Game theory in growth strategy: pricing, positioning and strategic interactions

Your price is only half of the decision

Most pricing advice treats your price as a dial you turn on your own. That works when nobody else is watching, and in a real market somebody always is. What you earn depends on what you choose and on what your rivals choose, at the same time.

The toolkit for this is game theory. A game is any situation where players each pick a move and each player's result depends on everyone's moves. Its central idea, the equilibrium John Nash defined in 1950, is why I write the game down before I touch a price.¹

This article teaches the handful of ideas I actually use, each with a worked example. The businesses and numbers are invented so that I can show every step. Nobody real, no client.

The running example is two online coffee roasters in Cork. Shandon Roasters, run by Aoife, sells a 1kg bag of beans for €15. Its rival, Mardyke Coffee, sells a very similar bag at the same price. Each sells about 1,000 bags in a normal November.

Worked example: why Aoife cannot decide alone

Aoife's bag costs her €6 in beans, packaging and postage. At €15 she keeps €9 a bag, so a normal November earns her $1{,}000 \times \text{€}9 = \text{€}9{,}000$ before fixed costs. She is thinking about a 20% Black Friday discount, which takes the price to €12 and her margin to €6 a bag.

Whether that pays depends on Mardyke. If it holds, her discount steals its customers. If it discounts too, both simply earn less on every bag. Same move, two outcomes. That is a game, and the first job is to write it down.

A payoff matrix writes the whole game on one page

A payoff matrix is a table. Your possible moves are the rows, the rival's possible moves are the columns, and each cell holds what each player earns if that pair of moves happens. By convention the first number in a cell is the row player's payoff and the second is the column player's.² It looks like a school exercise. It is the most useful pricing document I know, because it forces you to say what you think happens in every case.

Worked example: the Black Friday matrix

Aoife estimates four cases from her past Novembers. All figures are contribution for the month, price minus the €6 variable cost, times bags sold.

1. Both hold at €15. Each sells its usual 1,000 bags. Each earns $1{,}000 \times \text{€}9 = \text{€}9{,}000$.
2. Both discount to €12. The lower price brings a good few new buyers into the market, so each sells 1,100 bags. Each earns $1{,}100 \times \text{€}6 = \text{€}6{,}600$.
3. Shandon discounts, Mardyke holds. Shandon keeps its 1,000, takes 600 of Mardyke's customers and finds 100 new ones, 1,700 bags in total. It earns $1{,}700 \times \text{€}6 = \text{€}10{,}200$. Mardyke is left with 400 bags and earns $400 \times \text{€}9 = \text{€}3{,}600$.
4. Mardyke discounts, Shandon holds. The mirror image: Mardyke earns €10,200, Shandon €3,600.

In the matrix, Shandon chooses the row and Mardyke the column.

	Mardyke holds	Mardyke discounts
Shandon holds	€9,000, €9,000	€3,600, €10,200
Shandon discounts	€10,200, €3,600	€6,600, €6,600

Four cells, every assumption visible. If a number is wrong, Aoife can argue with the cell instead of with a feeling.

A dominant strategy is the move that wins whatever the rival does

A strategy is dominant when it gives you a better payoff than your alternatives against every move the other player could make.² Most pricing games do not hand you one. It is still the first thing to check, because when it exists it decides the game.

Worked example: checking each column

Aoife reads the matrix one column at a time, because a column is one thing Mardyke might do.

If Mardyke holds, Aoife compares €9,000 for holding with €10,200 for discounting. Discounting is better by €1,200.

If Mardyke discounts, Aoife compares €3,600 for holding with €6,600 for discounting. Discounting is better by €3,000.

Discounting wins in both columns, so for Shandon discounting is a dominant strategy. The game is symmetric, so Mardyke's owner, doing the same arithmetic row by row, finds the same answer. Each of them, reasoning carefully and alone, decides to discount.

