Grimmett and Stirzaker Exercise 1.4.5.2

In a game show you have to choose one of three doors. One conceals a car, 2 conceal goats. You choose a door but the door is not opened immediately. Instead the presenter opens another door, which reveals a goat. He offers you the opportunity to change your choice to the third door (unopened and so far unchosen ). Let $p$ be the conditional probability that the third door conceals the car. The presenter's protocol is:

(i) he is determined to show you a goat; with a choice of two, he picks one at random. Show that $p=2/3$

(ii)he is determined to show you a goat; with a choice of two goats (Billy and Nan), he shows Billy with probability b. Show that $p=\frac{1}{1+b}$

(iii) he opens a door at random irrespective of what is behind. Show that $p=1/2$

I understand (i) but not (ii).

For (i) my answer is:

Label the doors D1,D2,D3, the car C, the goats G1 and G2, a goat G

then

$P(D3=C|D2=G)=\frac {P(D3=C \ \cap\ D2=G)} {P(D2=G)}=\frac{P(D3=C\ \cap D2=G | D1=C) P(D1=C) + P(D3=C\ \cap\ D2=G | D1 \neq C) P(D1 \neq C) }{P(D2=G |D1=C)P(D1=C)+P(D2=G|D1 \neq C)P(D1\neq C)}=\frac{0*{1\over3}+1 * {2\over3}}{1*{1\over3}+1*{2\over3}}={2\over3}$

however similarly for (ii) my answer would be (calling Billy G1):

$P(D3=C|D2=G1)=\frac {P(D3=C \ \cap \ D2=G1)} {P(D2=G1)}=\frac{P(D3=C\ \cap D2=G1 | D1=C) P(D1=C) + P(D3=C \ \cap D2=G1 | D1 \neq C) P(D1 \neq C) }{P(D2=G1 |D1=C)P(D1=C)+P(D2=G1|D1 \neq C)P(D1\neq C)}=\frac{0*{1\over3}+{1\over2}*{2\over3}}{b*{1\over3}+1*{2\over3}}=\frac{1}{b+2}$

where is my mistake ?

[Note: this question has undergone some changes in wording over successive editions of the book, in an attempt at clarifying the problem statement, as seen in the comments below. I attempted to answer what I believe was the problem intended by the authors.]