I want to generate numbers $1$ to $10$ with uniform probability distribution. Will this work? I want to generate numbers $1$ to $10$ with uniform probability distribution.  So I write  the numbers $1$ to $10$ in the natural order.  I keep writing the next $10$ number block by permuting the first $10$ numbers.  Will this long string can be considered random with uniform distribution?  Is there a problem?
 A: Absolutely not.  The resulting distribution of numbers will not be random, in the sense that successive observations will fail to be independent and identically distributed:  if $X_i$ is the value of the $i^{\rm th}$ observation, then suppose we observed $${\boldsymbol X}_{1:19} = (X_1, X_2, \ldots, X_{19}) = (3, 5, 7, 10, 4, 6, 2, 1, 8, 9, 4, 7, 2, 1, 9, 8, 6, 10, 3).$$  Then $$\Pr[X_{20} = 5 \mid {\boldsymbol X}_{1:19}] = 1.$$  Moreover, you cannot even have $$X_1 = X_2 = 1,$$ which is possible if in fact the $X_i$'s are all independent and identically distributed discrete uniform random variables on $\{1, \ldots, 10\}$.
A: The distribution you will get will be far from uniformly distributed. For example, if I take 5 numbers between $1$ and $10$, there is a $0.001\%$ chance that all $5$ numbers will equal $3$. In your case, the probability is $0$. On the other hand, there will be exactly $100$ numbers that equal $5$ in the first $1000$ numbers you produce, while, if taking the truly random numbers, the probability of this happening is fairly low (something like $4.2\%$ if I am not mistaken).

So, where is the problem?
Your chain of numbers will be a chain of $10$ intertwined uniformly distributed variables, meaning that if you take your first, eleventh, twenty-first... number, the resulting requence will indeed be a sampling of a uniformly distributed random variable.