A Nash equilibrium is where nobody gains by moving alone

A Nash equilibrium is a combination of moves, one per player, in which no player could do better by changing their own move while everyone else keeps theirs.² It is not the best outcome. It is the stable one: once you are there, nobody has a reason to leave on their own. Nash showed in 1950 that every finite game has at least one such point, once you allow players to randomise.¹

Formally, a combination of strategies is a Nash equilibrium when, for every player:

$$u_i\left(s_i^{*}, s_{-i}^{*}\right) \geq u_i\left(s_i, s_{-i}^{*}\right) \quad \text{for every } s_i \in S_i$$

Here $i$ is any one player, $u_i$ is that player's payoff, $s_i^{*}$ is the move player $i$ makes in the equilibrium, $s_{-i}^{*}$ stands for the equilibrium moves of all the other players, $s_i$ is any alternative move player $i$ could make instead, and $S_i$ is the set of all moves open to player $i$. In plain words: holding everyone else still, player $i$'s equilibrium move pays at least as much as any other move. A payoff matrix is simply a grid of these values, which is why the Foundations: matrices section is the right background.

Worked example: testing every cell of the Black Friday matrix

For each cell, ask each player: if I switched and the other did not, would I earn more?

1. Both hold, €9,000 each. Shandon switching to discount earns €10,200. It would move. Not an equilibrium.
2. Shandon discounts, Mardyke holds. Mardyke switching to discount goes from €3,600 to €6,600. It would move. Not an equilibrium.
3. Shandon holds, Mardyke discounts. The mirror image, Shandon would move from €3,600 to €6,600. Not an equilibrium.
4. Both discount, €6,600 each. Shandon switching to hold drops to €3,600. Mardyke switching drops to €3,600 as well. Nobody moves. This is the equilibrium.

A confession. An earlier version of this article put the equilibrium of a matrix like this in the comfortable "both hold" cell. That was wrong. When discounting pays more against either move, both holding is exactly the cell each player wants to leave. The stable outcome is the one where both earn less.

Worked example: a positioning game with two equilibria

Suppose Shandon and a new Cork roaster are each choosing a position for next year, the premium single origin market or the value blend market. Aoife's estimates of yearly contribution:

	Rival goes premium	Rival goes value
Shandon goes premium	€14,000, €14,000	€40,000, €24,000
Shandon goes value	€24,000, €40,000	€8,000, €8,000

If the rival goes premium, Aoife's best reply is value (€24,000 against €14,000). If it goes value, her best reply is premium (€40,000 against €8,000). Her best move flips, so she has no dominant strategy.

Now the cells. In "both premium" either player gains by switching (€14,000 to €24,000), and in "both value" likewise (€8,000 to €40,000). In "Shandon premium, rival value", Aoife switching would drop from €40,000 to €8,000 and the rival switching would drop from €24,000 to €14,000. Nobody moves. The mirror cell is stable for the same reason. So this game has two Nash equilibria, and in both the two roasters split the market.

The theory says the two will probably end up in different segments, not which one Aoife gets. That depends on who commits first.

The prisoner's dilemma is why discount wars happen to sensible people

The Black Friday game has a famous shape. In the prisoner's dilemma, each player can cooperate or defect, and the payoffs are ordered so that the temptation to defect alone beats the reward for mutual cooperation, which beats the punishment for mutual defection, which beats the sucker's payoff for cooperating alone. Defecting is dominant for each player, so mutual defection is the only equilibrium, even though both would prefer mutual cooperation.³ The OpenStax economics text uses the same structure to explain why firms in a concentrated market end up producing more and earning less than they could, because each firm's dominant strategy is to expand.⁴

Written with the usual letters, a game is a prisoner's dilemma when:

$$T > R > P > S$$

Here $T$ is the temptation, what you earn defecting while the other cooperates, $R$ is the reward each earns when both cooperate, $P$ is the punishment each earns when both defect, and $S$ is the sucker's payoff, what you earn cooperating while the other defects. The ranking is all that matters, not the size of the gaps, and comparing values in order is the same skill as in Foundations: ratios and rates.

Worked example: the Black Friday game is a prisoner's dilemma

Map the matrix onto the letters, with "hold" as cooperate and "discount" as defect. $T = \text{€}10{,}200$, $R = \text{€}9{,}000$, $P = \text{€}6{,}600$, $S = \text{€}3{,}600$. Check the order: $10{,}200 > 9{,}000 > 6{,}600 > 3{,}600$. It holds, so this is a textbook prisoner's dilemma.

