Game theory: dominance, mixing and the indifference condition

Game theory questions at trading firms are rarely about proving existence theorems. They are about two mechanical skills — stripping out dominated strategies, and solving the indifference equations that define a mixed equilibrium — plus the judgement of when equilibrium play is actually the right choice.

Proposition 2.1

Start by deleting

A strategy is dominated when another does at least as well against every opponent action. Delete it, and then look again: removing one option can make a second one dominated that was not before. Iterated elimination solves a surprising share of the games you will be shown, and it costs nothing to try first.

Holds when

Strict dominance can always be eliminated safely. Weak dominance can delete equilibria, so say which you are using.
If elimination leaves one strategy each, that is the unique equilibrium and you are done.

Derivation 2.2

Mixed equilibrium by indifference

In a mixed equilibrium, each player randomises so that the *other* is indifferent. That is the equation to write down.

$\text{Let the row player play Up with probability } p$
You are solving for your own mix from the *opponent's* payoffs.
$\mathbb{E}[\text{col plays Left}] = \mathbb{E}[\text{col plays Right}]$
If the column player strictly preferred one, they would not mix — so they must be indifferent.
$\text{Solve for } p$

\text{your mix makes them indifferent; their mix makes you indifferent}

Why you solve with the wrong payoffs. The step candidates find strange is that your own mixing probability comes out of your opponent’s payoff matrix, not yours. The reason is that a mix is only stable if the other side has no reason to deviate — if they had a strictly better response, they would take it and your mix would stop being optimal. So your job is to leave them with nothing to prefer. It is the same logic as quoting a two-sided market: you set the price at which the other side is indifferent between hitting and lifting.

Example 2.3

A penalty kick

A kicker shoots left or right; the keeper dives left or right. The kicker scores $90\%$ shooting to their strong side if the keeper goes the wrong way, $30\%$ if the keeper guesses right; on the weak side it is $70\%$ and $20\%$ . What is the kicker’s equilibrium mix?

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Worked solution

Formula $\text{keeper indifferent: } \mathbb{E}[\text{dive strong}] = \mathbb{E}[\text{dive weak}]$
Substitute $\text{kicker shoots strong with prob } p: \quad 0.30p + 0.70(1-p) = 0.90p + 0.20(1-p)$
Solve $0.70 - 0.40p = 0.20 + 0.70p$
$0.50 = 1.10p$
Answer $p = \tfrac{5}{11} \approx 45\%$

Sanity check. The kicker shoots to their *weak* side more often than their strong side, which is the counter-intuitive result these questions exist to produce. The strong side is more valuable precisely when unguarded, so it has to be held back.

Proposition 2.4

Sequential games: work from the end

When players move in turn and can see what came before, solve the last decision first, then the one before it given that answer. Subgame perfection is nothing more than doing this consistently, and it is what makes the pirate-gold and centipede answers come out so lopsided.