Introduction to the problem
A set $C$ contains $c\in \mathbb{N}$ elements of three different kinds: There are $\alpha\in\mathbb{N}$ elements of kind $A$, $\beta\in\mathbb{N}$ elements of kind $B$, and $\gamma\in\mathbb{N}$ elements of kind $G$, and $|C|=c=\alpha+\beta+\gamma$. We suppose $\alpha,\beta,\gamma>0$ and we perform $n>0$ independent trials (i.e. with replacement) of one element at a time from $C$.
We define the event $\mathscr{L}_n^{\alpha_A}$ as "to get, in $n$ independent trials, at least one of the $\alpha$ elements of kind $A$", so that $P(\mathscr{L}_n^{\alpha_A})=1-\left(\frac{c-\alpha}{c}\right)^n$. Similarly, we define the events $\mathscr{L}_n^{\beta_B}$ and $\mathscr{L}_n^{\gamma_G}$, such that $P(\mathscr{L}_n^{\beta_B})=1-\left(\frac{c-\beta}{c}\right)^n$ and $P(\mathscr{L}_n^{\gamma_G})=1-\left(\frac{c-\gamma}{c}\right)^n$.
We finally define the event-intersection $\mathscr{I}_{n}^{C}=\mathscr{L}_{n}^{\alpha_A}\cap \mathscr{L}_{n}^{\beta_B}\cap \mathscr{L}_{n}^{\gamma_G}$.
By means of the principle of inclusion-exclusion, and observing that $P(\mathscr{L}_{n}^{\alpha_A}\cup \mathscr{L}_{n}^{\beta_B}\cup \mathscr{L}_{n}^{\gamma_G})=1$, we easily find that
$$P(\mathscr{I}_{n}^{C})=1-P(\mathscr{L}_{n}^{\alpha_A})-P(\mathscr{L}_{n}^{\beta_B})-P( \mathscr{L}_{n}^{\gamma_G})+P(\mathscr{L}_{n}^{\alpha_A}\cap\mathscr{L}_{n}^{\alpha_B})+P(\mathscr{L}_{n}^{\alpha_A}\cap\mathscr{L}_{n}^{\alpha_G})+P(\mathscr{L}_{n}^{\alpha_B}\cap\mathscr{L}_{n}^{\alpha_G}).$$
We can also explicit this expression in terms of the variables $\alpha,\beta,\gamma$, obtaining (with the help of Bayes' theorem),
$$ P(\mathscr{I}_{n}^{C})=1-\left(\frac{\alpha+\gamma}{c}\right)^n-\left(\frac{\beta+\gamma}{c}\right)^n-\left(\frac{\alpha+\beta}{c}\right)^n+\left(\frac{\alpha}{c}\right)^n+\left(\frac{\beta}{c}\right)^n+\left(\frac{\gamma}{c}\right)^n.$$
PROPERTY K: The probability of the event $\mathscr{L}_{n}^{\alpha_A}\cap \mathscr{L}_{n}^{\beta_B}\cap \mathscr{L}_{n}^{\gamma_G}$ is invariant by switching the number of elements of any two given kinds, i.e. performing the substitutions $\alpha\leftrightarrow \beta$, $\alpha\leftrightarrow \gamma$, and $\beta\leftrightarrow \gamma$.
The proof of this property (which is ultimately based on the symmetry of the binary operator $\cap$ and on the fact that $\mathscr{I}_{n}^{C}$ involves all the kinds of elements present in $C$) is trivial.
