Using the Hypergeometric Distributon notation from Wikipedia, if I treat $k$, $n$, and $\Pr(X\ >\ k)$ as constants and solve for $K$ as a function of $N$ in Mathematica, the relation appears to be linear. Here is my setup using $k\ =\ 3,\ n\ =\ 8,\ \Pr(X\ >\ 3)\ =\ .72$...
Similar results hold for other values of the constants. I find this result surprising, because, as an example, when we shuffle more and more copies of a $52$-card playing card deck together, the probability of drawing a Queen in a single draw remains constant at $\frac{1}{13}$, a reflection of the constant $\frac{K}{N}$ ratio, but as soon as we look for more than one Queen in multiple draws, the impact that drawing the first Queen has on the probability of drawing subsequent Queens depends on how many copies of the deck have been combined. This asymmetry arising from drawing without replacement should guarantee that the resulting probability is not a linear function of $\frac{K}{N}$, and thus when we hold that probability constant (and keep $k$ and $n$ constant as well), my intuition would suggest that the nonlinearity has to pop up in the relation between $K$ and $N$.
Is $K$ in fact linear (or at least affine?) in $N$, in all Hypergeometric Distributions? If so, then is there an intuitive explanation of why this is true? If not, then is there something wrong with my setup in the code?
Fix $0 \leq k < n$ and $\alpha \in (0, 1)$. Let $X$ have the hypergeometric distribution with parameters $N$, $K$, $n$, using the notation from Wikipedia. We consider the equation
$$ \mathbf{P}(X > k) = \alpha \tag{1} $$
in $N$ and $K$. For example, the blue curve in the figure below demonstrates the locus of all pairs $(N, K)$ when $n = 8$, $k = 3$, and $\alpha = 0.72$ (which is the same as in OP's example):
Note that both $N$ and $K$ are allowed to take values in $\mathbb{R}$ in the figure above. Also, the orange lines are asymptotes as $N \to \infty$.
To analyze the behavior of the curve as $N, K \to \infty$, consider the equation of the form
$$ f(p) := \mathbf{P}(X_p > k) = \alpha, $$
where $X_p \sim \text{Binomial}(n, p)$. Since $f(p)$ is a strictly increasing function of $p \in [0, 1]$ so that $f(0) = 0$ and $f(1) = 1$, it follows that the above equation has a unique zero in $(0, 1)$. Let us denote this unique zero by $p_0$. Now let $N$ and $K$ be large positive real numbers that solves the equation
$$ \sum_{j = k+1}^{n} \frac{\binom{K}{k}\binom{N-K}{n-k}}{\binom{N}{n}} = \alpha, \tag{2} $$
and define $c$ by the relation
$$ K = p_0 N + c. $$
By our choice of $p_0$ and the approximation to the binomial of the hypergeometric distribution, we know that $K/N \sim p_0$, i.e., $c = o(N)$ as $ N \to \infty$. Also, by plugging this to $\text{(2)}$, we get
\begin{align*} \alpha &= \sum_{j=k+1}^{n} \binom{n}{k} \frac{(K)_{k} (N-K)_{n-k}}{(N)_n} \\ &= \sum_{j=k+1}^{n} \binom{n}{k} \frac{\prod_{a=0}^{k-1}(p_0 N + c - a) \prod_{b=0}^{n-k-1}(N - p_0 N - c - b)}{\prod_{i=0}^{n-1} (N - i)} \\ &= \sum_{j=k+1}^{n} \binom{n}{k} \frac{\prod_{a=0}^{k-1}(p_0 + \frac{c - a}{N}) \prod_{b=0}^{n-k-1}(1 - p_0 - \frac{c + b}{N})}{\prod_{i=0}^{n-1} (1 - \frac{i}{N})}. \end{align*}
Then using the equality $\alpha = f(p_0) = \sum_{j=k+1}^{n} \binom{n}{k} p_0^j (1-p_0)^{n-j} $, we obtain
\begin{align*} 0 &= N(\alpha - \alpha) \\ &= N \sum_{j=k+1}^{n} \binom{n}{k} \Biggl[ \frac{\prod_{a=0}^{k-1}(p_0 + \frac{c - a}{N}) \prod_{b=0}^{n-k-1}(1 - p_0 - \frac{c + b}{N})}{\prod_{i=0}^{n-1} (1 - \frac{i}{N})} - p_0^j (1 - p_0)^{n-j} \Biggr] \\ &= N \sum_{j=k+1}^{n} \Biggl[ \sum_{a=0}^{k-1} p_0^{j-1} (1-p_0)^{n-j} \frac{c-a}{N} - \sum_{b=0}^{n-k-1} p_0^j (1-p_0)^{n-j-1} \frac{c + b}{N} \\ &\hspace{5em} + \sum_{i=0}^{n-1} p_0^j (1 - p_0)^{n-j} \frac{i}{N} + \o(N^{-1}) \Biggr] \\ &= \frac{k}{p_0} \left( c - \frac{k-1}{2} \right) - \frac{n-k}{1-p_0} \left( c + \frac{n-k-1}{2} \right) + \frac{(n-1)n}{2} + o(1). \end{align*}
The last expression then tells that $c$ is of the form
$$ \text{constant} + o(1),$$
and plugging this back to the equation above, we can in fact confirm that the error term is at most of the order $\mathcal{O}(N^{-1})$. This explains why $K$ looks like an affine function in $N$.