# Rabbits in the Hat Answer

Nothing is wrong with the calculation, but its interpretation is nonsense. When the various probabilities are combined, we are working out the probability of extracting a black rabbit, over all possible combinations of rabbits. It is fallacious to imagine that this probability is valid for any specific combination. The fallacy is glaring if there is only one rabbit in the hat. With one rabbit, a similar argument (ignoring adding and removing a black rabbit, which changes nothing essential) goes like this: the hat contains either B or W, each with probability $\frac{1}{2}$.

The probability of extracting a black rabbit is therefore

$\frac{1}{2} X 1 + \frac{1}{2} X 0$

which is $\frac{1}{2}$. Therefore (really?) half the rabbits in the hat are black, and half are white.

But there's only one rabbit in the hat...

As seen on page 19, Professor Stewart's Cabinet of Mathematical Curiosities