If my pot odds are 2.675 to 1 but my hand has 10 outs, how does my hand come up with a 4 to 1 odds pair?

For example a small pair to get a set on the turn has got 22-to-1 ratio. How can I count it? How the solution come from?

I don’t know what the “4 to 1 odds pair” refers to.

If you have pot odds of 2.575 to 1, you need card equity of at least 28% (1/3.575).
The equity equivalent to 10 outs depends on the street: about 40% on the flop and about 20% on the turn (2x,4x rule).

The 22 to 1 odds to hit a set on the turn is correct, equivalent to a 4.4% hit probability.

I’m just not sure what your question is and it may be a language issue.

I will try to help if you can be more specific –the question, street, cards, pot and bet sizes, stacks, etc. Whatever is relevant to your question.

I will give you an example. The odds of winning a hole straight from flop to turn are 11:1. How is this calculated? As well as 5:1 from flop to river?

A deck has 52 cards in it.
On the turn you have seen 6 cards: your 2 hole cards and the first 4 community cards.
That leaves 46 unseen cards.
The river will be one of the 46 unseen cards, and they're all equally likely. So you can calculate the odds of making your hand by counting how many of those unseen cards help you vs. how many don't.

Let's say you have an unimproved pocket pair on the turn. You have 2 outs to make a set. That's 2 cards out of 46 that help you, while 44 don't help you. Your odds of making a set on the river are 44-2. You can simplify that by dividing both sides by 2 to get 22-1.

Here are some other common scenarios on the turn:
15 outs (an open-ended straight flush draw): 31-15, or about 2-1
12 outs (a gutshot straight flush draw): 34-12, or about 3-1
9 outs (a flush draw): 37-9, or about 4-1
8 outs (an open-ended straight draw): 38-8, or about 5-1
5 outs (one pair hoping to improve to two pair or trips): 41-5, or about 8-1
4 outs (a gutshot straight draw): 42-4, or 10.5-1

How to calculate the odds of making your hand on either the turn or the river is more complicated. Basically, you have to figure your odds of not making your hand on the turn, multiply that by the odds of not making your hand on the river, and then subtract that product from 1.

For example, let's say you flop a flush draw.
9 out of 47 unseen cards will make your flush on the turn, so 38 won't. 38/47 is 81%.
Assuming you miss on the turn, 9 out of 46 unseen cards will make your flush on the river, so 37 won't. 37/46 is 80.5%.
81% x 80.5% = 65%—that's your chance of missing on both streets. So your chance of making your flush on one of those streets is the inverse of that, or 35%. You can convert that percentage back to odds and say that on the flop, your odds of making your flush by the river are about 2-1.

I want to refresh the basics. Thank you for your reply.
What I do not understand is the following:
If I have 6 clear outs, I have a 7:1 chance from flop to river, if I have a 3:1 chance from flop to river. Logically, the odds are higher if I have two streets left, but why 3:1? Why not 3.5 to 1?

Look at it in percentage terms, not in rounded for (correctly in most cases) simplicity odds terms. The question you are asking is "I have a 25% chance to hit with two streets to go, but only a 12.5% chance with one street to do. Why is it not 22% instead?"

Of course in reality it will not be exactly double for two streets instead of one, as you don't get paid twice if you hit twice, but you're fretting over complete nothings

You can't easily work mathematically with odds (at least I can't). Odds against an event is the following:

Odds against
= Number of ways event doesn't occur / Number of ways it does occur

= Pr(event doesn't occur)/Pr(event occurs)

Probabilities can be worked mathematically. For your case of 6 outs after the flop, there are 6 hit cards in a deck of 47. Thus, the probability of a hit on the turn is 6/47 = 12.8%. The odds are

Odds against a hit on turn = 41/6 = 6.8 or 7 to 1 approximately.

For the odds against a hit by the river, the hit probability is

P(hit on turn) + P(miss turn) * P(hit on river given miss turn) =6/47 + 41/47*6/46 = 24.1%

The equivalent odds = (1-0.241)/0.241 = 3.14

The reason the river hit probability is a bit less than half of the turn probability is that it's possible you hit on both the turn and river so by adding the turn and river hit probabies you are counting a double hit twice.

[Probability A or B) =P(A) +P(B) - P(AB)]

For the turn + river probability I find it intuitive to calculate the probability that I won’t hit an out. The probability that I will hit is just 1 minus this answer. This to me is intuitive because I find “and” probabilities easier than “or” ones. The probability I will miss on the turn and the river is simply the product of the probability of missing on each street.

Concrete example: I flop a combo flush/OE straight draw - 15 outs. I know my 2 cards plus the 3 flop cards, 47 unknown cards. I miss the turn 32/47 and the river 31/46 (assuming I miss turn), so the probability I miss both is 32*31/(47*46) =992/2162 or 45.88%. I am therefore a 54.12% favorite to hit my draw.

I also find it easier to convert pot odds to required win probability to break even. X:Y pot odds means you need Y/(X+Y) as a minimum win probability. 3:1 means 1/(3+1)=25%. 5:2 means 2/7 or 28.6%, and so on.