Are primes (ignoring $2$) equally likely to be $1~\text{or}~3\pmod 4$?

Question

For all primes $p\neq 2$, it's easy to see that $$p\equiv 1~\text{or}~3\pmod 4$$ I was wondering if it's equally likely ($50\%-50\%$) that prime modulo $4$ is $1$ or $3$. And if so, is there a simple proof?

Answer 1

There are infinitely many primes of forms $4n + 1$ and $4n +3$.

However, it is not clear which of the two are more abundant.

In 1853, Chebyshev in a letter indicated that he had a proof that the number of primes of the form $4n+1$ is less than the number of primes of the form $4k+3$. However, in 1914, Littlewood showed that Chebychev's assertion fails infinitely often; however, he did not specify where this first reversal occurs.

Nevertheless, some forty years later in a computer search, it was discovered that the first prime for which the $4n+1$ primes become more plentiful than the $4n+3$ primes is for the prime $26861$.

That situation is not reversed until the prime $616,841$.

Although every prime is either of one of these two types infinitely often, and despite Littlewood's proof, the density of each of these two prime types as far as I know has not been established. That being the case, it remains an open question that given a prime $p$, it is more likely to be of one type than another.

Answer 2

Summarizing the comments: They occur with equal probability, but being 3 mod 4 occurs more for the most part (but not by an overly large amount). More concretely, if $a$ and $b$ are the number of primes less than $N$ that are 1 mod 4 and 3 mod 4 resp., then $$\lim_{N\to\infty}\frac{a}{b}=1$$, but it's usually the case that $\frac{a}{b}<1$. These are due to the prime number theorem for arithmetic progressions and Chebyshev's bias respectively.