WEBVTT

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You already know
the Prisoner's Dilemma as a story.

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Our goal is to turn that kind

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of story into a calculation
you can carry out by hand.

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Two people choose actions,
their choices select one outcome,

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and we want every outcome where neither
person benefits by changing alone.

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The method has four stages.

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Read the payoff pairs,
mark each player's best replies,

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keep every cell with both
marks, and if no cell survives,

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ask whether randomized choices
can create an equilibrium.

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Start with the object
the whole method uses:

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a payoff matrix.

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Row has two actions, Top and Bottom.

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Column has two actions, Left and Right.

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One action from each player selects
exactly one of the four inner cells.

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Each inner cell contains
an ordered pair of numbers.

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The first number is Row's payoff,
and the second number is Column's payoff.

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In the upper-left cell, Row receives
four and Column receives three.

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A payoff can mean money, points, votes,
years of freedom, or simply a ranking.

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The scale depends on the story.

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For solving the game, each player
compares only that player's own numbers,

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and a larger number
means a preferred outcome.

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The players choose without first
observing the other's current choice.

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Still, to analyse incentives
we ask a conditional question.

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If I knew which action
the other player had chosen,

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which of my actions
would give me the largest payoff?

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That conditional answer
is called a best response.

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Begin with Row
and suppose Column chooses Left.

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Compare Row's first
numbers in the Left column:

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four from Top and two from Bottom.

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Four is larger, so Top
is Row's best response to Left.

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Now suppose Column chooses Right.

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Row compares zero from Top
with three from Bottom.

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Three is larger, so Bottom
is Row's best response to Right.

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We mark that first payoff in red as well.

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Now change viewpoints.

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Hold Row fixed on Top and compare
Column's second numbers across that row.

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Column gets three
from Left and one from Right.

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Left is better, so the second payoff
three receives Column's blue mark.

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Hold Row fixed on Bottom.

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Column compares zero
from Left with four from Right.

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Right is better, so the four
receives the second blue mark.

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Row's comparisons ran down columns;
Column's comparisons ran across rows.

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Now inspect whole cells.

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The upper-left cell carries Row's
red mark and Column's blue mark.

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The lower-right cell also carries both.

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At either outcome, each player's
chosen action is a best response

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to the other player's chosen action.

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A cell with both marks
is a pure-strategy Nash equilibrium.

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Pure means that each player
chooses one action with certainty.

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Nash equilibrium means that,

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holding the other player's action fixed,
neither player gains by changing alone.

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Check the upper-left outcome directly.

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If Column stays Left, Row falls
from four to two by switching.

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If Row stays Top, Column falls
from three to one by switching.

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Neither unilateral change helps.

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The same check works
at the lower-right outcome.

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Row would fall from three
to zero by switching,

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and Column would fall from four to zero.

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A game can therefore have more
than one pure equilibrium,

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and we must keep every double-marked cell.

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There is one small rule about ties.

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If two available actions give the same
maximum payoff against an opponent's

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action, both are best responses.

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Mark both.

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Never break a payoff tie
merely to force one answer.

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Now put the familiar Prisoner's
Dilemma into this formal language.

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Each prisoner chooses Cooperate or Defect.

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Mutual cooperation gives two each.

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A lone defector receives three
while the cooperator receives zero,

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and mutual defection gives one each.

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Mark Row's best responses first.

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Against Column's cooperation,
Row prefers three from defecting

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to two from cooperating.

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Against Column's defection,

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Row prefers one from defecting
to zero from cooperating.

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Both red marks land in the Defect row.

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Now mark Column's best
responses using the second numbers.

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Against Row's cooperation,
Column receives three by defecting.

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Against Row's defection,
Column receives one by defecting.

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Both blue marks land in the Defect column.

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Only the lower-right cell has both marks,

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so mutual defection
is the unique pure Nash equilibrium.

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Mutual cooperation
gives both players more,

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but either player can gain individually
by defecting while the other cooperates.

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That distinction matters.

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Equilibrium is a claim about
incentives against one-player deviations,

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not a claim that the outcome
is fair, cooperative, or jointly best.

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We now have a mechanical
test for pure equilibria.

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Next we need a game
where that test leaves no cell at all.

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Matching Pennies is the smallest
game in which the pure-equilibrium

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search fails completely.

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Each player secretly
chooses Heads or Tails.

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Row earns one when the faces match,
Column earns one when they differ,

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and the loser receives minus one.

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Apply the same marking procedure.

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If Column chooses Heads, Row wants Heads.

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If Column chooses Tails, Row wants Tails.

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Row's best responses are the two
matching cells, so mark Row's payoff

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in each of those cells red.

