Minimum defense frequency and its limits
A bet of 50 into a pot of 100 produces an MDF of two-thirds. That number describes the continuing frequency that makes a specified zero-equity bluff break even. Before applying it to a hand, check what counts as a continuation, what the bettor can win, and which decisions the model leaves out.
Minimum defense frequency is a range benchmark within a narrow payoff model. The desk below compares its two complementary frequencies using pot and bet size. It does not select calls or raises, estimate an opponent’s range, or tell you how to play a particular holding.

Ace: “The pot is 100 and the new bet is 50. I gave them separate parts of the bar.”
Valeria: “Keep the whole bar at 150. The pot contributes two-thirds; the new risk contributes one-third.”
Start with the bettor's two outcomes
Let P be the pot immediately before the bet and B the bettor’s new risk. There is one defender. When the defender folds, the bettor gains P net. Whenever the defender continues, the modeled bluff loses B. Every continuation defeats this bluff, with no ties, later betting, fees or additional risk.
The bettor’s uncalled bet is returned after a fold. It is not extra profit. With P = 100 and B = 50, a fold gains 100 from the decision and a continuation costs 50. Counting the returned 50 as winnings would change the reward to 150 and produce the wrong boundary.
Write f for the defender’s folding frequency. The bluff’s expected value is fP – (1 – f)B. At break-even, fP = B – fB, so f(P + B) = B. The folding boundary is therefore B/(P + B).
The remaining fraction continues: 1 – B/(P + B) = P/(P + B). That is MDF in this model. It is the smallest continuing frequency that leaves this particular bluff with no positive immediate expected value. More continuation makes the modeled bluff lose; less makes it profit.
| Quantity | Formula | P = 100, B = 50 |
|---|---|---|
| Maximum model fold frequency | B/(P + B) | 50/150 = 1/3 |
| MDF continuation benchmark | P/(P + B) | 100/150 = 2/3 |
| Combined frequency | (P + B)/(P + B) | 1, or 100% |
The single-street fold-equity lesson examines the bettor’s profit at a supplied folding frequency. Here the focus is the complementary defense boundary and the assumptions needed to interpret it. The same 100-and-50 example links the two calculations.
Read how the benchmark changes with sizing
With a fixed pot, a larger bet risks more to win the same amount. The zero-equity bluff therefore needs more folds to break even, and the complementary MDF falls. That relationship follows from the payoffs even before any cards or player tendencies enter the discussion.
| Pot before bet | New bet | MDF | Maximum model fold |
|---|---|---|---|
| 100 | 25 | 4/5 = 80% | 1/5 = 20% |
| 100 | 50 | 2/3 = 66.666667% | 1/3 = 33.333333% |
| 100 | 100 | 1/2 = 50% | 1/2 = 50% |
| 100 | 200 | 1/3 = 33.333333% | 2/3 = 66.666667% |
Count a fictional range without overlap
Suppose a fictional range contains 30 equally weighted combinations at the decision. Against the 50-into-100 bet, assign 14 combinations to call and six different combinations to raise. That gives 20 continues and 10 folds. Calls and raises occupy separate parts of the same total.
Under the stated payoff assumptions, the pure bluff wins 100 in 10 of the 30 cases and loses 50 in the other 20. Its average result is (10 x 100 – 20 x 50)/30 = 0. This verifies the two-thirds boundary by counting the outcomes.
If only 19 combinations continue, 11 fold and the bluff earns (11 x 100 – 19 x 50)/30 = 5 chips. With 21 continues, nine fold and the result is (9 x 100 – 21 x 50)/30 = -5 chips. One equal-weight combination moves this particular average by five chips.
A combination cannot count as both a full call and a full raise in the same allocation. If it calls half the time and raises half the time, its total continuing weight is one. Adding it once in each column would count two units where only one exists.

