Red Queen and her roses

Let $x_n$ and $y_n$ be the proportions of $AA$ and $Aa$ red roses present at the end of day $n$ . Then $x_0 = 0$ and $y_0 = 1$ , and the proportions of $AA$ , $Aa$ and $aa$ roses the next day (before culling) are $\begin{aligned} \hat{x}_{n+1} & = \; x_n \times x_n \times 1 + x_n \times y_n \times \tfrac12 + y_n \times x_n \times \tfrac12 + y_n \times y_n \times \tfrac14 \; = \; (x_n + \tfrac12y_n)^2 \; = \; (1 - \tfrac12y_n)^2 \\ \hat{y}_{n+1} & = \; x_n \times x_n \times 0 + x_n \times y_n \times \tfrac12 + y_n \times x_n \times \tfrac12 + y_n \times y_n \times \tfrac12 \; =\; x_ny_n + \tfrac12y_n^2 \; = \; \tfrac12y_n(2 - y_n) \\ \hat{z}_{n+1} & = \; y_n \times y_n \times \tfrac14 \; = \; \tfrac14y_n^2 \end{aligned}$ so that, after culling the $aa$ white roses, we have proportions $\begin{aligned} x_{n+1} & = \; \frac{(1 - \frac12y_n)^2}{1 - \frac14y_n^2} \; =\; \frac{2 - y_n}{2 + y_n} \\ y_{n+1} & = \; \frac{\frac12y_n(2 - y_n)}{1 - \frac14y_n^2} \; = \; \frac{2y_n}{2 + y_n} \end{aligned}$ and the Red Queen's dream will come true on Day $n$ when first $\hat{z}_{n+1} = \tfrac14y_n^2 < 10^{-6}$ , and hence when $y_n < \tfrac{1}{500}$ .

Since $y_{n+1}^{-1} = y_n^{-1} + \tfrac12$ , we deduce that $y_n^{-1} = 1 + \tfrac12n$ , so that $y_n = \frac{2}{n+2}$ for all $n \ge 0$ . Thus the Red Queen's dream will come true on Day $n$ , when first $\frac{2}{n+2} < \frac{1}{500}$ , and hence when first $n > 998$ , and so the Red Queen's dream comes true on day $\boxed{999}$ .