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Here is my question: do we have any kind of estimate about $p_{k, d}(n)$ the probability that there are at least $k$ prime numbers in a radius of $d$ around $n$?

Do you have any suggestions regarding related work?

For instance, we know that for $n$ there is a prime $p : n\leq p \leq 2n $ (Tchebychev, 1850), meaning:

$p_{1, n/2}(\frac{3n}{2}) = 1, \forall n>1$

Also since it has been shown that there are infinitely many prime gaps at most 246:

$p_{1, 246}(n) \neq 0, \forall n>1$

$^1$ I believe 246 is the smallest, though 2 is a well known conjecture

ted
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    Chebyshev means rather that $p_{1,n/2}(3n/2)=1$. What you have written means that there is a prime between $0$ and $2n$. – TonyK Mar 12 '19 at 20:51
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    I don't think this is true, it says: there is at least one prime between n and 2n which means there is at least 1 prime within a radius n from n does it not? – ted Mar 12 '19 at 20:55
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    I am not sure that $246$ is known in the sense that you have put - it is known that there are an infinite number of prime gaps less than or equal to this - but it would be possible for there to be a finite number of gaps of size $246$ provided there were a smaller number (say $214$) for which the number of gaps of this size is infinite. At least on my reading of what is known. – Mark Bennet Mar 12 '19 at 21:02
  • @MarkBennet In fact this is unknown. Even worse, for no even number $k$ it is known whether infinite many primes $p$ exist such that $p+k$ is prime as well. It is however (strongly) conjectured that for every even $k$, there are infinite many examples. – Peter Mar 12 '19 at 21:04
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    TonyK is saying that there is a prime within $n/2$ of $3n/2$ which halves the radius. – Mark Bennet Mar 12 '19 at 21:04
  • @Peter Indeed, but it is known that there is an even number $k\le 246$ for which this is true - we just don't know which one(s). – Mark Bennet Mar 12 '19 at 21:05
  • @MarkBennet Yes, of course, not all differences upto $246$ can produce only finite many examples. – Peter Mar 12 '19 at 21:06
  • "In a radius of $d$ around $n$" means from $n-d$ to $n+d$. So for $d=n$ that would be $0$ to $2n$, and you don't need Chebyshev for that: the prime $2$ would qualify as long as $n > 1$. – Robert Israel Mar 12 '19 at 21:11
  • Also, what do you mean by "the probability that..."? For any particular $n$ and $d$ there either is a prime in the interval or there isn't, no probability involved. – Robert Israel Mar 12 '19 at 21:13
  • I edited regarding the 246 matter and @TonyK you are right but I think my notation still holds, both are true. – ted Mar 12 '19 at 21:15
  • @RobertIsrael I'm saying it's not always the case that for given n and d there are k primes but it can happen so I'm interested in knowing if we have any kind of boundaries regarding this likelihood – ted Mar 12 '19 at 21:19
  • "I think my notation still holds, both are true": well, you could have written $2>1$ and it would have been true. But it wouldn't have been interesting. – TonyK Mar 12 '19 at 21:29
  • haha ok I just figured your point out about 2, makes sense to use your formulation instead I agree sorry about that – ted Mar 12 '19 at 21:31
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    There are arbitrarily large prime gaps, so there are some $n$ for which there are no primes in the interval $n-246, n+246$. For those $n$ your statement $p_{1,246}(n) \ne 0$ is false. For example, $n = 500! + 248$. – Robert Israel Mar 12 '19 at 21:38

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Which other unsolved problems, have necessary restrictions on the prime gaps? a related question I just got answered. As the comments on your question talk about though, there's not really a restriction. Primorials (products of all primes up to a number) have potentially massive gaps nearby, You can gaurantee all numbers from the primorial plus or minus 2, until the primorial plus or minus the first prime not in the primorial minus or plus 1, are composite for $30=2\cdot3\cdot5$ you get that all numbers in ranges 24-28 and 32-36 are necessarily composite (divisible by a prime in the factorization of 30). Unsolved conjectures, put bounds on d for all k values. Goldbach, has Bertrand's postulate as a necessary condition. Legendre, implies that two primes exists between $y^2$ and $(y+2)^2$ , $y$, a natural number. Grimm's, implies that $d<\pi(n)$ for $k=1$, for almost all ( all but finitely many) $n$. If not then we have a pigeonhole contradiction.