'In this hat are two rabbits,' he announced. 'Each of them is either black or white, with equal probability. I am now going to convince you, with the aid of my lovely assistant Gumpelina, that I can deduce their colours without looking inside the hat!'
He turned to his assistant, and extracted a black rabbit from her costume. ' Please place this rabbit in the hat.' She did.
Whodunni now turned to the audience. 'Before Grumpelina added the third rabbit, there were four equally likely combinations of rabbits.' He wrote a list on a small blackboard; BB,BW, WB and WW. 'Each combination is equally likely - the probability is .
'But then I added a black rabbit. So the possibilities are BBB, BWB, BBW and BWW - again, each with a probability .
'Suppose - I won't do it, this is hypothetical - suppose I were to pull a rabbit from the hat. What is the probability that it is black? If the rabbits are BBB, that probability is 1. If BWB or BBW, it is . If BWW, it is . So the overall probability of pulling out a black rabbit is;
which is exactly
'But - if there are three rabbits in a hat, of which exactly r are black and the rest white, the probability of extracting a black rabbit is r/3. Therefore r=2, so there are two black rabbits in the hat.' He reached into the hat and pulled out a black rabbit. 'Since I added this black rabbit, the original pair must have been one black and one white!'
The great Whodunni bowed, to tumultuous applause. Then he pulled two rabbits from the hat - one pale lilac and the other shocking pink.
It seems evident that you can't deduce the contents of a hat without finding what's inside. Adding the extra rabbit and then removing it again (was it the same black rabbit? Do we care?) is a clever piece of misdirection. But why is the calculation wrong?
As seen on page 19, Professor Stewart's Cabinet of Mathematical Curiosities
Answers next week!