Assuming Riemann Hypothesis Hardy showed primes $3\bmod 4$ are more common than primes $1\bmod 4$ <https://en.wikipedia.org/wiki/Riemann_hypothesis#Consequences_of_the_generalized_Riemann_hypothesis>. 


Is there any reason to believe there are asymptotically equal number of consecutive primes that are $1\bmod 4$ and $3\bmod 4$ with gap $4$? 


Is there a possibility that there are only finitely many primes of $1\bmod 4$ form with gap $4$ while there are infinite of them which are $3\bmod 4$? 


What is the intuitive reason why primes $3\bmod 4$ are more frequent?



Finiteness of number of consecutive primes with gap $4$
The answer ought to be 'no', that one should not expect there to only be finitely many prime pairs of the form $(4k+1, 4k+5)$ while expecting infinitely many pairs of the form $(4k+3, 4k+7)$. While we cannot prove this, it is known that weaker version of this is true.


Consider a $k$-tuple of linear functions $(L\_1(n), \cdots, L\_k(n)) = (a\_1 n + b\_1, \cdots, a\_k n + b\_k)$ with the property that there is no prime $p$ which divides $\prod\_{i=1}^k L\_i(n)$ for all integers $n$. Then it is known, from the seminal work of James Maynard (see here: <http://annals.math.princeton.edu/2015/181-1/p07>. This was independently discovered by Terry Tao), that there exist infinitely many $n$ such that a proportion of $(1/4 + o(1))\log k$ of the values $(L\_1(n), \cdots, L\_k(n))$ are prime. It is widely believed that the statement can be strengthened to assert that there are infinitely many $n$ (with a predicted density even) such that *all* terms in $(L\_1(n), \cdots, L\_k(n))$ are simultaneously prime. 


If we apply this to the tuple $(4n+1, 4n+5)$ and $(4n+3, 4n+7)$ which satisfy the hypothesis above, then the assertion is true. Notice that the theorem of Maynard (and Tao) does not require any additional hypotheses on the linear functions $L\_j(n)$ except what is absolutely needed in order to make the problem non-trivial.