Add it up. Both holding would earn $\text{€}9{,}000 + \text{€}9{,}000 = \text{€}18{,}000$ between them. Both discounting earns $\text{€}6{,}600 + \text{€}6{,}600 = \text{€}13{,}200$. The pair hands €4,800 to customers every November, and neither owner made a mistake in their own reasoning.

The horizontal axis shows the three kinds of outcome, and the vertical axis the November contribution of both roasters added together, in euro. The takeaway is the slope: every step away from both holding costs the pair money, and the stable outcome sits at the bottom. This is the arithmetic behind the sinking feeling of a price war, and it is the same story I tell from the retail side in beating price wars in a homogeneous product market.

Playing the same game every month changes the answer

The game above is played once. Pricing is not. In a repeated game, what you do today changes what the rival does tomorrow, and that can make holding your price the rational move after all.

Two strategies are worth knowing. The first, usually called GRIM or trigger, cooperates until the other player defects once and then defects for ever. John Nash himself noted that it is an equilibrium of the indefinitely repeated prisoner's dilemma.³ The second is tit for tat, which starts by cooperating and then copies whatever the other player did last round. In Robert Axelrod's computer tournaments of the early 1980s, tit for tat, submitted by Anatol Rapoport, won both rounds. Axelrod put its success down to four traits: it is nice (never the first to defect), retaliatory, forgiving and clear.³

Whether cooperation survives depends on how much the future matters. Game theorists capture that with a discount factor, a number between zero and one that shrinks next round's payoff, or equivalently the probability that there is a next round at all.³ Under GRIM, holding your price is the better plan when what you earn by holding for ever beats one month of temptation followed by punishment for ever:

$$\frac{R}{1 - \delta} \geq T + \frac{\delta P}{1 - \delta} \quad \Longleftrightarrow \quad \delta \geq \frac{T - R}{T - P}$$

Here $\delta$ is the discount factor, how much next month's euro is worth to you today, including the chance that the game continues, and $T$, $R$ and $P$ are the temptation, reward and punishment payoffs from the matrix. The left side, $\frac{R}{1 - \delta}$, is the value of earning $R$ every month for ever. The right side is $T$ once, then $P$ every month after that, discounted. Rearranging gives the threshold on the right. The idea of discounting a stream of future payments is set out in Foundations: discounting.

Worked example: how much the future must matter in Cork

Put in the Black Friday payoffs, read as a monthly game. Step one, multiply both sides of the left inequality by $1 - \delta$:

$$9{,}000 \geq 10{,}200\,(1 - \delta) + 6{,}600\,\delta$$

Here $\delta$ is still the discount factor, $9{,}000$ is the reward $R$ in euro, $10{,}200$ is the temptation $T$ and $6{,}600$ is the punishment $P$. Step two, multiply out: $9{,}000 \geq 10{,}200 - 10{,}200\,\delta + 6{,}600\,\delta = 10{,}200 - 3{,}600\,\delta$. Step three, rearrange: $3{,}600\,\delta \geq 1{,}200$, so $\delta \geq \frac{1{,}200}{3{,}600} = \frac{1}{3}$. The same answer comes straight from the short form, $\frac{T - R}{T - P} = \frac{1{,}200}{3{,}600}$. The arithmetic is plain Foundations: discounting again.

In practice: if both roasters expect to be trading next month, the future is worth far more than one third and holding is stable. The threshold is crossed when the future stops mattering: a rival about to close, a one off clearance, an investor demanding this quarter's numbers. In my experience that is when price wars actually start, not because someone misread the market but because someone's future shrank.

