A bag contains 2 counters, as to which nothing is known except that each is either black or white. Ascertain their colours without taking them out of the bag.
Carroll's solution: One is black, and the other is white.
Lewis Carroll's explanation:
We know that, if a bag contained $3$ counters, two being black and one white, the chance of drawing a black one would be $\frac{2}{3}$; and that any other state of things would not give this chance.
Now the chances, that the given bag contains $(\alpha)\;BB$, $(\beta)\;BW$, $(\gamma)\;WW$, are respectively $\frac{1}{4}$, $\frac{1}{2}$, $\frac{1}{4}$.
Add a black counter.
Then, the chances that it contains $(\alpha)\;BBB$, $(\beta)\;BBW$, $(\gamma)\;BWW$, are, as before, $\frac{1}{4}$, $\frac{1}{2}$, $\frac{1}{4}$.
Hence the chances of now drawing a black one,
$$= \frac{1}{4} \cdot 1 + \frac{1}{2} \cdot \frac{2}{3} + \frac{1}{4} \cdot \frac{1}{3} = \frac{2}{3}.$$ Hence the bag now contains $BBW$ (since any other state of things would not give this chance).
Hence, before the black counter was added, it contained BW, i.e. one black counter and one white.
Q.E.F.Can you explain this explanation?
I don't completely understand the explanation to begin with. It seems like there are elements of inverse reasoning, everything he says is correct but he is basically assuming what he intends to prove. He is assuming one white, one black, then adding one black yields the $\frac{2}{3}$. From there he goes back to state the premise as proof.
Can anyone thoroughly analyze and determine if this solution contains any fallacies/slight of hand that may trick the reader?