Let's make an example to understand how it applies: If we switch the $\alpha$ elements of kind $A$ with the $\gamma$ elements of kind $G$, we obtain a new set $C^*$, which contains $\gamma\in\mathbb{N}$ elements of kind $A$, $\beta\in\mathbb{N}$ elements of kind $B$, and $\alpha\in\mathbb{N}$ elements of kind $G$, and $|C^*|=c=\alpha+\beta+\gamma$. However, the probability of the event $\mathscr{I}_{n}^{C^*}=\mathscr{L}_{n}^{\gamma_A}\cap \mathscr{L}_{n}^{\beta_B}\cap \mathscr{L}_{n}^{\alpha_G}$ is still
$$ P(\mathscr{I}_{n}^{C^*})=P(\mathscr{L}_{n}^{\gamma_A}\cap \mathscr{L}_{n}^{\beta_B}\cap \mathscr{L}_{n}^{\alpha_G})=P(\mathscr{L}_{n}^{\alpha_A}\cap \mathscr{L}_{n}^{\beta_B}\cap \mathscr{L}_{n}^{\gamma_G})=P(\mathscr{I}_{n}^{C}). $$
In fact, the chance to get at least one element of each kind, in $n$ independent trials, does not depend on which elements are of a certain kind, but only on how many elements of the three different kinds are present in $C$, independently on the labels we use to distinguish the $c$ elements in the three different kinds.
A peculiar Urn
We study the Property K in case of an urn $C^\dagger$ such that, in $n$ independent trials, the probability to get at least one element of kind $A$ is equal to the probability not to get any element of kind $B$. In this urn, it must hold the constraint
$$ P(\mathscr{L}_n^{A})=P(\overline{\mathscr{L}_n^{B}}). $$
Clearly, this request will result in a particular choice of $\alpha,\beta,\gamma$, and $n$. In fact, if we explicit the probabilities of the events $\mathscr{L}_n^{A}$ and $\overline{\mathscr{L}_n^{B}}$, we find that the above constraint corresponds to the relation
$$ 1-\left(\frac{c-\alpha}{c}\right)^n=\left(\frac{c-\beta}{c}\right)^n, $$ which can be written $$ 1-\left(\frac{\beta+\gamma}{c}\right)^n=\left(\frac{\alpha+\gamma}{c}\right)^n $$ and $$ 1-\left(\frac{\alpha+\gamma}{c}\right)^n-\left(\frac{\beta+\gamma}{c}\right)^n=0. $$ If we substitute the constraint in the general expression of $P(\mathscr{I}_{n}^{C})$, we find
$$ P(\mathscr{I}_{n}^{C^\dagger})=-P( \mathscr{L}_{n}^{\gamma_G})+P(\mathscr{L}_{n}^{\alpha_A}\cap\mathscr{L}_{n}^{\alpha_B})+P(\mathscr{L}_{n}^{\alpha_A}\cap\mathscr{L}_{n}^{\alpha_G})+P(\mathscr{L}_{n}^{\alpha_B}\cap\mathscr{L}_{n}^{\alpha_G}), $$
and, in terms of the variables $\alpha,\beta,\gamma$,
$$ P(\mathscr{I}_{n}^{C^\dagger})=-\left(\frac{\alpha+\beta}{c}\right)^n+\left(\frac{\alpha}{c}\right)^n+\left(\frac{\beta}{c}\right)^n+\left(\frac{\gamma}{c}\right)^n. $$
My question is:
Does the Property K hold in the urn $C^\dagger$?
From one side I would say "no", because if we perform the substitution $\alpha\leftrightarrow\gamma$ in the expression of $P(\mathscr{I}_{n}^{C^\dagger})$ we do not obtain the same probability (unless $\alpha=\gamma$), although the meaning of the event $\mathscr{I}_{n}^{C^\dagger}$ is still the intersection of the three events $\mathscr{L}_{n}^{\alpha_A}$, $\mathscr{L}_{n}^{\beta_B}$, $\mathscr{L}_{n}^{\gamma_G}$. A similar argument applies if we perform the substitution $\beta\leftrightarrow\gamma$ (which would imply that the Property K holds only if $\beta=\gamma$). Since both these substitutions should not modify the probability of the event-intersection, it would follow that it must be $\alpha=\beta=\gamma$, a condition that is inconsistent with the constraint, unless $\alpha=\beta=\gamma=0$.
On the other hand, if we first perform the substitution $\alpha\leftrightarrow\gamma$ in the general expression of $P(\mathscr{I}_{n}^{C})$ and then we impose the constraint, we still have that $P(\mathscr{I}_{n}^{C^\dagger})$ is invariant by performing any switch of the variables, i.e. the Property K seems to hold.
This post is related to these other posts A weird problem of probability!, A problem of conditional probability, An urn with three kinds of balls... and a weird constraint!
