distribution of digits in prime numbers I am curious about this: suppose we consider all numbers in base $b$ such that the number of digits $n$ in this range is the same ( eg, in base $10$ it could be $10-to-99$ for $n=2$, or $100-to-999$ for $n=3$, etc; leading digit is non-zero), for the prime numbers in this range, if I were to choose a prime number at random can I expect the distribution of the digits of my prime to be uniform random? That is, $\frac{n}{b}$.
Thank you.

In an odd base, an odd number always has an odd number of odd digits.
Proof
$$\begin{eqnarray}(2n+1)+(2p+1)=2q
\\
(2r+1)+(2s)=(2t+1)
\\
(2u)+(2v)=(2w)\end{eqnarray}$$
By $$(x+y)+z=x+(y+z)\tag{4}$$ and $$a+c+d=a+d+c=c+a+d=c+d+a=d+a+c=d+c+a\tag{5}$$ we have $$\underbrace{(2h+1)+\cdots +(2h+1)}_{\text{2i+1 times}}=2e+1\implies (2i+1)(2h+1)=(2e+1)$$ $$ 123456789_b=1\cdot b^8+2\cdot b^7+3\cdot b^6+4\cdot b^5+5\cdot b^4 +6\cdot b^3+7\cdot b^2+8\cdot b^1 +9\cdot b^0$$ that in an odd (2h+1) base b, that all odd digits create odd summands. it follows from (1),(2)(3), that the even (2u) digits create even (2w) summands regardless of base. 
even (2v) bases luck out in that $b^0=1\quad b\neq 0$. Otherwise, they couldn't represent odd (2r+1) numbers at all.
There are other things like all primes greater than 3 being 1 or 5 on division by 6, that can play with things. in a $6k+1$ base, then the last digit being $6j+4$, will force the rest of the digits to represent a number of forms $6l+1$ or $6m+3$ etc. 
