Limit of conditional probabilities for increasing number of options

Question

Starting point

From a variant of the Boy and Girl Paradox.
Here, the assumption is made that whales have equal initial probabilities of being male and female. The only thing varying if the initial probabilities change are the values, the underlying reasoning should not be impacted, as far as I've observed and as long as I've not made any mistake.

1. You have two whales in your back pocket, one is a male, what is the probability of having 2 males?
   This yields the result $ \frac{1}{3} $. (In the case of the probabilities being 30%/70% male/female, the result yielded should be $ \frac{30}{170} (=\frac{3}{17}) $)
2. You have two whales in your back pocket, one is a male born on sunday, what is the probability of having two males? Following the same process, we get the result $ \frac{13}{27} $. (In the case of the probabilities being 30%/70% male/female, the result yielded should be $ \frac{90}{370} (= \frac{9}{37}) $)

I'll be calling male and female "states", while the additional choices (here the days of the week) will be called "options". Please, if there are more mathematical terms, tell me, I'll change them.

Generalization

Seeing as the result in the first case was completely different from the second case, my first impulse was to try and find a formula to get the probability of having 2 males, with the number of options as a parameter. This easily gives $ \frac{2*nbr\_options-1}{2*nbr\_options+(2*nbr\_options-1)} $ which gives us $ \frac{2*nbr\_options-1}{2*2*nbr\_options-1} $ (which can obviously be simplified further to $ \frac{2*nbr\_options-1}{4*nbr\_options-1} $, but we'll keep the former form). (In the case 30%/70%, this gives something more convoluted, at least what I've been able to find, $ \frac{(3+(nbr\_options-1)*6)}{(17+(nbr\_options-1)*20)} $. The fact that the two formulas are not in the same simple form upsets me a little, so if you have a more pertinent one, I'd happily take it! SEE EDIT)
These functions, when traced in $ \mathbb{N} $, resemble this:

Then I wanted to try and generalize it $ \mathbb{R}_+ $.
This is the result:

While we're at it, let's generalize to $ \mathbb{R} $.
This is the result:

Until there, I've tried to generalize with respect to the number of options, but we can also try to generalize with respects to the number of states. The generalized formula then becomes $ \frac{2*nbr\_options-1}{2*nbr\_states*nbr\_options-1} $.
In $ \mathbb(N)$:
We still find the shape of the precedent function on the x-z "face" of the shape:
We also find the generalization of the number of states can be observed better on the y-z "face":

The behavior is as expected, increasing the denominator of the fraction makes the result decrease.

From this, the next topic of interest is the shape of the function in $ \mathbb{R} $.
While this is quite hard to render in a way that makes sense, here it is:

And here is a zoomed out version:

Issue

I need help grasping the interpretations we can make for the following:

1. The fact that the limit of the conditional probability seems to be the initial state probability.
   Where does it come from? Obviously, the mathematical limit of the function tends to the initial state probability, but why? Also, is there any situation where this knowledge could be useful? Or is it a side-product of some trivial thing I haven't considered, rendering this interpretation useless (although interesting).
2. A number of options per state equal to 0.
   Although I can wrap my head around $ nbr\_options \in \mathbb{N_*} $, this is not the case when $ nbr\_options = 0 $. In this case, for a 50%/50% initial probabilities, we get a conditional probability of $ \frac{0-1}{0-1} = 1 $ of having two males.
3. A number of states equal to 0.
   Following the same process as in (2), the conditional probability is here equal to $ \frac{2*nbr\_options-1}{0-1} = -(2*nbr\_options-1) = 1-2*nbr\_options $. This is also not at all comprised between 0 and 1!!
4. A number of options per state which is not an integer.
   Similarly to (2), and now that we have cleared the "0 case", what does it mean to have $ nbr\_options \in \mathbb{R}_+ $?
5. A negative number of options per state.
   And now that we know (4), what does it mean to have $ nbr\_options \in \mathbb{R}_- $?
6. A number of states which is not an integer.
   Similarly to (4). This may or may not be redundant with (4).
7. A negative number of states.
   Similarly to (5). This may or may not be redundant with (5).
8. The values obtained for the cases 4 to 7.
   Now that we know what the input values mean, what does the output value represent in these cases, especially when these values go outside the $ [0, 1] $ interval.
9. In particular, the values obtained when $ nbr\_options = \frac{1}{2*nbr\_states} $.
   In this situation, the output value tends toward infinity. How to interpret this?

Although they are tending towards the initial probabilities, which is elegant (to me), the fact that the results can vary outside the $ [0, 1] $ interval is hinting me that the formulas I have may be incorrect, and thus the limit may just have been there "by construction".
Is this the case?
In which case, is there anything interesting to interpret from these still?
And is there correct ones and what would they be?
And in the particular case I've pointed out, what would be their real values? (This I can do on my own, but if there are notable things to understand from it, it could be interesting)

Additionally, in case there are more mathematical terms to describe concepts I've used here, or terms I've misused, please feel free to indicate which ones!

Thanks to everyone reading and taking the time to think about this!

EDIT

As per @Graham Kemp's comment, the "real" formula, obtained through the computation of the probabilities is $$ P(p_A, p_B) = \frac{p^2_A(1 - (1 - p_B)^2)}{1 - (1 - p_Ap_B)^2} $$ That being said, it doesn't change the results at all (sorry for the colors, I can't find ones that are less aggressive and yet are visible when superimposed)



The questions regarding the interpretation of the results still are the same.

Answer 1

Rather than counts of "states" and "options", focus on the probability weights.

So you have two items which may independently belong to two independent categories, $\rm A$ and $\rm B$.

Let $A_1$ indicate that item $1$ belongs to category $\rm A$, and so forth.

Let $p^{~}_{\rm A}$ be the probability at a particular item belongs to category $\rm A$, et cetera.

You seek the probability that both belong to category $\rm A$ given that at least one belongs to categories ${\rm A}$ and $\rm B$.

Expressing this as a function of $p^{~}_{\rm A}$ and $p^{~}_{\rm B}$ we have:

$$\begin{align}P(p^{~}_{\rm A},p^{~}_{\rm B})&=\mathsf P(A_1\cap A_2\mid (A_1\cap B_1)\cup (A_2\cap B_2))\\[1ex]&=\dfrac{\mathsf P(A_1\cap A_2\cap (B_1\cup B_2))}{\mathsf P((A_1\cap B_1)\cup(A_2\cap B_2))}\\[1ex]&=\dfrac{p_{\rm A}^2(2p^{~}_{\rm B}-p^2_{\rm B})}{2p^{~}_{\rm A}p^{~}_{\rm B}-p^2_{\rm A}p^2_{\rm B}}\\[1ex]&=\dfrac{p^2_{\rm A}(1-(1-p^{~}_{\rm B})^2)}{1-(1-p^{~}_{\rm A}p^{~}_{\rm B})^2}\\[3ex]P(1/2,1)&=\dfrac{(1/2)^2(2-1)}{2(1/2)-(1/2)^2}\\&=\dfrac{1}{3}\\[2ex]P(3/10,1)&=\dfrac{(3/10)^2(2-1)}{2(3/10)-(3/10)^2}\\&=\dfrac{3}{17}\\[2ex]P(1/2,1/7)&=\dfrac{(1/2)^2(2/7-1/7^2)}{2/14-1/14^2}\\[1ex]&=\dfrac{13}{27}\\[2ex]P(3/10,1/7)&=\dfrac{(3/10)^2(2/7-1/7^2)}{6/70-3^2/70^2}\\[1ex]&=\dfrac{39}{137}&&\bigstar!\end{align}$$