The line you must not cross

This edge matters more than any equilibrium. Deciding on your own to hold your price is ordinary business. Agreeing it with your rival is a cartel. Section 4 of the Competition Act 2002 prohibits agreements and concerted practices between businesses that restrict competition in the State, and the first example it names is directly or indirectly fixing selling prices.⁵ Article 101 of the Treaty on the Functioning of the European Union says the same wherever trade between member states may be affected.⁶ Since September 2023 the Competition and Consumer Protection Commission (CCPC) can also pursue administrative fines of up to €10 million or 10% of a company's annual worldwide turnover.⁷ Talking about prices without agreeing anything is risky too. The CCPC lists current and future prices, discounts and rebates as information competitors should, in most circumstances, not share, and warns that an individual convicted on indictment faces up to 10 years in prison.⁸ It has happened here. The Irish Ford Dealers Association circulated guide prices to its 53 members to limit deeper discounts, and in 2007 its secretary got a suspended 12 month sentence and a €30,000 fine for aiding and abetting the cartel.⁹ The repeated game is played through public prices, never over a coffee with the other owner.

Who moves first, and why some threats are empty

Often one player moves, the other sees it, then answers. For these sequential games the tool is backward induction: start at the last decision, work out what that player will do, then step back to the decision before, knowing the answer.² The outcome that survives this process at every step is called a subgame perfect equilibrium, and it throws out plans that depend on a threat nobody would carry out when the moment came. Game theorists call such a threat incredible: if carrying it out would hurt the person making it, the other side should ignore it, unless the threat has been made binding by a commitment in advance.²

Worked example: a half price entrant

A new subscription roaster, Brewline, is thinking about launching in Cork at half Shandon's price. Aoife is tempted to announce that she will match any price, whatever it costs. Her yearly estimates:

1. Brewline stays out. Brewline earns nothing, Shandon keeps €60,000.
2. Brewline enters and Aoife matches the half price. Both bleed. Brewline loses €10,000, Shandon earns €20,000.
3. Brewline enters and Aoife does not match. She keeps her price, adds value to her top subscription tier and lets Brewline take the most price sensitive buyers. Brewline earns €15,000, Shandon €40,000.

The diagram follows the order of the decisions, and the solution reads from the last one. After entry, matching earns Aoife €20,000 and not matching €40,000, so she will not match. Brewline knows this, compares €15,000 from entering with €0 from staying out, and enters. Her loud promise to match changes nothing, because Brewline can see she would not keep it.

The lesson: segmenting the customers and adding value at the top beats a price match, because the match costs the threatener more than the target. Making the threat credible would need a commitment that changes Aoife's own payoffs in advance, which is expensive and usually a poor trade.

Mixed strategies explain why sales look random

Some games have no stable pair of fixed moves at all. Then the equilibrium is a mixed strategy: each player picks among moves at random, with chosen probabilities, like flipping a weighted coin.² It sounds odd for a business, but Hal Varian's model of sales, published in 1980, showed that when some shoppers compare prices and others do not, shops in equilibrium randomise their prices, which is why the same product shows a scatter of prices that never settles into one.¹⁰ A study of a large price comparison site found exactly that: across 36 identical electronics products, prices did not converge on one level even after 18 months.¹¹

The way to find a mixed equilibrium is the indifference principle. Each player mixes in the proportions that leave the other player indifferent between their moves, because if the other player strictly preferred one move, they would always play it and the mix would collapse. For a player choosing between moves $X$ and $Y$ against a rival who plays $X$ with probability $q$:

$$q\,u(X, X) + (1 - q)\,u(X, Y) = q\,u(Y, X) + (1 - q)\,u(Y, Y)$$

Here $q$ is the probability that the rival plays $X$, $1 - q$ the probability that the rival plays $Y$, and $u(A, B)$ is the player's payoff from playing $A$ while the rival plays $B$. Each side of the equation is an expected payoff, a probability weighted average, explained from scratch in Foundations: expected value.

Worked example: when to run the promotion

Mardyke, the smaller roaster, can afford one promotion a month, either in payday week or mid month. Shandon can afford one counter promotion a month and wants to land it in the same week, to protect its customers. The payoffs for Mardyke: an unopposed payday promotion earns it €6,000 extra, an unopposed mid month promotion €3,000, and a promotion Shandon counters earns it nothing. Shandon gains €2,000 by countering in the right week and nothing if it guesses wrong.

If Mardyke always went for payday, Shandon would always counter payday, so Mardyke would switch, and round it goes. No fixed pair is stable, so the equilibrium is mixed.