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Column wants exactly the opposite pattern.

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If Row chooses Heads, Column wants Tails.

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If Row chooses Tails, Column wants Heads.

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Column's two best responses
are the mismatching cells,

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so mark those second payoffs blue.

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Inspect all four cells.

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Every cell carries one player's
mark, but no cell carries both.

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At every deterministic outcome,

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one player is losing
and can reverse the result by switching.

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Therefore Matching Pennies
has no pure Nash equilibrium.

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The incentives form a cycle.

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Start at Heads, Heads.

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Row wins there, so Column
wants to switch to Tails.

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That change carries
the outcome to Heads, Tails.

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Now Row is losing,
so Row switches to Tails.

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At Tails, Tails, Column
is losing and switches to Heads.

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Then Row is losing
and switches back to Heads.

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We return to the starting outcome

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without ever reaching a cell
where both players want to stay.

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This cycle is another
way to check the matrix.

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A pure equilibrium would be a stopping
point with no profitable arrow leaving it.

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Here every one of the four outcomes
has an escape for exactly one player.

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The practical problem is predictability.

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Suppose Row always chooses Heads,
or follows a pattern Column has learned.

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Column chooses Tails, forces
a mismatch, and wins every round.

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The same vulnerability runs
in the other direction.

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If Column always chooses Heads,
Row copies Heads and wins every round.

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In this game, any deterministic pattern

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that an opponent can predict
reveals the pure reply that defeats it.

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Randomizing does not mean alternating
according to a visible schedule.

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Heads, Tails, Heads,

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Tails is deterministic
and therefore exploitable once noticed.

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A mixed strategy assigns probabilities

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and uses genuine random choice so that
past actions do not reveal the next one.

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The next question is precise.

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Can we choose probabilities

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that leave the opponent
indifferent between Heads and Tails?

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If both pure replies give
the same expected payoff,

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the opponent has no profitable
way to exploit one of them.

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That indifference condition is the key
to a mixed-strategy equilibrium.

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We will write each pure action's expected
payoff as a function of the opponent's

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probability, plot the two functions,
and find exactly where they cross.

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Let q be the probability
that Column chooses Heads.

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We are choosing Column's probability,

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but the equations we
compare are Row's payoffs.

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Column's mix must remove Row's preference
between Row's Heads and Row's Tails.

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If Row chooses Heads,

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Row earns one when Column chooses Heads
and minus one when Column chooses Tails.

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Weight those two payoffs
by q and one minus q.

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Simplifying gives two q minus one.

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The red line on the plot rises with q
because Heads becomes more attractive

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as Column chooses Heads more often.

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If Row instead chooses Tails,

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matching Column's Heads loses one and
differing from Column's Tails wins one.

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The weighted payoff is q times
minus one plus one minus q times one.

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That simplifies to one minus two q.

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The blue line falls as q rises

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because Tails becomes less attractive
when Column chooses Heads more often.

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At q equals zero,
Column always chooses Tails.

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Row strongly prefers Tails,

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so the blue payoff is one
and the red payoff is minus one.

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Move q upward and that advantage shrinks.

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Indifference occurs
where the two payoff lines cross.

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Set two q minus one
equal to one minus two q.

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Adding two q and adding one gives
four q equals two, so q equals one half.

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At the yellow crossing, Row's Heads
and Tails both have expected payoff zero.

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This calculation found Column's
probability from Row's payoffs.

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That cross-player
direction is not a trick.

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A player's mix is chosen
to control the opponent's incentives.

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Now let p be the probability
that Row chooses Heads.

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To find p, compare Column's two
pure actions using Column's payoffs.

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Row's probability
must make Column indifferent.

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If Column chooses Heads,

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Column loses one when Row also chooses
Heads and wins one when Row chooses Tails.

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The expected payoff is p times
minus one plus one minus p times one.

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That simplifies to one minus two p.

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If Column chooses Tails, Column wins
against Row's Heads and loses against

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Row's Tails, giving two p minus one.

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Set the two expressions equal.

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One minus two p equals two p
minus one, so p equals one half.

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The symmetry of Matching Pennies
produced the same half-and-half

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probability for both players.

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That symmetry is special.

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The method, comparing
the opponent's pure-action payoffs,

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is the part that generalizes.

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The mixed-strategy equilibrium has each

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player choose Heads with probability one
half and Tails with probability one half.

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Each of the four outcome cells then
occurs with probability one quarter.

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Verify Row first.

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Against Column's half-and-half mix, Heads
wins one half the time and loses one half,

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so its expected payoff is zero.

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Tails has the same
calculation and also gives zero.

