If $p$ is the largest prime less than $2n$, what is the probability that $2n-p$ is a prime? Let $p$ be the largest prime not exceeding an even number $2n$. I observed that $2n - p$ has a high density of primes and it decrease very slowly. More specifically, for $2n \le 10^6$ the density is $0.6212$ and it drop down to $0.58427$ for $2n \le 3.65 \times 10^{10}$.
Question 1: What is the density of primes in the sequence $2n-p$?
Some observation: The divisors of $n$ do not divide $2n-p$ for $n \ge 4$. Thus, the  more divisors $n$ has, the more likely it is for $2n-p$ to be a prime. This was observed in the data. Let $a_{n,d}$ be the sequence of positive integers which have $d$ or more divisors. I observed that as $d$ increases, the density of primes in the sequence $2a_{n,d} - p$ increase. Given below are the densities of primes in the first $10^6$ terms of $2a_{n,d} - p$ for different values of $d$.
(d, density)
(2, 0.621)
(3, 0.625)
(5, 0.635)
(10, 0.647)
(20, 0.662)
(50, 0.695)
(100, 0.720)
(150, 0.723)
(200, 0.744)
(230, 0.747)

 A: I would like to say that this probability to goes to zero if $n$ goes to inifity... and the speed, I guess, will be REALLY SLOW.
Let $\pi(n)$ to be the number of primes smaller than $n$. Let $m=\pi(2n)$ and denote $g_{m}=2n-p_m$ Let the gaps $g_t$ to be $p_{t+1}-p_t$ for $t\le m-1$, where $p_{t}$ is the $t$th prime.  We can use binomial coefficients to proof that $\pi(n)<\frac{10n}{\ln n}$. This is a really rough bound, but it is enough. Let $m=\pi(2n)$ and denote $g_{m}=2n-p_m$.
Now consider the $2n$'s between $p_k$ to $p_{k+1}$ for $k\ge 2$. So the $2n-p_k$ between them will be $1,3,\dots,p_{k+1}-p_k-1$, and at most $\pi(p_{k+1}-p_k)\le 10 g_k/\ln g_k$ primes inside them.
So the total number of even $l\le n$ such that $l$ minus the largest prime small then $l$ is also a prime will be $\le \sum_{i=1}^m 10g_i/\ln g_i$.
Notice that $f(x)=10x/\ln x$ is a concave function when $x\ge 8$. So it is largest when some of them are equal, and others are $2,4,6$. So we can over estimate them: $m$ of them are $2$, $m$ of them are $4$, $m$ of them are $6$, and the else are averaged by $m$ numbers with the total sum of $2n$. So we have:
$$\sum_{i=1}^m 10g_i/\ln g_i \le 10m(2/\ln 2+4/\ln 4+6/\ln 6)+10m(2n/m)/\ln (2n/m)\le 120m+20n/\ln(n/m)\le 2400n/\ln n+20n/\ln(0.1\ln n)\le 2400n/\ln n +200n/\ln \ln  n$$ if $n\ge20$.
Therefore, the probability is upper bounded by $\frac{2400}{\ln n}+\frac{200}{\ln \ln n}$, which tends to zero if $n\to\infty$.
