Counting zeros and ones in binary representations of prime numbers If you write down binary representations of all prime numbers starting from 3 up to some (very big) $N^{th}$ 
prime number and denote with $S_1(N)$ the total number of ones (1) and with $S_0(N)$ the total number of zeros (0) used in all binary representations, what kind of relation would you expect between $S_1$ and $S_0$?
I did not expect to see anything like 50:50 distribution. All binary representations start with 1 and all primes end with 1. So it looks like digit 1 is favored: binary representation of every odd prime is guranteed to have at least two ones. There's no such guarantee for zeros. 
But what is the ratio between ones and zeros if you remove the first and the last digit from all binary representations? I meen, what happens if you look only at the "central" part of the number?
I might be too naive but I expected to see that, with the first and the last digit removed from every binary presentation, the ratio between ones and zeros would be close to 50:50. In other words, I expected to see something like:
$$S_1(N)\approx S_0(N) + 2 N\tag{1}$$
But in reality it's not like that and I would say not even close. I did an experiment with the first 10 million primes and got the following results (BTW, 10,000,000th odd prime turned out to be 179,424,691):
$$N=10,000,000$$
$$S_1=139,605,415$$
$$S_0=124,501,052$$
Proper relationship seems to be:
$$S_1(N) \approx S_0(N) + \frac32 N\tag{2}$$
The margin of error in my experiment is less then 0.1%. 
Is there a simple explanation why the correct factor seems to be equal to $\frac32$, not 2?
(I can also share the Java code if someone wants to validate it.)
EDIT: Here is the graph of function:
$$\frac{S_1(N)-S_0(N)}{N}$$
for the first 20 million primes:

It turns out that the value 1.5 is reached only from time to time. 
 A: If you consider the primes between $2^n$ and $2^{n+1}-1$, inclusive, you would expect there to be approximately $\frac{c2^n}{n}$ of them for some fixed $c$ (I believe it's $\frac{1}{\ln 2}$), and each should have "on average" $\frac{n+3}{2}$ ones and $\frac{n-1}{2}$ zeroes (assuming the middle bits are chosen uniformly at random. This gives, up through $2^n$, 
$$c\sum_{k=1}^{n-1} \frac{2^k}{k}\cdot\frac{k+3}{2}$$
ones, and
$$c\sum_{k=1}^{n-1} \frac{2^k}{k}\cdot\frac{k-1}{2}$$
zeroes, in 
$$c\sum_{k=1}^{n-1} \frac{2^k}{k}$$
primes. Letting
$$f(n)=c\sum_{k=1}^{n-1} 2^k,\ \ g(n)=c\sum_{k=1}^{n-1} \frac{2^k}{k},$$
we have, out of $g(n)$ primes, $\frac{f(n)+3g(n)}{2}$ ones and $\frac{f(n)-g(n)}{2}$ zeroes; the difference is thus $2g(n)$, as expected.

Let's say, however, that you count up to $3\cdot 2^{n-1}$ instead. You still have the same number of $0$s and $1$s you did before, but now you have all the numbers from $2^n$ to $3\cdot 2^{n-1}$ as well. We expect there to be about
$$c\frac{2^{n-1}}{n}$$
of them, and they have two ones, a zero, and $n-2$ "free digits," for "on average" $\frac{n+2}{2}$ ones and $\frac{n}{2}$ zeroes. This gives us, across our
$$g(n)+\frac{c2^{n-1}}{n}$$
primes, approximately
$$\frac{f(n)+3g(n)}{2}+\frac{n+2}{2}\cdot\frac{c2^{n-1}}{n}$$
ones and
$$\frac{f(n)-g(n)}{2}+\frac{n}{2}\cdot\frac{c2^{n-1}}{n}$$
zeroes, for a difference of 
$$2g(n)+\frac{c2^{n-1}}{n}.$$
The ratio between this and the number of primes we have is 
$$\frac{2g(n)+\frac{c2^{n-1}}{n}}{g(n)+\frac{c2^{n-1}}{n}};$$
since $g(n)\sim \frac{c2^n}{n-1}$, this gives us a ratio of
$$\frac{4+1}{2+1}=\frac{5}{3}.$$
This isn't exactly the $\frac{3}{2}$ you got, but it's certainly not $2$ either. As such, we should expect this "difference ratio" to differ from $2$ significantly depending on where your cutoff is. It's possible that if you stop at $\frac{5}{4}\cdot 2^n$ instead, you'll get a different number; the bottom line is that your heuristics can still be right without $\frac{S_1(N)-S_0(N)}{N}\to 2$.
