Before starting, I would like to specify a few things:

  1. The deck has 52 cards. From 1 (Ace) to 10, J, Q, K, all in four different suits: spades, diamonds, club and hearts. I don't add jokers because that is far beyond my possibilities of calculations.
  2. For better precision, I won't use chance, instead I would use combinations or the frequency (it is quite easy to then divide that frequency by the total number possible of combinations to get a chance).
  3. This is based on poker, but it isn't exactly poker. I will completely ignore things like straights or flushes. Additionally, if I say "at least a pair" it means that not only "a pair" counts, but also "a three of a kind" (if you have 3 cards of the same kind, you at least have two), "a four of a kind", "a three of a kind + a pair", etc. Another example would be considering in "at least two pairs" a "four of a kind" (if you have 4 cards of the same kind, you have two groups of two cards of the same kind also).

What do I want?

The title is a bit misleading because such a general formula is impossible. I used that title because I can't explain what I want in just a few words. I want to know the frequency of having:

  1. At least a pair.
  2. At least a three of a kind.
  3. At least a four of a kind.
  4. At least three pairs.
  5. At least a four of a kind and a pair.
  6. At least two three of a kind.
  7. At least a three of a kind and a pair.
  8. At least two of a pair.

All the previous frequencies I want to calculate them with:

  1. A 2-card hand.
  2. A 3-card hand.
  3. A 4-card hand.
  4. A 5-card hand.
  5. A 6-card hand.

How would I calculate them?

There are many different forms of making the calculations. These forms might include:

  1. Calculating the total possible combination and then subtracting all combinations that aren't what I want.
  2. Calculating the total combinations of at least what I want and then subtracting all repeated options. Doing that is tempting, but with hands with many cards, it starts to become complicated. Also, it is sometimes tricky to know how many times you must subtract something.
  3. Calculating the total combinations of exactly what I want and then adding the combinations of things that can also be included in the "at least". For example, sum the frequency of a pair, two pairs, three of a kind and a four of a kind to get the chance of "at least a pair".

I will use this last method. It is probably the method that requires more calculations, but also is the simplest and straightforward method. To do so, I will create two tables. The first table will contain the exact things, for example exactly two cards of the same kind and then two other different cards (that do not form a second pair) for the chance of "a pair in a 4-card hand". Then, the second table will contain the "at least" frequencies summing multiple frequencies of the previous table.

Exact Frequencies

To make it shorter, I will name pairs as "2K", three of a kind as "3K", four of a kind as "4K" and then the combinations such as a three of a kind and a pair (normally "full house") "3K2K".

To refer to the number of cards drawn, I will add the number at the end of the name, for example, If I want to tell the frequency of two pairs in a 5-card hand, I will say 2K2K5

Hands/Draw ...K2 ...K3 ...K4 ...K5 ...K6
4K - - 13 624 13.728
3K - 52 2.496 54.912 732.160
2K 78 3.744 82.368 1.098.240 9.884.160
2K2K2K* - - - - 61.1776
4K2K - - - - 936
3K3K - - - - 1.248
3K2K - - - 3.744 164.736
2K2K* - - 2.808 123.552 2.471.040

*The pairs are different. That means that for example that in 2K2K2K, (AA)(AA)(QQ) doesn't count. Instead, that would fall into 4K2K.


Below I will show the formulas. I won't explain each of them because it would take a lot of space. However, if there are some formulas that you would like to know how I made them, I can explain them, editing this post and putting the explanations at the end of the post.

