How To Calculate Poker Hand Combinations

Poker Hand Combinations (Combinatorics)

Does it seem that you are dealt some poker hand combinations more frequently than others? At first, this may not seem logical. Let’s see why this is true. (We briefly discussed hand combinations in a previous article)

Learning to calculate Hand Combinations, or “Combinatorics”, as it’s often called, will help us understand why this happens. Also, by calculating Poker Hand Combinations, at the table, we will be able to narrow the possibility of some hands from the range of each player, after the flop, of course.

In order to approach Combinatorics from an easy-to-understand viewpoint, we are going to narrow the starting hands into three categories.

• Pocket Pairs
• Suited Cards
• Non-Suited Cards

Pocket Pairs

Let’s first look at pocket pairs. We all know that we get dealt pocket pairs 6% of the time or once every 17 hands. But how many ways can you get a pocket pair, of say deuces? As demonstrated below, there are six ways to be dealt any pocket pair.

Suited and Unsuited Hand Combinations

We are dealt suited cards 24% of the time or once every four hands. But what about specific suited cards, like AK suited? You will get AK, AJ, AQ, etc. suited about once every 331 hands or 0.30% of the time. There are four combinations of AK, etc. suited. See below.

This brings the combination of AK (or any exact two card combinations) to 16 total, 4 suited, 12 unsuited. You will be dealt AK Suited or Unsuited 1.10% of the time or once every 90 hands.

So how does this help us narrow an opponent’s range? Let’s look at the math.

Hand Combinations – Narrowing The Range

Let’s say we hold:

The Flop comes:

How many combinations of AK and TT are out there that our opponent could hold? (4,4 would be the same as TT, as it regards the calculation process)

• C = Total Hand Combinations
• A1 = Available Cards For The First Card
• A2 = Available Cards For The Second Card
• Therefore, C Will Equal A1 X A2

Process: Multiply the numbers of available cards for each of the two cards, which will give you the total combinations available.

If we hold KQ on a KT4 flop, how many combinations of AK are there?

There are 4 Aces and 2 Kings (4 minus the 1 on the flop and minus the 1 in our hand) available in the deck.

C = A1 x A2
C = 4 x 2
C = 8

C = 8, so there are 8 remaining combinations of AK if we hold KQ on a flop of KT4.

Pair Combinations

How many combinations of TT (or 4,4) are there on a KT4 flop?

Well, on a flop of KT4, here are 3 Tens and 3 Fours left in the deck, so…

C = [(A) x (A-1)] / 2
C = [(3) x (3-1)] / 2
C = [3 x 2] / 2
C = 3

C = 3, which means there are 3 possible remaining combinations of TT and 4,4.

Why is Combinatorics Useful

As I mentioned earlier, by determining possible hand combinations, we can further define a player’s range. For example, let’s say that a player’s 3-betting range is 2%. Then this would mean that they are only 3-betting AA, KK and AK. So we know they are a tight player.

Just looking at the range of hands this player would 3-bet with, we would think, they have a big pocket pair when they 3-Bet. Both AA and KK are in his range, along with the single unpaired hand of AK. Void of Combinatorics, you might conclude that the probability of him holding each hand is 33%. (33% for AA, 33% for KK and 33% for AK)

But is that mathematically accurate? Let’s see.

• AA has 6 combinations. 21.5% Of 3-Betting Hands
• KK has 6 combinations 21.5% Of 3-Betting Hands
• AK has 16 combinations. 57% Of 3 Betting Hands
• Total of 28 Potential 3-Betting Hands

So that means that both AA and KK make up 21.5% of his raising hands, or 43% combined. However, AK represents 57% of his 3-betting hands, meaning he is more likely to have AK when 3-betting, versus AA or KK combined.

Let’s Play A Hand Using Combinatorics

You are holding:

The pot is $80 and you bet $50 into it. Your opponent moves all in for $200, making the pot $330. Now you have to call $150 to win $330.

You believe your opponent has either a set or two pair, probably with an Ace (AJ, A8, A5 or A2).

You are getting 2.2 to 1 pot odds. If you make the call, which means you must win the pot how often in order for this to be a break even call? In order to get that ratio, we must look at the total pot, if you call, which would be $480. Divide the amount you have to call, by the total pot. In this case, it’s 31.25%. Therefore, in order to breakeven on this call, you must win the hand 31.25% of the time.

Hands You Beat

• AJ – 3 Aces/3 Jacks = 9 Combinations
• A8 – 3 Aces/3 8s = 9 Combinations
• A5 – 3 Aces/1, 5 = 3 Combinations
• A2 – 3 Aces/ 3 2’s = 9 Combinations
• 22 – 3 X 2/2 = 3 Combinations
• Total of 33 Combinations that you beat

Hands You Lose To

• AA – 3 x 2 /2 = 3 Combinations
• JJ – 3 x 2 /2 = 3 Combinations
• 88 – 3 x 2 / 2= 3 Combinations
• Total of 9 Combinations that beat you

That’s a total of 42 Combinations. You beat 33 of these combinations or 79% and lose to 9 or 21%. You only need to win the hand 31.25% of the time. Therefore, it’s an easy call, with a positive EV.

I hope this brief description of How To Calculate Poker Hand Combinations In Real Time has been helpful. Please leave your comments below and let me know.

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