Finiteness of number of consecutive primes with gap $4$ 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?
 A: 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.
A: Since there are several questions posed, here is my go at an intuitive explanation of Chebyshev's bias. Consider the two arithmetic progressions $4 n + 1$ and $4 n + 3$:  
\begin{matrix}
4n+1& 4n+3\\
\hline
1   & 3\\
5  & 7\\
9       & 11 \\
\vdots  & \vdots
\end{matrix} 
Now let's pick any two numbers to generate a new composite: There are three possibilities: 
1) A product of two elements from column 1 ends up in column 1. 
2) A product of one element from column 1 and one from column 2 ends up in column 2. 
3) A product of two elements from column 2 ends up in column 1.   
So, on average there are more composites pushed into column 1 and we should therefore expect less primes in this column than in column 2. 
In general, consider the arithmetic progressions $dn + a_i$, where $a_i$ and $d$ are coprime, $1\leq i \leq \phi(d)$, and $\phi(d)$ is the Euler totient function. Then we can set up the matrix $(dn + a_i)(dn + a_j) \mod d$, telling us in what arithmetic progression a composite ends up. In the case above we get the matrix
$$
\left(
\begin{array}{cc}
 1 & 2 \\
 . & 1
\end{array}
\right),
$$
so we see directly that the ratio is 2:1 in favour of sending composites to $4n+1$. 
Another example is the case of $5n + a_i$. The resulting matrix is then 
$$
\left(
\begin{array}{cccc}
 1 & 2 & 3 & 4 \\
 . & 4 & 1 & 3 \\
 . &  . & 4 & 2 \\
  . &  . &  . & 1 \\
\end{array}
\right).
$$
Here we see that more composites are sent to the progressions $5n+1$, and $5n+4$, so we should expect $5n+2$, and $5n+3$ to have more primes. The ratio this time is 3:2:2:3, in the order of $a_i$. The larger $\phi(d)$ is, the closer the ratios will be to 1, and in the infinity limit of $\phi(d)$ we should not be able to separate out a winner of any prime race. 
None of this give any proof of anything, but as a starting point to think about this problem, i quite like this approach. For more meaty stuff, one should check out the article Chebyshev's bias