4K4 $={13 \choose 1}{4 \choose 4}$

4K5 $={13 \choose 1}{4 \choose 4}{13-1 \choose 1}{4 \choose 1}$

4K6 $={13 \choose 1}{4 \choose 4}{13-1 \choose 2}{4 \choose 1}^2$

3K3 $={13 \choose 1}{4 \choose 3}$

3K4 $={13 \choose 1}{4 \choose 3}{13-1 \choose 1}{4 \choose 1}$

3K5 $={13 \choose 1}{4 \choose 3}{13-1 \choose 2}{4 \choose 1}^2$

3K6 $={13 \choose 1}{4 \choose 3}{13-1 \choose 3}{4 \choose 1}^3$

2K2 $={13 \choose 1}{4 \choose 2}$

2K3 $={13 \choose 1}{4 \choose 2}{13-1 \choose 1}{4 \choose 1}$

2K4 $={13 \choose 1}{4 \choose 2}{13-1 \choose 2}{4 \choose 1}^2$

2K5 $={13 \choose 1}{4 \choose 2}{13-1 \choose 3}{4 \choose 1}^3$

2K6 $={13 \choose 1}{4 \choose 2}{13-1 \choose 4}{4 \choose 1}^4$

2K2K2K6 $={13 \choose 3}{4 \choose 2}^3$

4K2K6 $={13 \choose 1}{4 \choose 4}{13-1 \choose 1}{4 \choose 2}$

3K3K6 $={13 \choose 2}{4 \choose 3}^2$

3K2K5 $={13 \choose 1}{4 \choose 3}{13-1 \choose 1}{4 \choose 2}$

3K2K6 $={13 \choose 1}{4 \choose 3}{13-1 \choose 1}{4 \choose 2}{13-2 \choose 1}{4 \choose 1}$

2K2K4 $={13 \choose 2}{4 \choose 3}^2$

2K2K5 $={13 \choose 2}{4 \choose 3}^2{13-2 \choose 1}{4 \choose 1}$

2K2K6 $={13 \choose 2}{4 \choose 3}^2{13-2 \choose 2}{4 \choose 1}^2$

At Least Frequencies

Hands/Draw ...K2 ...K3 ...K4 ...K5 ...K6
4K - - 13 624 14.664
3K - 52 2.509 59.280 912.808
2K 78 3.796 87.685 1.281.072 13.329.784
2K2K2K - - - - 62.712
4K2K - - - - 936
3K3K - - - - 1.248
3K2K - - - 3.744 166.920
2K2K - - 2.821 127.920 2.713.464


Whenever I refer to a frequency of another formula, for example, 4K4, I refer to the Exact Frequency and not the At least Frecuency.

4K4 $=4K4$

4K5 $=4K5$

4K6 $=4K6+4K2K6$

3K3 $=3K3$

3K4 $=3K4+4K4$

3K5 $=3K5+4K5+3K2K5$

3K6 $=3K6+4K6+3K2K6+3K3K6+4K2K$

2K2 $=2K2$

2K3 $=2K3+3K3$

2K4 $=2K4+3K4+4K4+2K2K4$

2K5 $=2K5+3K5+4K5+2K2K5+3K2K5$

2K6 $=2K6+3K6+4K6+2K2K6+3K2K6+3K3K6+4K2K+2K2K2K$

2K2K2K6 $=2K2K2K6+4K2K6$

4K2K6 $=4K2K6$

3K3K6 $=3K3K6$

3K2K5 $=3K2K5$

3K2K6 $=3K2K6+4K2K6+3K3K6$

2K2K4 $=2K2K4+4K4$

2K2K5 $=2K2K5+4K5+3K2K5$

2K2K6 $=2K2K6+4K6+3K2K6+3K3K6+4K2K6+2K2K2K6$


I would like to know if my math is correct. Probably there might be more than one error at high card drawn.

What would be considered an answer?

In case it is correct, I don't need any demostration. In case something is incorrect, which is quite possible, I would like to know what is wrong, why it is wrong, which formula would correct that, and what that new formula means.

As there are quite a lot of calculations, it is quite possible that only one person won't find all error, in case there were many. The accepted answer would be the one that correct the most number of errors.

