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Sunday, 21 April 2019
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How Your Opponents’ Cards Affect Drawing Odds

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By Tony Guerrera

I love getting feedback from my readers. No, it’s not just because it’s mostly positive. It’s also because their feedback gives me easy access to how various types of players think. Sure, I’ve acquired lots of data through all the battling I’ve done. But when it comes to a game that’s as information-dependent as poker, each piece of additional perspective-now matter how minute-goes a long way (know thy opponents).

One important lesson I’ve learned is that players have varying ideas when it comes to factoring opponents’ hole cards into the probability of hitting a draw. I occasionally get emails that look something like:

Tony, I’ve been reading Killer Poker by the Numbers, and it rocks! You da man! But I have a question. Why does poker math assume the best-case scenario? For example, I’m dealt A♠5♠ in a 9-handed ring game. The flop is T♠8♠2♦, giving me the nut flush draw. The general consensus is that I have 9 outs, but this assumes that no spades were dealt to other players. Don’t you effectively have fewer than 9 outs? Isn’t 9 outs simply best-case scenario for me? In general, aren’t my true odds LESS than what poker math says since I really have fewer outs than what the best-case scenario assumes?

This example email gets at the heart of what probability calculations are all about: information. Assume that all your opponents are on random hands (or “were” for the players that folded preflop-and yes, I know it’s unrealistic to assign random folding distributions because your opponents won’t fold hands like AA, but bear with me for a bit…). Because all your opponents are on random hands, you only know five cards in the deck: your 2 hole cards and the 3 cards on the flop. No matter how many spades you think were opponents were dealt, you have 9 outs in a deck of 47 cards.

The fundamental concept at work is that there’s a difference between the physical deck and the theoretical deck used for probability calculations. For example, I open a brand-new deck of cards. We inspect the deck and see that it’s a complete deck. I shuffle the deck, and I light 51 cards on fire. Only 1 card for the deck physically exists, but the probability of it being an ace is . (I don’t want to cause unnecessary air pollution-especially since I live in Southern California-so don’t go out and start lighting decks on fire!)

If you still aren’t convinced, the following calculation shows that assuming 9 outs in a deck of 47 cards is mathematically equivalent to considering all the possible ways spades could be distributed among your opponents (if your opponents’ playing and folding distributions are random):


When you have 9 outs in a 47-card deck, the probability of hitting your flush with two cards to come is .

Now, let’s account for all the card combinations that could have been dealt to your nine opponents. Your opponents hold a total of 18 cards. Of those 18 cards, 0, 1, 2, 3, 4, 5, 6, 7, 8, or 9 are spades. The spades are drawn from a pool of 9 spades, and the nonspades are drawn from a pool of 38 nonspades. The 18 cards held by your opponents are drawn from a pool of 47 cards. Let represent the number of combinations that exist if you take r objects from a pool of n objects, meaning that where k! means :

The probability of you hitting your flush is therefore:

Accounting for all the possible ways that spades can be distributed among your opponents’ hands, we see that you get the same result as when you simply consider yourself to have 9 outs out of 47 cards. Again, poker probabilities have nothing to do with the physical deck in the dealer’s hand; instead, they have to do with a theoretical deck that’s based entirely off your knowledge of the situation.

Probability is a Pure Function of Your Knowledge

The calculations above assumed that all your opponents had random hands. Your opponents’ cards can begin to matter when you account for their nonrandom distributions into account-but they only when the outs you’re counting aren’t evenly distributed among the hands in your opponents’ distributions.

You have 5♥4♥, and the board is T♥8♥2♣ . Your opponent goes all-in on the flop, and his prior play suggests that he’s on [AA-88], a distribution comprising 36 combinations (6 combinations each of AA, KK, QQ, JJ, and 99; 3 combinations each of TT and 88). Of these combinations, 15 contain 1 heart, and 21 contain 0 hearts. Therefore, the probability that you’ll hit your flush with two cards to come is:

Instead of having 9 outs in a deck of 47 cards, you effectively have something like 8.5 outs in a deck of 45 cards. But the probabilities are only off by .0012. Generally speaking, you won’t notice a big change in the probabilities when considering 8+ outers unless your opponents’ distributions consist entirely of cards that interfere with your draw. For example, you have an ace-high flush draw against one player with AA and another player with a lower flush draw. You’re drawing to 7 outs from a 43-card deck. The probability of you hitting your draw with two cards to come is only .

When playing poker, you always need to account for your opponents’ hand distributions. The work in this article was far from exhaustive, but it suggests that your opponents’ hand distributions don’t affect the probabilities much unless all the hands in an opponent’s distribution contain some of your outs. But even when their entire hand distributions count against you, you can make an on the fly estimate and subtract something like 3%-5% from the winning percentage you’d get if you put your opponents on random hands. So go out there, read your opponents, and be confident in the math governing your drawing hands!

Tony Guerrera is the author of Killer Poker By The Numbers and co-author of Killer Poker Shorthanded (with John Vorhaus)

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