Roulette complex question I have a complex situation that I don't know how to solve.
So I made a program that simulates roulette spins, and makes statistics from the numbers.
For example:
From 25 000 spins the program calculates that the biggest gap for the first 12 numbers is for example 190. Now my question is, that if I spin 25 000 more, what is the probability that from the second 25 000 spins the biggest gap for the first 12 numbers is 190 again?
 A: You have a sequence of $N = 25000$ independent Bernoulli trials, where I'll consider a "success" to be a spin not from 1 to 12.  In each trial the probability of success is $p$, and the probability of failure is $q=1-p$.  Consider the distribution of (success) runs of length $r$, where we say a run of length $r$ occurs at trial number $j$, $r \le j \le N$, if trials number $j-r+1$ to $j$ are successes but a run of length $r$ 
does not occur on any of $j-r+1$ to $j-1$.  Thus in the sequence $SSFSSSS$ (where S denotes success and F failure) we have runs of length $2$ occurring at trials $2$, $5$ and $7$.  Let  $f_r(n)$ the probability that the first run of length
$r$ occurs at trial $n$.  The probability that there is no run of length $r$ in
$N$ trials is thus $Q_r = 1 - \sum_{n=1}^N f_r(n)$.  Since if there is 
a run of length $r$ there are also runs of all lower lengths, $$Q_r - Q_{r-1}
= \sum_{n=1}^N \left(f_{r-1}(n) - f_r(n)\right)$$ 
is the probability that the longest run has length $r$.
Now  the generating function of the $f_r(n)$ is
$$ F_r(s) = \sum_{n=r}^\infty f_r(n) s^n = \dfrac{p^r s^r(1-ps)}{1-s+qp^r s^{r+1}}$$
Thus the generating function of $f_{r-1}(n)-f_r(n)$ is
$ G_r(s) = F_{r-1}(s) - F_r(s)$.
From this we can calculate the probability of the longest run having length $r$
as $\sum_{n=1}^N g_n$ where $G_r(s) = \sum_{j=0}^\infty g_j s^j$.
As 
KSmarts remarked, for your case with $r = 190$ this probability is going to be extremely small.  If you continued your trials indefinitely, the expected number of trials until the first run of length $r$ would be
$(1 - p^r)/(q p^r)$, and with $r = 190$ and $p = 13/19$ this is
approximately $6.52 \times 10^{31}$.
A: Assuming that you are using an American-style Roulette wheel, with 38 slots numbered 1-36 and two zeroes, then the chance of getting a number between 1 and 12 (which I'll call "low numbers") on any particular spin is $\frac{12}{38}=\frac{6}{19}\approx0.316$. Then the number of spins between low numbers is a geometrically distributed random variable with parameter $p=\frac{6}{19}$.
Taking a single trial, where you spin until you get a low number, the chance of it taking exactly $190$ spins is
$\left(1-\frac{6}{19}\right)^{190}\left(\frac{6}{19}\right)$,
which is, in technical terms, really friggin' small. It's approximately $1.5\times10^{-32}$. If we were each to randomly select one kilogram of matter from the (discretely-divided) sun, it would be more likely that we picked the same $1$kg than that this roulette test would fail $190$ times in a row. Granted, you spun $25000$ times, during which you would have gotten thousands of low numbers, and thus thousands of chances to wait, but it is still really unlikely. This makes me suspect that either your program is flawed, or I have misunderstood your question.
As to the general problem of the largest number of consecutive failures in $n$ trials, I don't know the solution offhand. I will continue working on and searching for it and will update this answer, but I would like to hear your response to my above observation as well.
