Distribution of primitive roots mod p Let $p$ be a prime number. I am interested in knowing how many primitive roots mod $p$ there are; at least, gaining some insight into the distribution of primitive roots mod $p$. 
If I need to go looking for a primitive root, how far down the list of integers should I expect to look before I find one?
I know there are $\phi(p-1)$-many primitive roots mod $p$. Therefore the ratio of primitive roots mod $p$ is given by $\phi(p-1)/(p-1)$. 
I could not find any theorems talking about bounds on this value for any primes of a certain form. So I plotted it for the first 100000 primes 
I appreciate that there are infinitely many primes and that the behaviour of the first 100000 need not tell us anything about the overall behaviour. That being said, I am hoping someone could explain some of the features of this plot that stand out to me. For example: 


*

*The Number of primitive roots is bounded between 1/5 and 1/2. It seems some might sneak lower than 1/5.

*There are a number of dense lines.  For example: It seems there are lots of primes with 1/3 of integers being primitive roots. 
If anyone can point out any references about the distribution of primitive roots. Or say anything about what might be happening here, that would be great. 
 A: One can certainly find integers $n$ with
$$\frac{\varphi(n)}{n} \sim \frac{e^{-\gamma}}{\log \log(n)}$$
(for example, the product of the first $k$ primes), and this is best possible. By Dirichlet's theorem, there exists a prime 
$$p \equiv 1 \mod n.$$
Linnik proved there exists such a prime $p < n^{C}$ for some absolute fixed constant  $C$ whose value will not be important. (Allowing an extra constant factor, I think the best known bound has $C$ roughly aroung $5$.)
Since
$$\frac{\varphi(p-1)}{\varphi(n)} = \frac{p-1}{n} \prod_{q|n}^{q \nmid p-1}
\left(1 - \frac{1}{q} \right) \le \frac{p-1}{n},$$
it follows that, for such a $p < n^C$ (so $n > p^{1/C}$),
$$\frac{\varphi(p-1)}{p-1} \le \frac{\varphi(n)}{n} \sim  \frac{e^{-\gamma}}{\log \log(n)}
\le \frac{e^{-\gamma}}{\log \log(p^{1/B})} \sim \frac{e^{-\gamma}}{\log \log(p)}.$$
Hence
$$\liminf \frac{ \log \log p \cdot  \varphi(p-1)}{p-1} 
= e^{-\gamma}.$$
On the other end, standard sieving certainly shows you can find primes $p$ such
that $p-1 = 2 q_1 \ldots q_k$ where all the $q_k$ are greater than (say) $p^{1/100}$.
 For these primes, you certainly have
$$\frac{\varphi(p-1)}{p-1} \sim \frac{1}{2}.$$
Random example:
$$p = 106696591 = 2 \cdot 3 \cdot 5 \cdot 7 \cdot 11^2 \cdot 13 \cdot 17 \cdot 19 + 1,$$
$$\frac{\varphi(p-1)}{p-1}  \sim 0.17\ldots$$
$$\frac{e^{-\gamma}}{\log \log p} = 0.19 \ldots$$