Step one, Shandon's mix, $q$ the probability it counters in payday week. It must leave Mardyke indifferent: $q \times \text{€}0 + (1 - q) \times \text{€}6{,}000 = q \times \text{€}3{,}000 + (1 - q) \times \text{€}0$. That gives $6{,}000 - 6{,}000\,q = 3{,}000\,q$, so $q = \frac{6{,}000}{9{,}000} = \frac{2}{3}$.

Step two, Mardyke's mix, $p$ the probability it promotes in payday week. It must leave Shandon indifferent: countering payday earns $p \times \text{€}2{,}000$, countering mid month earns $(1 - p) \times \text{€}2{,}000$. Setting them equal gives $p = \frac{1}{2}$.

Step three, the value. Mardyke expects $(1 - \frac{2}{3}) \times \text{€}6{,}000 = \text{€}2{,}000$ from a payday promotion and $\frac{2}{3} \times \text{€}3{,}000 = \text{€}2{,}000$ from mid month, the same either way, as the principle requires.

Two results surprise people. Payday is worth twice as much to Mardyke, yet it promotes there only half the time. And Shandon's mix is set entirely by Mardyke's payoffs. The practical point: a promotion calendar a rival can predict is one a rival can counter.

Zero sum is much rarer than it feels

A zero sum game is one where one player can only be made better off by making the other worse off.² What one wins, the other loses, and the total never changes. Competition often feels like that. Most of it is not, and that matters, because only a game whose total can change leaves room for anything except beating the other side.

The old version of this article called bidding on the same advertising slot zero sum. Strictly, it is not. When two shops bid up the same keyword, both can end up paying more for the same clicks and the money leaves the game for the platform. Both can lose together, which is closer to the prisoner's dilemma. The mechanics of the auction itself are in auction theory and bidding strategy.

Worked example: adding up the cells

In a zero sum game every cell of the matrix adds up to the same total. Add up the Black Friday cells: both hold, $\text{€}9{,}000 + \text{€}9{,}000 = \text{€}18{,}000$. One discounts, $\text{€}10{,}200 + \text{€}3{,}600 = \text{€}13{,}800$. Both discount, $\text{€}6{,}600 + \text{€}6{,}600 = \text{€}13{,}200$. Three different totals, so this is not a zero sum game. The pie shrinks as the discounts spread.

It can grow too. If each roaster runs its own stall at a Cork food festival, costing €1,500, and each wins 150 new customers buying three bags a year at €9 contribution, each gains $150 \times 3 \times \text{€}9 - \text{€}1{,}500 = \text{€}2{,}550$ without the other losing anything. Growing the category needs no agreement on price at all.

A reaction function turns a rival's habits into a forecast

The matrices so far assume each player picks from two moves. Prices are continuous, and rivals rarely jump in one step. A best response function, or reaction function, describes a rival's likely next price as a function of yours.² In practice I often start with the simplest one that fits a rival's history: each period they close a fixed share of the gap between their price and mine.

$$P_{R}(t + 1) = \alpha\,P_{M}(t) + (1 - \alpha)\,P_{R}(t)$$

Here $P_{R}(t)$ is the rival's price in period $t$, $P_{M}(t)$ is my price in period $t$, $\alpha$ is the share of the gap the rival closes each period (between 0 and 1), and $t + 1$ is the next period. Because the gap shrinks by the same factor $1 - \alpha$ every period, it behaves like compound growth in reverse, which is set out in Foundations: compounding.

Worked example: six months after a price cut

Aoife wants to know what happens if she cuts her bag from €15 to €12 and leaves it there. Mardyke's price history suggests it closes about 30% of any gap each month, so $\alpha = 0.3$, and it starts at €15.

The gap starts at $\text{€}15 - \text{€}12 = \text{€}3$. Each month it is multiplied by $1 - 0.3 = 0.7$.

1. Month one: gap $\text{€}3 \times 0.7 = \text{€}2.10$, rival at €14.10.
2. Month two: gap $\text{€}2.10 \times 0.7 = \text{€}1.47$, rival at €13.47.
3. Month three: gap $\text{€}1.47 \times 0.7 = \text{€}1.03$, rival at €13.03.
4. Month four: gap €0.72, rival at €12.72.
5. Month five: gap €0.50, rival at €12.50.
6. Month six: gap €0.35, rival at €12.35.