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Verify Column in the same way.

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Either pure action wins half the time
and loses half, so both give zero.

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Because neither player
has a better pure reply,

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neither can improve by changing
to any other mixture either.

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Notice what equilibrium does not require.

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The realized actions
can differ from round to round,

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and after any particular round one player
may wish the coin had landed differently.

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Equilibrium says the probability rule
itself cannot be profitably replaced

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while the opponent
keeps the equilibrium mix.

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We now know the general shape
of a mixing calculation.

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Assign a probability
to one player's first action,

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compute the other player's two pure-action
payoffs, set them equal, and solve.

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The asymmetric example next
will show why remembering whose payoffs

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to use is essential.

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Now take a genuinely asymmetric game.

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A worker chooses Work or Shirk,

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while an inspector chooses
Inspect or Do Not Inspect.

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The roles differ, the available
actions differ, and the first payoff

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in each cell belongs to the worker.

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Read the worker's incentives.

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If inspection occurs, Work gives two while
Shirk gives minus two, so Work is better.

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Without inspection, Shirk gives three
while Work gives two, so Shirk is better.

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Now read the inspector's
incentives using the second payoffs.

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If the worker Works,

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avoiding the inspection cost
gives zero instead of minus one.

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If the worker Shirks, inspection
gives one instead of minus two.

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No cell carries both marks.

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Work makes Do Not Inspect attractive,
which then makes Shirk attractive.

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Shirk makes Inspect attractive,
which then makes Work attractive.

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The pure incentives cycle,
so there is no pure equilibrium.

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A natural guess says
that if the worker Works most of the time,

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the inspector must also
Inspect most of the time.

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Perhaps inspection needs
probability one half or more.

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That guess confuses the frequency

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of an action with the strength
of the incentive created by it.

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Let q be the probability
that the inspector inspects.

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To find q, use the worker's payoffs.

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The inspector must choose q

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so that the worker
is indifferent between Work and Shirk.

00:14:39.506 --> 00:14:43.687
Work pays the worker two
if inspected and two if not inspected.

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Its expected payoff is two q plus two
times one minus q, which is simply two.

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The blue line is flat.

00:14:52.262 --> 00:14:56.628
Shirk pays minus two if inspected
and three if not inspected.

00:14:56.732 --> 00:15:01.400
Its expected payoff is minus two
q plus three times one minus q.

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Simplifying gives three minus five q.

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The red line slopes downward because more
inspection makes shirking less attractive.

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At an inspection probability of zero,
Shirk pays three and beats Work's two.

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Move q to one tenth and Shirk
still pays two point five,

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so the worker still prefers Shirk.

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Set the two worker payoffs equal.

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Two equals three minus five q.

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Solving gives q equals one
fifth, or twenty percent.

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At exactly twenty percent inspection,

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Work and Shirk both give
the worker an expected payoff of two.

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Below that crossing, Shirk is better.

00:15:41.043 --> 00:15:43.227
Above it, Work is better.

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The mixing point is the boundary
between those strict preferences.

00:15:48.745 --> 00:15:52.112
Now let p be the probability
that the worker Works.

00:15:52.425 --> 00:15:55.642
To find p, switch
to the inspector's payoffs.

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The worker must choose p

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so that Inspect and Do Not Inspect give
the inspector the same expected payoff.

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Inspect gives the inspector minus one
against Work and one against Shirk.

00:16:08.431 --> 00:16:12.226
Its expected payoff
is minus p plus one minus p,

00:16:12.309 --> 00:16:15.085
which simplifies to one minus two p.

00:16:15.881 --> 00:16:21.025
Do Not Inspect gives zero against
Work and minus two against Shirk.

00:16:21.338 --> 00:16:25.783
Its expected payoff is zero times
p minus two times one minus p,

00:16:25.866 --> 00:16:29.153
which simplifies to minus two plus two p.

00:16:30.065 --> 00:16:32.655
Set the two inspector payoffs equal.

00:16:32.968 --> 00:16:36.637
One minus two p equals
minus two plus two p.

00:16:36.834 --> 00:16:41.804
Rearranging gives three equals
four p, so p equals three quarters.

00:16:42.600 --> 00:16:45.666
Move to p equals zero point seven five.

00:16:45.863 --> 00:16:49.207
The two payoff lines
cross at minus one half.

00:16:49.404 --> 00:16:55.047
Inspect and Do Not Inspect are equally
good, so the inspector can genuinely mix.

00:16:55.844 --> 00:16:59.290
The equilibrium has the worker Work
with probability three quarters

00:16:59.373 --> 00:17:02.927
and the inspector Inspect
with probability one fifth.

