Proof by Lewis Carroll that an urn cannot contain two balls I am self-learning probability theory and solving some problems from the book Probability and Random processes by Grimmett and Stirzaker. Exercise 1.4.4 poses a nice problem. I would like someone to verify, if my solution is correct and thought process is correct.

What do you think of the following 'proof' by Lewis Carroll that an urn cannot contain two balls of the same color? Suppose that the urn contains two balls, each of which is either black or white. In the obvious notation $P(BB)=P(BW)=P(WB)=P(WW)=\frac{1}{4}$. We add a black ball, so that $P(BBB)=P(BBW)=P(BWB)=P(BWW)=\frac{1}{4}$. Next, we pick a ball at random, the chance that the ball is black is (using conditional probabilities)

$$1 \cdot \frac{1}{4}+\frac{2}{3}\cdot \frac{1}{4} + \frac{2}{3}\cdot \frac{1}{4} + \frac{1}{3}\cdot \frac{1}{4}=\frac{2}{3}$$
However, if there is a probability $\frac{2}{3}$ that a ball, chosen randomly from three, is black, then there must be two black and one white, which is to say that originally there was one black and one white ball in the urn.
Solution.
I think, the fallacy in this proof has to do with mistaking the unconditional probability $P(BBB)$ for the conditional probabilities $P(BBB|\text{atleast one ball is black})=\frac{1}{4}$.
\begin{align*}
P(\text{Ball drawn at random is black}) = 1\cdot\frac{1}{8} + \frac{3}{8}\cdot\frac{2}{3} + \frac{3}{8}\cdot \frac{1}{3} + \frac{1}{8}\cdot 0 = \frac{1}{2}
\end{align*}
This makes perfect intuitive sense, because without any knowledge of what ball was added to the urn, the unconditional probability that a ball drawn at random is black is simply $\frac{1}{2}$; the ball is equally likely to be black or white.
Surely, the urn can have two balls of the same color! I'm not sure, how to argue this. But, what I can tell for sure is,
\begin{align*}
P(BBB|\text{atleast one ball is black}) = P(BWW |\text{atleast one ball is black})= \frac{1}{4}
\end{align*}
 A: This is equivalent to this problem.  Suppose there are two families each with two children one has two girls, the other has two boys.  If you pick child at random the probability is $\frac 12$ it's a girl.  But the probability is $\frac 12$ then that means that family had to have a boy and a girl.  Do you see the error there?
There is no "that family".  There are two possible families with different probabilities.  We can't assume a global probability for any possible family will determine anything about the family we eventually pick.  That's bait and switch.
So the urn problem is exactly.  Of the many possible urns we get the probability of drawing a black ball is $\frac 23$.  But that doesn't say anything about the condition of one particular urn arrangement which we had no guarentee of being "the" urn.
A: Probability of drawing a black ball, after adding a black ball, is $\frac{1}{3} p + \frac{2}{3} q + 1 (1-p-q) = \frac{3-2p-q}{3}$, where $p$ is the probability the urn had no black ball and q the probability it had one black ball at the start. This is equal to $\frac{2}{3}$ for $p = \frac{1-q}{2}$ and $p=0, q=1$ is one of the solutions.
The “proof” in the question has this embedded as an assumption.
The famous Monty Hall problem is in some ways related to this.
