ohhh, I did a confucious, my correct conclusions were

If the number I imagine starts with one "on" bit then the probability of winning or losing is the same.

If the number I imagine starts with two "on" bits then the probability of losing is two times the probability of winning.

do you understand why "which means it can be one of the 1/4 of the total numbers inside the set"?

lets say I have a number represented by an infinite length bitarray, if the only thing I know about this number is that the first bit is "off" I can say this number is even, which means this number can one of the 1/2 of the total numbers inside the set.

Now If I know two bits inside array I know this number can one of the 1/4 of the total numbers inside the set

If I know three bits inside array I know this number can one of the 1/8 of the total numbers inside the set

If I know four bits inside array I know this number can one of the 1/16 of the total numbers inside the set

... etc

here the chances of winning or losing are measured in x/y of the total numbers inside the set...

Each bit has a probability of 1/2 of being on and 1/2 off regardless of the state of the other bits and its position in the binary representation. You problem is thus (if I understand it correctly) equivalent to the following:

Choose a number between 0 and 3. You win if the number is 1 or 3, and you lose if the number is 3. It doesn't really make sense as a problem since the winning event and the losing events are not disjoint. Moreover, what happen in all other cases?

I wrote " I imagine a number inside this set, it starts with two "on" bits "

after reading this people will think the number I am thinking is "3" and this is the mistake, so I edited and added more words to the main problem to clarify...

some explanation...

lets change this... " I imagine a number inside this set, it starts with ONE "on" bit " <--- I CHANGE TWO TO ONE

Here the number I am thinking is not "1", it is an odd number...

" I imagine a number inside this set, it starts with ONE "on" bit "

that is the same I saying " I imagine a number inside this set, and it is an odd number "

So how I select the odd number?...whut?....what is the problem with selecting a random odd number? the probability of selecting a random odd number is the same for all odd numbers in the set

the probability of selecting a random odd number is the same for all odd numbers in the set, it is 1/(total number in the set), so this makes no sense because 1/(total number in the set) = 0,

but does it makes sense to say 1/(total number in the set) ? 0 (it is just close but not equal) and continue solving the problem?

While I agree that the probability of selecting a single number is zero, I think it is possible to make sense of something like "the probability of a natural number to be odd" by using a limit, i.e. P({n in {0, 1, 2, ..} : n is odd}) = lim_{M to infinity} P({n in {0, 1, 2, .., M-1} : n is odd}). This is a well defined limit and its value is 1/2. But even if we use a definition like this one, the rest of the sentences in the problem still does not make any sense to me.

When I said the problem was equivalent to the one where we limit ourself to the first two bits I used the definition above. Indeed, if we take M = 2^m, we may represent the numbers from 0 to M-1 as strings of m bits. The set we are sampling from is then equal to {0, 1}^m and the set of odd numbers is {1} x {0, 1}^{m-1}. Since the remaining (m-1) bits can be chosen at random without limitations, this is equivalent to the choice of just the first bit value (the rest does not matter for the probability). For this reason, we have that

P({n in {0, 1, 2, ..} : n is odd}) = lim_{M to infinity} P({n in {0, 1, 2, .., M-1} : n is odd}) = lim_{m to infinity} P({n in {0,1, .. 2^(m-1)} : n is odd} = lim_{m to infinity} P({n in {0, 1} : n == 1}) = P({n in {0,1} : n == 1}).