Texas Holdem Probability - Preflop Fold Equity

By Tony Guerrera

This is part of a series of articles covering Texas holdem probability.

Fold equity refers to the number of chips you expect to win as a result of your opponents folding. Preflop fold equity is important in all forms of Texas holdem, but it’s particularly important in the tournament setting where the difference between a huge payout and the bubble is often the number of blinds you can successfully steal. Given that you know your opponents’ folding distributions, you can explicitly calculate the probability that all your opponents will fold to a preflop raise (a folding distribution is the set of hole cards that an opponent will fold).

We can’t walk until we can crawl, so we'll open the discussion by calculating the probability that you'll be dealt hole cards from a well-defined distribution. From there, we'll move on to consider your opponents’ holdings.

Texas Holdem Probabilities Involving Being Dealt A Hand From A Defined Distribution

When you're dealt a hand in Texas holdem, you're given 2 cards from a deck of 52 cards. The total possible number of 2 card combinations from a 52-card deck is .

To find the probability of being dealt a hand from a certain distribution, you just need to find the number of combinations that compose the distribution.

What’s the probability of being dealt a specific pocket pair?

All the pocket pairs are equivalent, so let's take AA as our example.

Six AA combinations exist (A§A¨, A§A©, A§Aª, A¨A©, A¨Aª, and A©Aª), so on any deal, the probability of being dealt AA is .

The math governing AA is identical to the math governing all pocket pairs. Therefore, the probability of being dealt a specific pocket pair is .

Suppose you want to know the probability of being dealt AA, KK, QQ, JJ, or TT. AA, KK, QQ, JJ, and TT. Each have 6 possible combinations, so the distribution comprises 30 combinations in total.

The probability of you being dealt AA-TT is ; simply find the number of combinations composing the distribution and divide by 1,326, the total number of possible combinations. 

Let's move on to the probabilities associated with being dealt unpaired hole cards. Unpaired hole cards are either suited or unsuited. 4 combinations exist for a specific set of suited unpaired hole cards.

Consider AKs. You can either have A§K§, A¨K¨, A©K©, or AªKª.

The probability of being dealt AKs is therefore .

In general, the probability of being dealt a specific set of unpaired hole cards is .

Meanwhile, 12 combinations exist for a specific set of unsuited, unpaired hole cards.

Consider AKo. The deck contains 4 aces, and each of those 4 aces can be paired with 3 possible kings.

For example, the A§ can be paired with the K¨, the K©, or the Kª. (4)(3) = 12, meaning that 12 AKo combinations exist. In general, the probability of being dealt a specific set of unsuited, unpaired hole cards is .

Sometimes we're concerned with suitedness. Other times we aren't. Suppose you just want to know the probability of being dealt AK. 4 suited combinations and 12 unsuited combinations make 16 total combinations.

Therefore, the probability that you'll be dealt AK on any given hand is ; the probability that you'll be dealt a specific set of unpaired hole cards is .

Having considered probabilities involving being dealt specific hole cards, let’s move on to probabilities involving being dealt a hand from a defined distribution. Actually, we did this earlier when we calculated the probability of being dealt AA-TT, but it'll be instructive to go through another example.

Suppose you raise all unraised pots from the button with {AA-22, AK-A2, KQ-K8, QJ-QT, JT, J9s-53s, T9s-54s}.

What percentage of unraised pots will you raise from the button?

The first step is to break things down in terms of combinations:

{AA-22}: (13 hands)(6 combinations per hand) = 78 combinations

{AK-A2}: (12 hands)(16 combinations per hand) = 192 combinations

{KQ-K8}: (5 hands)(16 combinations per hand) = 80 combinations

{QJ-QT}: (2 hands)(16 combinations per hand) = 32 combinations

{JT}: (1 hand)(16 combinations per hand) = 16 combinations

{J9s-53s}: (7 hands)(4 combinations per hand) = 28 combinations

{T9s-54s}: (6 hands)(4 combinations per hand) = 24 combinations

In total, you're raising with 450 combinations. Therefore, the probability that you'll raise preflop from the button in an unraised pot is …you’ll raise about 33.9% of unraised pots from the button.

That's it. It doesn't get any more complicated than that that when it comes to calculating preflop probabilities involving your own hole cards.

Texas Holdem Probabilities Involving Your Opponents' Preflop Hand Distributions

When thinking about your own hole cards, you’re getting 2 cards from a deck of 52 cards. 1,326 possible combinations exist. 6 combinations are possible for each pocket pair, and 16 combinations are possible for each unpaired set of hole cards. When thinking about your opponents' hole cards, important adjustments to these numbers are necessary.

Suppose action folds to you, and you're in the small blind. You hold Q7. What's the probability that your opponent will fold to a raise if he'll stay in the hand with {AA-22, AK-A7, KQ-KT, QJ-QT, JT}?

First, we need to figure out how many possible combinations exist for your opponent’s hole cards. You know your 2 hole cards, so only 50 unknown cards remain in the deck.

That means that your opponent holds 1 of  combinations.

Second, we need to enumerate all the combinations in your opponent's playing distribution. Paired hole cards not containing a Q or a 7 have 6 combinations each, and unpaired hole cards not containing a Q or a 7 have 16 combinations each. That leaves us with QQ, 77. AQ, A7, KQ, QJ, and QT to deal with. 

For QQ, there are only 3 queens left in the deck since you hold 1.

QQ involves choosing 2 queens from a pool of 3 queens, meaning that there are  available QQ combinations.

