I got the numbers {0,1,2,3....infinity} in my table

I take half of this numbers and put them in a set(even numbers) and the other half in another set(odd numbers)

how many different pairs can I make using a number from each set?

How is the answer represented mathematically?

As it's been years since I've messed about with sets, this is a guess based on vaguely remembered (or understood) theorems:

1. The pairs are uncountable.

2. The cardinality of that set of pairs is Aleph-One.

As it's been years since I've messed about with sets, this is a guess based on vaguely remembered (or understood) theorems:

1. The pairs are uncountable.

2. The cardinality of that set of pairs is Aleph-One.

I think the pairs are actually countable. See this link for a proof.

So to answer the OP's question, the number of different pairs is countably infinite, so essentially there are as many pairs as there are integers.

The set is countable and thus have the same cardinality of natural numbers. If you define the following order on the set:

(a, b) < (d, e) IFF a + b < d + e OR (a + b = d + e AND b < e)

This is well defined and you can use this order to create a sequence of pairs. Given each pair you can determine its position in the sequence (which is finite) and you thus have a surjective, injective and invertible map between the set of the pairs and natural numbers.

I think the pairs are actually countable.

The set is countable and thus have the same cardinality of natural numbers.

You are correct. They are countable - implying Aleph-Null, not Aleph-One.

ok, now I have the same set {0,1,2,3...} each number represented using bitarrays, so the length of a number's bitarray is infinite.
I imagine a number inside this set, the only thing I know about this number is it starts with two "on" bits, I don't know the values of the next bits in the array, which means it can be one of the 1/4 of the total numbers inside the set.

there are two different numbers inside the set "a" and "b", the only things I know about those numbers

"a" number starts with one "on" bit
"b" number starts with two "on" bits

In "a" and "b" I don't know the values of the next bits in the array

if the number I am thinking is the same as "a" I win, if it is "b" I lose.

I will want an expert in this kinds of mathematics to tell me the probability of winning or losing,
I came to the conclusion that the chances of losing is two times the chances of winning
Am I right?

I am sure I know enough mathematics to solve the problem, but you are going to have to pose it correctly first.

ok, now I have the same set {0,1,2,3....infinity} each number represented using bitarrays, so the lenght of a number's bitarray is infinite.

No, that's not right. For starters, it's not clear to me what set is {0,1,2,3,...infinity}. If you had just said {0,1,2,3,...}, I would imagine you mean the set of natural numbers, but then "infinity" shouldn't be there. Then, if you represent each number using bitarrays, the length of a number's bitarray is finite.

So please try again, and be as precise as possible.

I didn't know about the first thing you pointed out, thanks.
But about the second thing, it is obvious that the number zero can be represented by an infinite bit array where all it's bits are "off" the same for all the other numbers, why are you wronging that?

I am not "wronging" anything. I am just trying to understand what you are saying.

so is the problem clear now?