00:17:03.239 --> 00:17:04.424
Frequent work is supported

00:17:04.424 --> 00:17:08.766
by infrequent inspection because
being caught while shirking is costly.

00:17:09.563 --> 00:17:11.166
Verify the worker.

00:17:11.270 --> 00:17:14.475
At q equals one fifth, Work gives two.

00:17:14.671 --> 00:17:18.608
Shirk gives three minus five
times one fifth, also two.

00:17:18.804 --> 00:17:20.558
Verify the inspector.

00:17:20.755 --> 00:17:26.642
At p equals three quarters, both Inspect
and Do Not Inspect give minus one half.

00:17:27.438 --> 00:17:29.492
Here is the rule worth carrying away.

00:17:29.575 --> 00:17:33.509
The worker's payoff numbers determined
the inspector's mixing probability.

00:17:33.592 --> 00:17:37.911
The inspector's payoff numbers determined
the worker's mixing probability.

00:17:38.108 --> 00:17:42.404
Your probability is chosen to erase
the opponent's strict preference.

00:17:43.201 --> 00:17:48.159
That is why intuition based only on how
often an action appears can be misleading.

00:17:48.356 --> 00:17:52.081
Mixed equilibrium probabilities
are not direct measures of effort,

00:17:52.164 --> 00:17:54.301
importance, or virtue.

00:17:54.498 --> 00:17:58.492
They are the probabilities that balance
the other player's expected payoffs.

00:17:59.539 --> 00:18:02.733
Let's finish by solving a new
game from beginning to end.

00:18:03.185 --> 00:18:05.519
Row chooses Up or Down.

00:18:05.716 --> 00:18:08.132
Column chooses Left or Right.

00:18:08.584 --> 00:18:12.427
The question asks for every
equilibrium, so we must check pure

00:18:12.427 --> 00:18:16.932
cells first and then ask whether
a mixed equilibrium also exists.

00:18:17.729 --> 00:18:20.713
Pause at the matrix and begin with Row.

00:18:21.026 --> 00:18:26.019
Against Column's Left, compare
Row's first payoffs three and zero.

00:18:26.471 --> 00:18:30.292
Up is the unique best
response, so mark the three red.

00:18:31.344 --> 00:18:35.350
Against Column's Right,
Row compares zero with one.

00:18:35.663 --> 00:18:40.285
Down is the best response, so mark
the one in the lower-right cell red.

00:18:41.499 --> 00:18:44.693
Now analyse Column
using the second numbers.

00:18:45.145 --> 00:18:50.022
Against Row's Up, Column compares
two from Left with zero from Right.

00:18:50.218 --> 00:18:53.075
Left is better, so mark the two blue.

00:18:53.988 --> 00:18:58.865
Against Row's Down, Column compares
zero from Left with three from Right.

00:18:59.062 --> 00:19:01.965
Right is better, so mark the three blue.

00:19:02.762 --> 00:19:04.864
Inspect complete cells.

00:19:05.060 --> 00:19:10.529
Up, Left carries both marks,
and Down, Right carries both marks.

00:19:10.726 --> 00:19:14.129
Those are two
pure-strategy Nash equilibria.

00:19:14.833 --> 00:19:18.374
Do not stop merely
because pure equilibria exist.

00:19:18.571 --> 00:19:23.446
Some two-by-two games, including
this one, also have a mixed equilibrium.

00:19:23.529 --> 00:19:27.477
We find it with exactly the same
indifference method used before.

00:19:28.873 --> 00:19:32.949
Let p be the probability that Row
chooses Up, and q the probability

00:19:32.949 --> 00:19:34.842
that Column chooses Left.

00:19:35.294 --> 00:19:37.454
Remember the cross-player rule.

00:19:38.069 --> 00:19:43.004
Use Row's payoffs to find q,
because q must make Row indifferent.

00:19:43.800 --> 00:19:48.538
If Row chooses Up, the payoff is three
against Left and zero against Right.

00:19:48.734 --> 00:19:51.614
The expected payoff is therefore three q.

00:19:52.318 --> 00:19:57.636
If Row chooses Down, the payoff is zero
against Left and one against Right.

00:19:57.833 --> 00:20:00.817
Its expected payoff is one minus q.

00:20:01.614 --> 00:20:03.818
Set Row's two payoffs equal.

00:20:03.901 --> 00:20:09.753
Three q equals one minus q, so four q
equals one and q equals one quarter.

00:20:09.950 --> 00:20:15.129
Column chooses Left with probability
one quarter to keep Row willing to mix.

00:20:16.525 --> 00:20:19.217
Now use Column's payoffs to find p.

