Gold/silver ring problem

The three-box problem (Section ) seems to be intuitive for some, but not for everybody. Let us label the three boxes: $A$, Golden-Golden; $B$, Golden-Silver; $C$, Silver-Silver. The initial probability (i.e. before having checked the first ring) of having chosen the box $A$, $B$, or $C$ is, by symmetry, $P_\circ(A)=P_\circ(B)=P_\circ(C) = 1/3$.

This probability is updated after the event $E=$`the first ring extracted is golden' by Bayes' theorem:

$$P(A\,\vert\,E) = \frac{P(E\,\vert\,A)\cdot P_\circ(A)} {P(E\,\vert\,A)\cdot P_\circ(A)+P(E\,\vert\,B)\cdot P_\circ(B)+ P(E\,\vert\,C)\cdot P_\circ(C)} = 2/3$$

$$P(B\,\vert\,E) = \frac{P(E\,\vert\,B)\cdot P_\circ(B)} {P(E\,\vert\,A)\cdot P_\circ(A)+P(E\,\vert\,B)\cdot P_\circ(B)+ P(E\,\vert\,C)\cdot P_\circ(C)} = 1/3$$

$$P(C\,\vert\,E) = \frac{P(E\,\vert\,C)\cdot P_\circ(C)} {P(E\,\vert\,A)\cdot P_\circ(A)+P(E\,\vert\,B)\cdot P_\circ(B) +P(E\,\vert\,C)\cdot P_\circ(C)} =0\, ,$$

where $P(E\,\vert\,A)$, $P(E\,\vert\,B)$ and $P(E\,\vert\,C)$ are, respectively, 1, 1/2 and 0.

Finally, calling $F=$`the next ring will be golden if I extract it from the same box', we have, using the probability rules:

$$P(F\,\vert\,E) = P(F\vert A,E)\cdot P(A\vert E) + P(F\vert B,E)\cdot P(B\vert E)+P(F\vert C,E)\cdot P(C\vert E)$$

$$= 1 \times 2/3 + 0 \times 1/3 + 0 \times 0 = 2/3\, .$$