autosomal recessive population numbers
In population genetics, with a simple one-locus two-allele model, we consider the frequency of the two alleles to be $p$ and $q$ where $p + q = 1$ (and thus $1-p=q$ and $1-q=p$). When an allele is recessive it is only expressed at the phenotypic level when the individual carries two copies of the allele.

Following from your example, gene $A$ might be responsible for hair colour, with alleles $A_1$ and $A_2$ where $A_2$ is a recessive allele responsible for purple hair. I will denote the frequency of $A_1$ with $p$ and frequency $A_2$ with $q$. This means that the frequency of individuals with purple  hair ($f_{A_2A_2}$) is $q^2$ and the frequency of people with not-purple hair is $p^2 + 2pq$ (assuming an idealised population). As such, a population of 500 diploid individuals* where there is 100 copies of the $A_2$ allele would have the following genotypic (&amp; phenotypic) frequencies:

$$f_{A_1 A_1} = p^2 = 0.9^2 = 0.81$$ 
$$f_{A_1 A_2} = 2pq = 2 \times 0.9 \times 0.1 = 0.18$$
$$f_{A_2 A_2} = q^2 = 0.1^2 = 0.01$$ 

So the allele for purple hair would be present in 19% of the population, but the trait would only manifest in 1% (only ~5.3%, $\frac{1}{19}$, of the people that carry the purple hair allele exhibit the trait). Therefore selection would be quite inefficient at removing this allele from the population, even if you could prevent all purple-haired individuals from reproducing. To find the frequency of $q$ required to give a genotypic/phenotypic frequency where 1 in $x$ individuals have purple hair just take the square root of $1/x$. E.g. to get 1 in 1000 individuals expressing purple hair 

$$q = \sqrt{q^2} = \sqrt{(1/1000)}$$

What does this mean for evolution? Alleles that are recessive deleterious are very hard to remove from the population because, at low frequencies, they rarely produce the detrimental effect. Assuming complete dominance ($A_1 A_2$ = exactly like $A_1 A_1$), denoted by $h = 0$, selection only acts on $A_2 A_2$ homozygotes. The frequency of the $A_1$ allele increases at the rate $1/(1-sq)$ where $s$ is the relative fitness advantage. Mutation can also introduce more copies of the $A_2$ allele to the population such that there will be an equilibrium, balanced by the rate of loss due to selection, and rate of gain by mutation. Furthermore, to further build on your analogy, a neighbouring population, where purple-hair is not selected against, may bring copies of the allele in by migration.



Suggested further reading on evolution here and reading on eugenics would likely be of interest to you.

* 500 diploid individuals = 1000 gene copies. Thus $q = 100/(500 \times 2) = 0.1$