00:20:19.300 --> 00:20:23.260
If Column chooses Left,
the expected payoff is two p.

00:20:23.364 --> 00:20:28.357
If Column chooses Right, the expected
payoff is three times one minus p.

00:20:29.153 --> 00:20:30.675
Set those equal.

00:20:30.872 --> 00:20:37.606
Two p equals three minus three p, so five
p equals three and p equals three fifths.

00:20:37.803 --> 00:20:42.749
Row chooses Up with probability three
fifths to keep Column indifferent.

00:20:43.662 --> 00:20:46.808
Notice that the probabilities
are not mirror images of the most

00:20:46.808 --> 00:20:48.469
attractive payoffs.

00:20:48.782 --> 00:20:52.879
Column's one-quarter probability
came from Row's payoff comparison,

00:20:52.962 --> 00:20:57.676
while Row's three-fifths probability
came from Column's payoff comparison.

00:20:58.979 --> 00:21:02.347
Always verify a mixed
result before accepting it.

00:21:02.544 --> 00:21:04.632
At q equals one quarter,

00:21:04.715 --> 00:21:08.825
Row's Up payoff is three times
one quarter, or three quarters.

00:21:09.022 --> 00:21:13.795
Row's Down payoff is one minus
one quarter, also three quarters.

00:21:14.591 --> 00:21:19.338
At p equals three fifths, Column's
Left payoff is two times three fifths,

00:21:19.421 --> 00:21:20.815
or six fifths.

00:21:21.128 --> 00:21:25.645
Column's Right payoff is three
times two fifths, also six fifths.

00:21:25.841 --> 00:21:28.245
Both indifference conditions hold.

00:21:28.856 --> 00:21:31.551
A directional check catches sign errors.

00:21:31.654 --> 00:21:35.962
If q rises above one quarter,
Up becomes better for Row;

00:21:36.159 --> 00:21:40.188
if q falls below one
quarter, Down becomes better.

00:21:40.385 --> 00:21:44.240
The crossing is exactly
where Row changes preferred actions.

00:21:45.037 --> 00:21:50.054
Likewise, if p rises above three
fifths, Left becomes better for Column;

00:21:50.250 --> 00:21:53.478
below three fifths, Right becomes better.

00:21:53.675 --> 00:21:57.797
At the crossing, each player
is willing to use either pure action.

00:21:58.594 --> 00:22:01.370
This game therefore has three equilibria:

00:22:01.566 --> 00:22:03.585
the two double-marked pure cells

00:22:03.668 --> 00:22:06.989
and the interior mixed
equilibrium we just verified.

00:22:07.185 --> 00:22:09.670
Finding one equilibrium
is not permission to stop

00:22:09.670 --> 00:22:12.132
when the question asks for all of them.

00:22:13.528 --> 00:22:16.478
Here is the complete
hand method in six steps.

00:22:16.675 --> 00:22:19.206
First, label the payoff order.

00:22:19.403 --> 00:22:23.058
Second, for each column
mark every largest Row payoff.

00:22:23.141 --> 00:22:26.869
Third, for each row mark
every largest Column payoff.

00:22:27.480 --> 00:22:31.660
Fourth, every cell with both
marks is a pure equilibrium.

00:22:31.857 --> 00:22:36.779
Fifth, if mixing is relevant, introduce
probabilities and use each probability

00:22:36.779 --> 00:22:38.765
to make the opponent indifferent.

00:22:39.078 --> 00:22:41.888
Your mix controls
the opponent's incentives.

00:22:42.685 --> 00:22:44.822
Sixth, verify.

00:22:45.134 --> 00:22:48.456
Probabilities must lie
between zero and one.

00:22:48.652 --> 00:22:52.960
Pure actions used with positive
probability must give equal expected

00:22:52.960 --> 00:22:56.861
payoffs, and any unused
action must not give more.

00:22:57.913 --> 00:23:00.305
The shortest memory aid is this:

00:23:00.502 --> 00:23:04.613
your mixing probability is solved
from the opponent's payoff comparison.

00:23:04.809 --> 00:23:07.234
It is chosen to make
the opponent indifferent,

00:23:07.317 --> 00:23:10.290
not to make your own two
payoffs equal directly.

00:23:11.203 --> 00:23:12.281
With that discipline,

00:23:12.364 --> 00:23:17.461
a two-by-two game becomes a small sequence
of comparisons and two linear equations.

00:23:17.913 --> 00:23:21.998
Read the pairs, mark the best
responses, keep every double mark,

00:23:22.081 --> 00:23:25.449
balance the opponent
when mixing, and check the result.