This problem, perhaps more than any other, started the *Bohr Group*.

Before the guard returns, **A** knows that he will return, **A**
knows that he will say either that he gave the note to **B** or that he gave
the note to **C**.
So **A** hasn't learned anything new about *her*
chances of being executed, and that probability
remains what it was before: One third.

This can be shown formally using Bayes law

P(a|b) = P(b|a) × P(a) / P(b)

That is, the probability of a given that we know b
equals the probability of b given a
times the probability of a, all divided by the
probability of b.
Let a mean “A will be executed.”
Let b mean “B was given the note.”

P(b|a)=1/2.

P(a)=1/3.

P(b)=1/2.

(1/2)×(1/3)/(1/2) = 1/3.

The probability of **B** getting the note if **A**
is to be executed is 1/2 (could be **B** or **C**).
The *prior* probability that **A** is to be executed is 1/3.
The probability of **B** getting the note is
1/2.
This is intuitive because of symmetry, or it can be calculated
as
P(**A** is executed, **B** gets the note) +
P(**B** is executed, **B** gets the note) +
P(**C** is executed, **B** gets the note) = (1/3 × 1/2) + (1/3 × 0) + (1/3 × 1) = 1/2.