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

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