## Introduction

What’s the probability of getting dealt, or your opponent holding, a particular hand/range? Use combinatorics, and probability to answer these questions. If you think you know it all, try betzoo.uk’s interactive combinatorics and probability quiz, which will give you instant feedback.

There are 1,326 different starting hands you could be dealt (assuming the order of your two cards don’t matter), and you need to have strategies on how to proceed for all of these. If you are wondering how this is calculated you could be dealt any of the 52 cards in the deck as your 1st card, and any of the remaining 51 for your 2nd card. When performing these calculations, it doesn’t matter how many other players have also been dealt cards. Whether, and how many of, the remaining cards have been dealt to the other players or are still in the dealer’s hand make no difference. You then multiply 52 x 51 = 2,652. There is one final step. As the order of the cards in your hand makes absolutely no difference, e.g. if the dealer gave you black aces, whether you got the ace of clubs first and ace of spades second (or the other way around) doesn’t matter, so we divide by 2. So 2,652/2=1,326 unique starting hands.

If however you count all hands with cards of the same rank as equal (for example 2[h]2[s] = 2[c]2[d], or A[s]K[c] = A[h]K[h], then there are only 91 possibilities of hands a player can hold. 13 of these are pairs, and 78 of these are unpaired hands.

In many situations, ranges tend to be linear – this means if an opponent is playing a hand they are usually playing all better hands.

Note, that even though there are 91 possibilities of hands, it is not true that a player has a 1/91 chance of getting each one.

This is because for any pair, say JJ there are 6 ways of being dealt the pair:

J[s]J[d]

J[s]J[h]

J[s]J[c]

J[h]J[d]

J[h]J[c]

J[d]J[c]

For any unpaired hand, say KQ there are 16 ways of being dealt that unpaired hand:

K[s]Q[s], K[s]Q[h], K[s]Q[d], K[s]Q[c]

K[h]Q[s], K[h]Q[h], K[h]Q[d], K[h]Q[c]

K[d]Q[s], K[d]Q[h], K[d]Q[d], K[d]Q[c]

K[c]Q[s], K[c]Q[h], K[c]Q[d], K[c]Q[c]

As you can see out of these 16 ways, 4 ways are suited, and 12 are unsuited. This will be important in range analysis, as you will might think your opponent might be playing just the suited combos of say AK (or another hand), or all the combos, depending on the situation.

To check our sums, for the:

13 pairs, each has 6 combinations: 13 x 6 = 78 combos

78 unpaired hands, each has 16 combinations: 78 x 16 = 1,248 combos

Therefore, all hands (paired + unpaired) = 78 combos + 1,248 combos = 1,326 combos

You are in a position to work out the chance of getting dealt any specific hand, or range of hands using combinatorics and probability. You can also read a full introduction to blockers – and how this affects combinatorics and probability here at betzoo.uk.

## What about blockers?

We have a separate post at betzoo.uk about the effect of blockers on combinatorics and probability,

## Combinatorics and probability examples

**What’s the probability of getting dealt a pocket pair preflop (if it doesn’t matter which one):**

78 combos of pocket pairs divided by 1,326 total combos equals ~5.9%

When you are playing Texas Holdem, you will expect to see a pocket pair dealt to you, ~5.9% of the time.

**A beautiful way to get to this answer in a matter of seconds is to understand that to get a pocket pair, it doesn’t matter what card the dealer deals you first. All that matters is your second card matches the first card dealt. How often will your second card match the first card dealt? There will be 3 remaining cards in the deck that match your first card (regardless of what it was), and 51 cards left in the deck (cards dealt to our opponents do not have to be considered, as we don’t know what they are), so the answer is 3/51 =~5.9%**

**What is the probability of getting dealt a suited hand?**

There are 78 unpaired hand types, and each has 4 suited combos (spades, hearts, diamonds, clubs), e.g. AK has the following suited combos: A[s]K[s], A[h]K[h], A[d]K[d], A[c]K[c]

So, there are 78 x 4 = 312 combos of suited hands

312 combos of suited hands / 1,326 total combos = ~23.5%

**A quick way to work this out. It doesn’t matter what the first card the dealer deals you is. There will be 12 remaining cards in the deck that have the same suit as your first card (regardless of what it was), and 51 cards left in the deck (cards dealt to our opponents do not have to be considered, as we don’t know what they are), so the answer is 12/51 =~23.5%**

**What is the probability of getting dealt an unpaired unsuited hand?**

There are 78 unpaired hand types, and each has 12 unsuited combos

So, there are 78 x 12 = 936 combos of unsuited hands

936 combos of unsuited hands / 1,326 total combos = ~70.6%

**A quick way to work this out. It doesn’t matter what the first card the dealer deals you is. There will be 36 remaining cards in the deck that are a different suit that also don’t pair your hand as your first card (regardless of what it was), and 51 cards left in the deck (cards dealt to our opponents do not have to be considered, as we don’t know what they are), so the answer is 36/51 =~70.6%**

You can also work out the probability of getting dealt a particular hand:

**What is the probability of getting dealt AA?**

Each of the 4 aces, can be dealt with one of other three aces to make AA, thus there are (4 x 3)/2 = 6 combos of AA (we divide by 2 as order doesn’t matter, A[h]A[s] is exactly the same as A[s]A[h])

6 combos of AA / 1,326 total combos =~0.45%

This is the probability of getting dealt any specific pair, be it AA or 66.

**What is the probability of getting dealt AKs?**

Obviously, there are 4 combos of any suited hand (both cards must be spades, or both cards must be hearts, or both cars must be diamonds, or both cards must be clubs).

4 combos of AKs / 1,326 total combos =~0.3%

This is the probability of getting dealt any given suited hand, be it AKs or 74s

**What is the probability of getting dealt JTo?**

Each of the 4 jacks, can be dealt with one of three tens (the fourth ten will be the same suit as the jack in question) to make JTo, thus there are 4 x 3 = 12 combos of JTo

12 combos of JTo / 1,326 total combos =~0.9%

This is the probability of getting dealt any given offsuit hand, be it JTo or 92o

**What is the probability of getting dealt KQ (either offsuit, or suited)?**

Each of the 4 kings, can be dealt with one of four queens to make KQ, thus there are 4 x 4 = 16 combos of KQ.

16 combos of KQ / 1,326 total combos =~1.2%

This is the probability of getting dealt any given unpaired hand, be it KQ or 72