’s dilemma, next lesson), can be one of several with no way to say which (the standards game), or can require randomising to exist at all (matching pennies). Read a Nash equilibrium as “where rational play gets stuck,” never as “where everyone ought to want to be.”
When to use it
Reach for the Nash equilibrium whenever there’s no dominant strategy and you have to predict where a strategic situation will actually settle — a negotiation, a price war, a standards battle, a standoff. Run the best-response method: mark each player’s best reply, and read off every cell where all the marks coincide. If there’s one such cell, that’s your prediction. If there are several, the live question becomes which one — a coordination problem. If there are none in pure moves, expect randomisation. And keep the warning close: the equilibrium tells you where play gets stuck, not where anyone deserves to be.
The most important case of all is the one we’ve been teasing — a unique, perfectly stable equilibrium that leaves everyone worse off than they could be, with no lone escape from the trap. It’s so important it gets its own lesson. That’s the prisoner’s dilemma — Stable Isn’t Good: The Prisoner’s Dilemma — next.