Lewis Carroll's Pillow Problem via Bayes' Theorem

Question

In Pillow Problems and a Tangled Tale Lewis Carroll presents the following problem:

A bag contains a counter, known to be either white or black. A white counter is put in, the bag is shaken, and a counter is drawn out, which proves to be white. What is now the chance of drawing a white counter?

One can easily reason that in the end, the bag must be in one of three states:

• Black remains, new white taken
• New white remains, original white taken
• Original white remains, new white taken

In two of those states, a white counter remains to be drawn. Therefore, the chance is 2/3.

How can one arrive to this answer using Bayes' theorem?

Answer 1

Let A be the event that the original counter is white.
Let B be the event that you pulled out a white counter.

Bayes' Theorem is:

$$P(A|B) = \frac{P(B|A)P(A)}{P(B)}$$

Then $P(B|A)=1, P(A)=\dfrac12, P(B)= P(A)P(B|A) + P(A')P(B|A')= \dfrac34$.

So $P(A|B)=\dfrac12\dfrac43=\dfrac23$.