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.]
Your notation is a bit confusing (at least to me), since I would usually expect $D_1$ to $D_3$ to refer to specific doors (e.g. the first, second and third door from the left), whereas if I understand your calculations correctly, you're using them to refer, respectively, to "the door I picked", "the door the presenter opened" and "the door I could switch to". That's fine, but you should have explained it.
With that interpretation, your calculations, corrected as per your comment, are correct. You could have saved yourself a lot of trouble in case (i), though, since in that case it's clear from the protocol that $D_2$ will always have a goat, so you could have dropped $D_2=G$ everywhere and just written
\begin{align} P(D_3=C)&=P(D_3=C\mid D_1=C)P(D_1=C)+P(D_3=C\mid D_1\ne C)P(D_1\ne C)\\ &=0\cdot\frac13+1\cdot\frac23\\ &=\frac23\;. \end{align}