G-Spot: Quickly Estimating Your Opponents’ Folding Percentages
By Tony Guerrera
Getting opponents to fold to your aggression is important in all forms of poker hands, but particularly the shovefest that defines the endgame of most tournaments. The methodology in this article will allow you to estimate your probability of winning uncontested to within a few percent.
Assign Distributions
To estimate your fold equity, first assign hand distributions and calling distributions to your opponents. Suppose action folds to you in the cutoff, and you have 75s. Since your opponents haven’t acted yet, their hand distributions are [All]. If you push all-in, you believe that all your opponents will call with [AA-55, AK-AT].
You know your 2 hole cards, so 50 cards remain. There are 1,225 total combinations of 2 hole cards from a 50 card deck (commit this number to memory). For pocket pairs in your opponents’ distributions, there are 6 combinations, 3 combinations, or 1 combination, depending on if the pocket pair has 0, 1, or 2 cards in common with your hole cards. In this example, [AA-88,66] represent 6 combinations each. And [77,55] represent 3 combinations each. For unpaired hole cards, take the number of the first card available and multiply by the number of the second card available. In this example, your 75s doesn’t counterfeit any of the AK-AT hands, so each of those hands has (4)(4) = 16 combinations.
Multiply
This isn’t exact, but to find the probability that all your opponents will fold, simply multiply all your opponents’ folding probabilities. For example, if P(A) is probability that Player A will fold and P(B) is the probability that Player B will fold, the probability that both players will fold is P(A)P(B).
In this example, each player has the same calling distribution, so P(Fold) is be the same for each. To find P(Fold), first find the total number of calling combinations. In this example, that number is 118. Since there are 1225 combinations total, this means that your opponents are folding 1225 – 118 = 1,107 combinations. P(Fold) for each poker player is 1,107/1,225. The probability that all three players will fold is (1,107/1,225)(1,107/1,225)(1,107/1,225).
Of course, these really big numbers are tough to multiply in your head. So to get a quick estimate, say that 1,107/1,225 is close to 1,100/1,200 which is close to 11/12 which is about .92 which is about .9. (.9)(.9)(.9) is .729, meaning that you’ll win uncontested about 73% of the time.
Be Comfortable With Estimating
73% might not be the true percentage that all your opponents will fold. But at most, it’s only off by a few percent. When you’re in the heat of battle, you need to make educated decisions in compressed time frames. Using estimates is a great way to make sure that you’re at least in the ballpark. Play smart, and enjoy the rewards!
Tony Guerrera is the author of Killer Poker By The Numbers and co-author of Killer Poker Shorthanded (with John Vorhaus).