The horizontal axis is the month after the cut, and the vertical axis the price of a 1kg bag in euro. Mardyke's line falls fast, then flattens towards Shandon's flat €12. The takeaway: the price advantage is real in month one and nearly gone by month six, while the margin Aoife gave up is permanent.

The model is an assumption to be tested, not a law. I estimate $\alpha$ from a rival's own price history, and the regression for that is the same one I use in modelling demand and price elasticity with regression. Watching public prices is fine. Asking the rival about their plans is not.

How I use game theory without fooling myself

Game theory has limits. The payoffs are estimates. Real people in long repeated games behave quite differently from what strict backward induction predicts.³ Most markets have more than two players. That makes the method a discipline rather than an oracle, and the value is in the order of questions I ask before any price change.

Each step fills in the matrix or tests it. The last matters most to me: knowing which estimate, if wrong, would change the decision tells me what to watch after the move.

Worked example: the whole sequence for the Black Friday decision

Aoife runs the list. Players: Shandon and Mardyke. Moves: hold at €15 or discount to €12. Payoffs: the matrix above. Dominant strategy: discounting, for both, if the game is played once. Nash equilibrium: both discount, €6,600 each. Repeated: yes, both expect to keep trading, so the future is worth well over the threshold of one third and holding is stable. Her decision: hold at €15, make the November offer about something other than price, and watch Mardyke's public price. What would change her mind: a Mardyke discount, or news that it is closing, either of which shrinks the future.

That is the honest promise of game theory in pricing. It will not tell you the future. It tells you which outcomes are stable, which threats are empty, and which number, if you have it wrong, turns a calm November into a price war.

Footnotes

1. John F. Nash Jr., Equilibrium Points in N Person Games, Proceedings of the National Academy of Sciences, volume 36, issue 1, pages 48 to 49, 1950, https://pmc.ncbi.nlm.nih.gov/articles/PMC1063129/

2. Don Ross, Game Theory, The Stanford Encyclopedia of Philosophy, revised 2023, https://plato.stanford.edu/entries/game-theory/

3. Steven T. Kuhn, Prisoner's Dilemma, The Stanford Encyclopedia of Philosophy, revised 2025, https://plato.stanford.edu/entries/prisoner-dilemma/

4. OpenStax, Principles of Microeconomics 3e, section 10.2 Oligopoly, Rice University, 2022, https://openstax.org/books/principles-microeconomics-3e/pages/10-2-oligopoly

5. Law Reform Commission, Competition Act 2002, revised, section 4, https://revisedacts.lawreform.ie/eli/2002/act/14/revised/en/html

6. European Union, Consolidated version of the Treaty on the Functioning of the European Union, Article 101, EUR-Lex, 2008, https://eur-lex.europa.eu/legal-content/EN/TXT/?uri=CELEX:12008E101

7. Competition and Consumer Protection Commission, New competition law enforcement powers come into effect today, CCPC, September 2023, https://www.ccpc.ie/business/new-competition-law-enforcement-powers-come-into-effect-today/

8. Competition and Consumer Protection Commission, Sharing information with competitors is a risky business, CCPC, 24 August 2016, https://www.ccpc.ie/news-and-media/news/article/2016/08/24/sharing-information-with-competitors-is-a-risky-business

9. Competition and Consumer Protection Commission, Ford Dealer Cartel, competition case, CCPC, 2007, https://www.ccpc.ie/enforcement-and-regulation/competition-law/competition-cases/details/ford-dealer-cartel-conviction

10. Hal R. Varian, A Model of Sales, American Economic Review, volume 70, issue 4, pages 651 to 659, 1980, https://ideas.repec.org/a/aea/aecrev/v70y1980i4p651-59.html

11. Michael R. Baye, John Morgan and Patrick Scholten, Persistent Price Dispersion in Online Markets, working paper, 2002, https://faculty.haas.berkeley.edu/rjmorgan/ppd.pdf