TLDR: How does one have 1, 2 and 3 holding?
Let $N$ be a large number. By the PNT we have $n \in [N, 2N)$ is prime with probability $\frac{1}{\log N}.$ I don't understand how the following then applies:
- $\sum_{h=1}^{k}\mathbb{P}(n+h\text{is prime}) > 1$ if $k$ is of size (roughly) $\log N$. In particular I don't understand why it is $> 1$ and not $= 1.$
- $\mathbb{P}(\text{at least two of}n+h\text{is prime}) > 0$ by a variant of the pigeonhole principle.
The above gives a prime gap that is of size $\ll \log N$. However because we don't have to choose $n$ uniformly, we could hope to find weights, $w_{n}$ that are positive and determine a probability distribution (probability density?) $\frac{w_{n}}{\sum_{N\leq n < 2N}w_{n}}$ such that if $N$ is large enough the following holds, $ \sum_{h=1}^{k}\mathbb{P}(n+h \text{is prime}) = \sum_{h=1}^{k} \frac{\sum_{N \leq n < 2N}w_{n}\chi_{n+j_{m} \text{is prime}}}{\sum_{N \leq n < 2N}w_{n}} > \rho$, for some $\rho > 1$ and $j_{m}$ belongs to some admissible set $J$. This would then prove bounded gaps between primes.
The weights are allowed to be a bit worse than optimal (since otherwise it is as difficult as showing the prime k-tuples conjecture). How does 3 hold?
- For all except a finite number of $n \in [N, 2N), P(n) = \prod_{j \in J}(n+j)$, the product of integers are all smaller than $N^{2}.$ Therefore $\chi_{\text{all} n+j_{i}\text{prime}} = \chi_{p|P(n) \Rightarrow p > N}$