This same analysis holds true for 77.

Let's move on to AQ. Normally, 16 AQ combinations exist; however, you hold a queen. This means that you can form AQ combinations from a pool of 4 aces and 3 queens. (4)(3) = 12, meaning that only 12 AQ combinations exist. Note that there are also 12 each for A7, KQ, QJ, and QT. Your opponent's playing distribution, broken down with respect to available combinations, is below:

{AA-KK, JJ-88, 66-22}: (11 hands)(6 combinations per hand) = 66 combinations

{QQ,77}: (2 hands)(3 combinations per hand) = 6 combinations

{AK,AJ-A8}: (5 hands)(16 combinations per hand) = 80 combinations

{AQ,A7}: (2 hands)(12 combinations per hand) = 24 combinations

{KQ}: (1 hands)(12 combinations per hand) = 12 combinations

{KJ-KT}: (2 hands)(16 combinations per hand) = 32 combinations

{QJ-QT}: (2 hands)(12 combinations per hand) = 24 combinations

{JT}: (1 hands)(16 combinations per hand) = 16 combinations

In total, your opponent's playing distribution consists of 260 combinations. This means that your opponent will fold 1,225 - 260 = 965 combinations.

The probability that your opponent will fold to your preflop raise is therefore .

Instead of having just one player remaining to act behind you, suppose that a number of players remain to act behind you...each with a different folding distribution. Calculating your fold equity precisely in such a scenario is really tough. Fortunately, we can make an approximation that gets us results close enough to be useful.

Assume that two players remain to act behind you. The probability that the first player will fold is A, and the probability that the second player will fold is B. The probability that both players will fold is AB. In general, to find the probability that all your opponents will fold, simply multiply their individual folding probabilities.

Suppose a somewhat liberal limper calls from early position. Action folds to you, and you have J9s on the button with 8 big blinds. You put the limper on {AA-22, AK-A8, KQ-KT, QJ-QT, JT, J9s-86s, T9s-76s}. If you push all-in, the limper will call with {AA-66, AK-AJ}. Meanwhile, the blinds will call you with {AA-99, AK-AQ}. If you push all-in, what’s the probability that everyone will fold?

The limper’s hand distribution, broken down in terms of combinations, is as follows: 

{AA-QQ, TT, 88-22}: (11 hands)(6 combinations per hand) = 66 combinations

{JJ, 99}: (2 hands)(3 combinations per hand) = 6 combinations

{AK-AQ, AT, A8}: (4 hands)(16 combinations per hand) = 64 combinations

{AJ, A9}: (2 hands)(12 combinations per hand) = 24 combinations

{KQ, KT}: (2 hands)(12 combinations per hand) = 24 combinations

{KJ}: (1 hand)(12 combinations per hand) = 12 combinations

{QT}: (1 hand)(16 combinations per hand) = 16 combinations

{QJ}: (1 hand)(12 combinations per hand) = 12 combinations

{JT}: (1 hand)(12 combinations per hand) = 12 combinations

{J9s}: (1 hand)(3 combinations per hand) = 3 combinations
{T8s, 86s}: (2 hands)(4 combinations per hand) = 8 combinations

{97s}: (1 hand)(3 combinations per hand) = 3 combinations

{87s, 76s}: (2 hands)(4 combinations per hand) = 8 combinations

{T9s, 98s}: (2 hands)(3 combinations per hand) = 6 combinations

In total, the limper’s hand distribution comprises 264 combinations. Meanwhile, his playing distribution, broken down in terms of combinations, is:

{AA-QQ, TT, 88-66}: (7 hands)(6 combinations per hand) = 42 combinations

{JJ, 99}: (2 hands)(3 combinations per hand) = 6 combinations

{AK-AQ}: (2 hands)(16 combinations per hand) = 32 combinations

{AJ}: (1 hand)(12 combinations per hand) = 12 combinations

In total, the limper’s playing distribution comprises 92 combinations. Therefore, his folding distribution comprises 264 – 92 = 172 combinations: 

The blinds have random hands. We’re making an approximation and assuming that knowledge of the limper’s hand distribution has a negligible impact on the blinds’ hand distributions. Therefore, each of the blinds’ hand distributions comprises the 1,225 available hand combinations. The blinds’ playing distributions, broken down in terms of combinations, are as follows:

{AA-QQ, TT}: (4 hands)(6 combinations per hand) = 24 combinations

{JJ, 99}: (2 hands)(3 combinations per hand) = 6 combinations

{AK-AQ}: (2 hands)(16 combinations per hand) = 32 combinations

In total, the blinds’ playing distributions comprise 62 combinations. Therefore, their folding distributions comprise 1,225 – 62 = 1,163 combinations:

The probability that your all-in wins the pot uncontested is:

An exact calculation of winning the pot uncontested would account for the conditional probabilities associated with your opponents’ hand and calling distributions. And while exact calculations are nice, sometimes they are too long. Fortunately, the approximation I used in this article is accurate enough for most purposes. When simply dealing with random hands, it will be accurate to within 3% (with fewer than four players remaining to act behind you, it’ll be much more accurate). And as long as limpers are on hand distributions wider than just a few hands, you’re also good.

If you’re at the table, you won’t be able to do exact calculations like those I did here in this article. But understanding these calculations is important: you can do your homework away from the table and make intelligent estimates at the table. You only have two ways to win a hand: make your opponents fold or win in showdown. Therefore, having a firm grasp on your fold equity is essential.