Indifference & equilibrium frequencies
Why does the reference page say a pot-sized river bet carries about one-third bluffs? Why must a defender keep half their range against one? These aren't conventions to memorize — they fall out of one principle: indifference. At equilibrium, each player mixes so that the OPPONENT'S pure options break even. If calling were strictly profitable, you'd call always and the bettor would adjust; if folding were strictly better, you'd fold always and bluffs would print. The equilibrium sits exactly where neither side can do better by picking one action and sticking to it — and algebra pins those mixing frequencies down completely.
- Name the victim.Decide whose decision you are neutralizing — usually the weakest hand class on either side.
- Write both EVs.Express continue-vs-give-up for that hand as functions of the opponent's unknown frequency.
- Set them equal.The frequency making EV(call) = EV(fold) is the equilibrium mix — solve for it.
- Sanity-check both seats.Every frequency you compute belongs to ONE seat's range; the mirror question has a different answer.
Worked example: deriving both river frequencies from scratch
Pot 10bb, villain bets 10bb (pot-sized) with a polarized range: fraction x value, 1−x pure bluffs. You hold a pure bluff-catcher — beats every bluff, loses to every value bet.
| Question | Indifference | Solution |
|---|---|---|
| How often may VILLAIN bluff? | Your call EV equals fold EV: x·(30) − 10 = 0 | x = 10/30 = 33.3% bluffs |
| How often must YOU defend? | Villain's bluff EV equals zero: y·10 − (1−y)·10 = 0 | y = MDF = 10/20 = 50.0% |
The two rows are the two seats of the same bet — the 33% belongs to the BETTOR's range, the 50% to the DEFENDER's. Confusing them is the classic error the Bluffs-vs-value reference warns about.