Ace: “I entered this combination once under calls and once under raises.”
Valeria: “Split its action weight between those columns. Its total can still be only one, and folds complete the ledger.”
Use combinations and action weights
A hand label is not always one equally likely item. Before any known cards are removed, a pocket pair has six combinations, a particular suited non-pair hand has four, and its offsuit version has 12. Counting those three labels equally would ignore how many physical holdings each represents.
Known cards remove incompatible combinations. Earlier actions can give the remaining combinations different weights. The denominator for a defense frequency is the total relevant range weight at this decision, after those conditions, rather than a count of names copied from a chart.
For a small example, suppose four combinations each reach the decision with weight one and four others each have weight one-half. The total weight is 4 + 2 = 6. Continuing with three full-weight combinations and two half-weight combinations gives weight 3 + 1 = 4, or two-thirds of the total.
That allocation includes five of the eight listed combinations, yet its continuing frequency is 4/6 rather than 5/8. Both fractions can be calculated correctly; only the first uses the weights we supplied. Record the weights before comparing a frequency with MDF.
An action mix applies another weight. A combination that reaches the decision with weight one-half and then continues half the time contributes one-quarter to continuing weight. Its reach weight still contributes one-half to the denominator.
Fractional totals are ordinary frequency arithmetic. With 10 equally weighted combinations, the two-thirds benchmark is 6 and 2/3 combinations. Six could always continue and another could continue two-thirds of the time. This illustrates the fraction without identifying suitable hands.
Keep calling equity on a separate line
The two-thirds MDF for P = 100 and B = 50 is a frequency over the defending range. A separate calculation asks how often one hand must win for a call to break even. Use a closing-action toy with no later decisions, no fees and no ties for that comparison.
Before the call, the available pot is P + B = 150. Calling costs another 50, and the final pot would be 200. A win gains 150 net from the calling decision; a loss costs 50. If the hand wins a fraction e of the time, its call EV is e x 150 – (1 – e) x 50.
Setting that expression to zero gives e = 50/200 = 25%. In general, the closing-action call threshold is B/(P + 2B). Its denominator includes the original pot, the bettor’s bet and the defender’s call.
| Question | Relevant quantity at P = 100, B = 50 | What the percentage measures |
|---|---|---|
| How much continuation makes the toy bluff break even? | 2/3, about 66.666667% | Weight of the defending range that continues |
| How often must one hand win in the closing-call toy? | 1/4, or 25% | Winning outcomes for that hand against the betting range |
A 25% win threshold does not mean calling 25% of a range. Likewise, an MDF of about 66.67% does not mean an individual hand needs 66.67% equity. Ties require fractional pot shares, and future betting requires a fuller payoff calculation.
Check the range and remaining decisions
On the flop or turn, a continuing hand may face another bet before showdown. It may have to fold later or pay more to reach the river. Position affects which information and actions become available along that path. Pot size and one bet amount cannot describe those future costs.
A draw used as a bluff can improve after a call. Once the bettor sometimes wins the continuation branch, that branch no longer always loses B. The MDF formula on this page keeps the zero-equity assumption; it does not adjust for a semi-bluff’s cards or possible runouts.
A bettor may also gain from checking. Making a bet worth zero is then different from making betting and checking equally attractive. Comparing those choices requires the value of the alternative action, which these two inputs do not supply.
Range strength matters even when there is no next card. Suppose a river bettor has only hands that beat a particular bluff-catcher, with no ties. In that fictional situation, calling loses the call amount every time. The MDF boundary cannot create bluffs in a range defined to contain none.
Card removal can change that evidence for a particular holding. A card in your hand may rule out some value combinations or some bluff combinations. The balance depends on the proposed betting range. The MDF desk contains no cards and makes no conditional range count.

Ace: “The frequency is filled in, but the range and later-street files are still blank.”
Valeria: “Keep those blanks visible. The pot and bet establish the toy boundary; those files are needed to assess a real continuation.”
Treat multiple defenders and raises separately
With several defenders, the bettor’s immediate fold win requires all of them to fold. One player’s fold frequency is not that joint probability. A later player’s response can depend on what an earlier player did, so multiplying separate fold estimates requires an independence assumption.
For example, if one defender folds in 60% of cases and the second folds in half of those cases, the joint fold frequency is 0.60 x 0.50 = 30%. The second figure is conditional on the first fold. It is not automatically the second player’s overall fold frequency.
A multiway review needs the response sequence and payoffs when one or several players continue. This desk has one defender and cannot assign a share of the model’s continuation frequency to each player.
A raise counts as a continuation in the binary model only because it defeats the bluff at a loss of B. That can describe the bettor giving up immediately to a raise. If the bettor calls or re-raises, there is additional risk and another outcome to evaluate.
The B input is the new bet in the stated model. It is not a universal raise-to amount. A real raise may involve money already committed, a call portion and further action; establish its actual risk and reward before applying any separate calculation.
Compare the model frequencies
Enter the pot before the bet and the bettor’s new bet in the same chip units. Both must be positive, no greater than 1,000,000, and have at most two decimal places. The default example is P = 100 and B = 50. Select Compare the model frequencies.
Compare the complementary model boundaries
Binary bluff model: a fold gains the bettor the existing pot; every modeled continuation defeats a zero-equity bluff at the loss of its new bet. No fees, ties, later betting or additional risk. Enter a positive pot and new bet in the same unit, up to 1,000,000 with at most two decimal places. Real response choices require further analysis.
For a pot of 100 before a bet of 50, model continuation frequency is 100/150 = 2/3, about 66.666667%. Its complementary fold frequency is 50/150 = 1/3, about 33.333333%. These exact fractions sum to one. The exercise does not select individual hands.
Read the exact fractions first: MDF 2/3 and maximum model fold 1/3. Their rounded percentages are about 66.666667% and 33.333333%. A rounded decimal is a display of the fraction, not a replacement for the exact boundary.
Editing either input clears the previous output. Calculate again after a change. Reset example restores 100 and 50. The tables remain available without the controls.
Keep a review note with the amounts and assumptions: “P = 100, B = 50, one defender, every continuation defeats the zero-equity bluff, no additional costs or action.” Add actual range weights and future decisions in a separate analysis. The Info & Tools lessons provide related probability and betting exercises.