Introduction

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

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

Formulas

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

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

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

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

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

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

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

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

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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

Formulas

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 $=4K6+4K2K6$

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3K3 $=3K3$

3K4 $=3K4+4K4$

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

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

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2K2 $=2K2$

2K3 $=2K3+3K3$

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

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

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

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2K2K2K6 $=2K2K2K6+4K2K6$

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4K2K6 $=4K2K6$

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3K3K6 $=3K3K6$

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3K2K5 $=3K2K5$

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

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2K2K4 $=2K2K4+4K4$

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

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

Question

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.