  • $\begingroup$ Note that it should be easy to deduce these probabilities (or frequencies) from the standard Poker Probability computations (at least for standard hand sizes). Of course, the usual computations for say, three of a kind, would exclude four of a kind or full houses, so you'll have to combine the appropriate cases. $\endgroup$
    – lulu
    Feb 19, 2022 at 19:59
  • $\begingroup$ +1: to your question for outstanding work shown + outstanding discussion of the various methods. Minor criticism specific to MathSE. You are asking a reviewer to study your work from top to bottom, including every single question that you asked. This isn't really reasonable. You discussed $3$ different methods for attacking an individual question. I suggest that you edit your question to pick out two of the more complicated questions, each of which uses a different one of the $3$ tools, and ask a MathSE reviewer to examine only those $2$ questions. ...see next comment $\endgroup$ Feb 19, 2022 at 20:16
  • $\begingroup$ As you have surmised, you would typically have an answer that looks like $$\frac{N\text{(umerator)}}{D\text{(enominator)}},$$ where $~\displaystyle D = \binom{52}{5} \approx 2.6~$ million. So, when you present the questions, you don't need to present the tables. Simply present the $2$ questions, along with your computations for $N$ and $D$, in each question. ...see next comment $\endgroup$ Feb 19, 2022 at 20:21
  • $\begingroup$ Further, there is another (better) approach, assuming that you are reasonably proficient at some PC programming language (e.g. Python, C, Java, ...). In my experience, a typical PC can handle up to $(10)^8$ simulations. So, for a specific problem, let the computer work through the $2.6$ M cases, and programmatically note the number of satisfying cases.The programming approach is often referred to as sanity checking. $\endgroup$ Feb 19, 2022 at 20:22
  • $\begingroup$ @user2661923 I mentioned the methods to make it known that there exist multiple ways of doing this. However, I specified that I only used one of them. In my calculations, I started using the 2nd, then I moved to the 1st after a suggestion of N. F. Taussig from this answer. Finally, I moved to the 3rd method and started over again my math because I thought that it was easier and cleaner. All my calculations were made with that last method. $\endgroup$
    – Enderluck
    Feb 19, 2022 at 20:25

1 Answer 1


I will try to corroborate my own calculations using the 2nd method suggested in my question. I won't calculate all the possibilities. Instead, I will only calculate the most complex of the calculations (complex for the method that I used in the question, because they require quite a lot of different calculations): at least 2K2, 2K3, 2K4, 2K5, 2K6.

The reason for only calculating them is because those frequencies that I have already calculated in the question requires all other "exact" frequencies. This means that if I'm able to reach that numbers using the 2nd method, it is highly possible that the "exact" frequencies that were used to calculate them are correct. Of course, there could be the possibility that more than one of those frequencies are wrong and when summing them to calculate the at least frequency they compensate themselves leading to a correct value, but I think that it is highly improbable. Also, the other "at least" frequencies calculated in my question could be wrong, summing the exact frequencies that aren't required or not summing the exact frequencies that are required, but there are only sums and I think that I haven't mistaken any of them.


The probability of 2Kn can be calculated as the difference between all possible combinations drawing n cards, and the combinations that have all those cards of different ranks.

$${52 \choose n}-{13 \choose n}{4 \choose 1}^n$$

This formula is based on one suggestion from this answer.

${52 \choose n}$ represents all possible combinations of n cards.

${13 \choose n}$ represents drawing n cards of different ranks.

${4 \choose 1}^n$ represents that all those cards previously drawn can be of any of the four different suits.

Explained that, we only need to replace n to make the different possibilities.

$$2K2={52 \choose 2}-{13 \choose 2}{4 \choose 1}^2=78$$

$$2K3={52 \choose 3}-{13 \choose 3}{4 \choose 1}^3=3.796$$

$$2K4={52 \choose 4}-{13 \choose 4}{4 \choose 1}^4=87.685$$

$$2K5={52 \choose 5}-{13 \choose 5}{4 \choose 1}^5=1.281.072$$

$$2K6={52 \choose 6}-{13 \choose 6}{4 \choose 1}^6=13.329.784$$


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