What is the probability you guess the number I am thinking of? Probability is defined as the likely number of outcomes over all total outcomes.  In this case, 1 over infinity; which would equate to zero.
But, there is a chance you can guess the number I am thinking of.
Furthermore, would it depend on the number of guesses you have?
Let's say you have 2 guesses? 100 guesses? Infinity guesses?
I'm looking for any insight as to the answer to this problem whatsoever.  Is there even a way to answer this question?
Let me know.
Thanks!
 A: You assume your brain is a uniform random number generator, which it is clearly not (because you are more likely to say $3$ rather than $3.68394234918923483904582\ldots$). Your argument that the probability is $0$ assumes that the brain is.
Without any further knowledge it is impossible to deal with the question, and even with further knowledge the question is certainly not a trivial one. You would have to determine a way to see what random numbers humans generate the most often, and your answers may vary wildly depending on who you ask.
A: Pick a number between $0$ and $1$ inclusive.  What is the chance you pick $0$?  
Well, the chance that you pick something less than $.5$ is $1/2$.  Likewise, the chance you pick something less than $.25$ is $1/4$, etc.  Continuing this logic, the probability you pick $0$ must be less than any positive number.  But negative probability doesn't make sense, so the answer must be that the probability of picking $0$ is exactly $0$.
You can make a similar argument and show that the probability of picking any number in the range $[0,1]$ is exactly $0$.
If you look past $[0,1]$ and allow picking any real number, the answer remains exactly the same for similar reasons.
What if you picked $n$ random numbers?  Would it make any difference?  No - $n\times0 $ doesn't make your odds any better.
What if you picked an infinite number of numbers?  It turns out that any list $x_1, x_2, x_3, \ldots$ is, in a sense, infinitely smaller than the set of all real numbers, or even the set $[0,1]$.  For example, I could start listing all of the rational numbers in $[0,1]$:
$$0,1,\frac{1}{2},\frac{1}{3},\frac{2}{3},\frac{1}{4},\frac{3}{4}\ldots$$
and I would hit any rational value less than one eventually.  You cannot list all of the real numbers in this way, you will always leave infinitely many out of your list.  You can read about this more here.
The probability of picking a number from $[0,1]$ and getting a rational number is $0$, and any infinite list is in one-to-one correspondence with the rational numbers.  Because of this, you have $0$ probability of picking the number in question even with an infinite list of guesses.
How can it be that you have $0$ probability of doing something that is clearly possible?  Say you pick a number $x$ randomly from $[0,1]$.  How could you have done that if the chance of picking $x$ was zero to start with?  It turns out this is not inconsistent, probability in infinite spaces is just counter-intuitive.  Mathematicians use the term almost certainly to describe things that happen with probability $1$.
A: I suspect that you are getting more than you asked for in some of the answers.  The key is in the third paragraph of @julien's answer: for the probability of picking each number to be finite (greater than zero, informally), the probabilities must be distributed in a way that gives a finite sum.
So, you cannot give an "equal" weight to "every" number, because the weights sum to infinity and cannot be normalized.  However, there are plenty of distributions which do work, and if we are talking about a real human choosing a number, it seems valid to assume that very large numbers are unlikely to be chosen.
The simplest example would be to weight the integers to have probability $\frac1{2^n}$.  Thus 1 has probability $\frac12$ of being chosen, 2 has $\frac14$, and so on; this sums to one for the total probability of a positive integer being chosen, but every number has a finite probability.
Just to show there are other possibilities, we could say that the probabability of choosing a d-digit integer was $\frac9{10^d}$, giving 1-digit numbers a 90% chance in total, 2-digit numbers 9%, 3-digit .9%, etc.  Again, the total probability sums to one but the individual probabilities are all finite: 10% each for 1 through 9, .1% each for 10 through 99, etc.
So, it all comes down to the assumed distribution of chosen numbers, which ultimately depends on a model of human thinking.
A: To talk about probabilities, you first need to define a probability space, that is a measure space $\Omega$ such that $\mu(\Omega)=1$. The measurable subsets of $\Omega$ are your events.
For instance, if you are to choose a number in $\{1,2\}$, you could set $\Omega:=\{1,2\}$ and put the counting measure on it, divided by $2$. There are four events: $\emptyset$ (you neither pick $1$ nor $2$, $\{j\}$ (you pick $j$) and $\Omega$ (you pick $1$ or $2$). In this case, the probability you guess a randomly generated number in $\Omega$ is $1/2$. 
Back to your case, the problem is to define a probability measure on $\mathbb{N}$. You can't do that with the counting measure anymore: that is you can't put an equal nonzero probability on each atom $\{n\}$, which is what you've done implicitly. Indeed, the measure of $\mathbb{N}$ would then be$+\infty$. So that would not be a probability space.
Because I choose the number and I am a limited human being, there is intuitively a good chance that I pick a number smaller than, say, $1000$ billions. So a more realistic model would say that the weight of $\{n\}$ should tend to $0$ as $n$ tends to $+\infty$. Now this still leaves you with a lot of choices since putting a probability measure on $(\mathbb{N},\mathcal{P}(\mathbb{N}))$ amounts to choosing a converging series with nonnegative general term and sum equal to $1$. Which series do you choose?
