What is the probability of occurrence of natural numbers? What is the probability of a number $n ∈ ℕ$ to occur when someone uses a natural number in daily life?
(You can assume that humankind had a number system in which no psychologically distinguished numbers existed like 10000, 9999, 250, or 333. E.g. if humankind would use an ∞-ary number system, then this assumption would be fulfilled.)
 A: A distribution like $P(n) = 2^{-n}$ isn't going to give very good predictions.  According to that I'm $8$ times more likely to say "$997$" than "$1000$".  You mention that base $10$ is sort of a cultural thing that should maybe be ignored, but plenty of numbers distinguish themselves for other reasons and should be more common than their neighbors.
One way of improving this situation is to fix some description language and then define $P(n) = c \cdot 2^{-2^{K(n)}}$ where $K(n)$ is the Kolmogorov complexity of some number $n$ and $c$ is a constant chosen to make all the probabilities add up to $1$.  This will give higher probabilities to numbers with shorter descriptions, so "googolplex" will have a much higher probability than some random number with a googol digits.  Like base $10$, whatever description language you choose will come with some cultural baggage but some are definitely more natural than others (for example English vs. combinator calculus).
A: I wondered about thinking about it in the following way. Suppose we consider a subset of $\mathbb{N}$ which only goes up to $N$. Then suppose, in daily life there arises the situation which will use a number no higher than $X\leq N$ where  $X$ is a random variable uniformly taking a value in $\{1,2,...,N\}$. Then, assuming the probability of drawing each number from this set is equal, the probability of a number $k\leq N$ being used in daily life is 
$$\begin{align}
P(k) &= \sum_{i=k}^N \frac{1}{i}P(X=i)\\
     &= \frac{1}{N} \sum_{i=k}^N \frac{1}{i}.
\end{align}$$
I was then planning on taking the limit of this as $N\rightarrow\infty$, however I think this goes to zero... I wonder if some variant of this would be the right approach?
A: I once did an analysis on the "popularity of numbers" by just googling them and then recording how many "google search results" they obtained. 
The result was a pretty consistent relationship where $$\text{num_results}(n) \sim \frac{1}{n}$$.
I know this is only one case of your questions in general, but the internet happens in my daily life at least.
A: 0, If the distribution is uniform (if you allow any Natural number in the list from which you choose), which I assume you mean when you say there is no distinguished numbers. Otherwise, for an infinite collection, there is no uniform distribution and you need to specify the distribution used. 
EDIT: If you want more of a real-life distribution, you could consider, e.g., Benford's law : https://en.wikipedia.org/wiki/Benford%27s_law
A: I'm not sure this question has any kind of assessable answer, but one common assertion that seems to hold some water is the notion that numbers in discourse are distributed roughly logarithmically; this is the basis behind such patterns as Benford's Law.  Granted, that usually applies to real numbers (ignoring sign), but it doesn't take too much work to apply it to the natural numbers.
Of course, the sum of the logs (or the integral of the log function) is unbounded, so using that to directly arrive at a probability distribution won't work.  Ultimately, we will have to set an upper limit to the numbers we consider.  We might use any of a number of very large numbers used in various proofs, but since you mention "daily life" and not "abstruse mathematics," I think it's reasonable to set an upper limit at about a trillion, the largest number that most people are likely to come across in everyday discourse (in connection with government budgets, let's say).
I would also aggregate all numbers with the same first three or so significant digits; that is to say, the number $345217469$ would be assigned a probability of essentially zero, as would most nine-digit numbers beginning $345$, but $345000000$ would be given the sum of all of their probabilities.
That's not quite ideal, but given that the logarithmic distribution is already an idealization (!), that's probably close enough to get in the neighborhood.
