Monty Hall - 1000 Exercises in Probability The following problem is exercise 1.4.5(a) in One Thousand Exercises in Probability:

5. The Monty Hall problem: goats and cars. (a) Cruel fate has made you a contestant in a game show; you have to choose one of three
  doors. One conceals a new car, [sic] two conceal old goats. You
  choose, but your choosen door is not opened immediately. Instead, the
  presenter opens another door to reveal a goat, and 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 value of $p$ depends on the presenter's
  protocol. Devise protocols to yield the values $p = \frac{1}{2}$, $p =\frac{2}{3}$.
  Show that, for $\alpha \in [\frac{1}{2}, \frac{2}{3}]$,
  there exists a protocol such that $p = \alpha$. Are you well advised
  to change your choice to the third door?

This is the solution given later in the book:

5. (a) One cannot compute probabilities without knowing the rules governing the conditional probabilities. If the first door chosen
  conceals a goat, then the presenter has no choice in the door to be
  opened, since exactly one of the remaining doors conceals a goat. If
  the first door conceals the car, then a choice is necessary, and this
  is governed by the protocol of the presenter. Consider two 'extremal'
  protocols for this latter situation.
(i) The presenter opens a door chosen at random from the two
  available.
(ii) There is some ordering of the doors (left to right, perhaps) and
  the presenter opens the earlier door in this ordering which conceals a
  goat.
Analysis of the two situations yields $p = \frac{2}{3}$ under (i), and
  $p = \frac{1}{2}$ under (ii).
Let $\alpha \in [\frac{1}{2}, \frac{2}{3}]$, and suppose the presenter
  possesses a coin which falls with heads upwards with probability
  $\beta = 6\alpha - 3$. He flips the coin before the show, and adopts
  strategy (i) if and only if the coin shows heads. The probability in
  question is now $\frac{2}{3} \beta + \frac{1}{2}(1 - \beta) = \alpha$.
You never lose by swapping, but whether you gain depends on the
  presenter's protocol.

This solution seems completely wrong to me for multiple reasons:


*

*The deterministic protocol adds information, and the probability shouldn't be reduced with more information.

*By switching, the contestant is guaranteed the car as long as he does not select the car first, and this outcome is independent of the chosen protocol, since the protocol only applies when the contestant selects the car first.

*This question appears to have already been answered on StackExchange, and the answer there is not in agreement with this given solution.
As far as I can tell, the authors are confusing "the conditional probability that the third door conceals the car" with the conditional probability of winning by switching when the location of the car is unknown.
To see what I mean, suppose we label the doors 1, 2, and 3, and we let $C^{(d)}$, $G^{(d)}$, and $S^{(d)}$ be the events that the $d$th door contains a car, contains a goat, or is first selected by the contestant, respectively. Let $d_1$, $d_2$, and $d_3$ be the doors first selected by the contestant, revealed by the host, and then leftover, respectively.
With this notation, the conditional probability $p$ is
$P(C^{(d_3)} | G^{(d_2)}, S^{(d_1)}) = \dfrac{P(C^{(d_3)}, G^{(d_2)} | S^{(d_1)})}{P(G^{(d_2)} | S^{(d_1)})} = \dfrac{\frac{2}{3}}{1} = \dfrac{2}{3}$
However, we can construct a specific situation where the conditional probability is $\frac{1}{2}$: Suppose the host always reveals the smallest numbered door containing a goat. If we take $d_1 = 3$ and $d_2 = 1$, then we have
$P(C^{(2)} | G^{(1)}, S^{(3)}) = 1 - P(G^{(2)} | G^{(1)}, S^{(3)}) = 1 - \dfrac{P(G^{(2)}, G^{(1)} | S^{(3)})}{P(G^{(1)} | S^{(3)})} = 1 - \dfrac{\frac{1}{3}}{\frac{2}{3}} = \dfrac{1}{2}$
But this is a specific situation that relies on more than just the protocol of the host, since there is a $\frac{1}{3}$ chance that the car could be behind the first door, so that, by picking the third door and having the second door revealed, we would know with certainty that the car is behind door one before we switch. Thus, the conditional probability of $\frac{1}{2}$ is only attained when the location of the car is unknown after the reveal. (And our sanity check of $\frac{1}{3} \cdot 1 + \frac{2}{3} \cdot \frac{1}{2} = \frac{2}{3}$ agrees with our conditional probability $p$ that the third door contains a car, not knowing which specific doors are opened).
That the authors made a mistake is further cemented in the last sentence of their solution: "whether you gain depends on the presenter's protocol." In fact, regardless of the presenter's protocol, there is always an event of non-zero probability where you benefit by swapping. It should read, "whether you have the possibility of not gaining by switching after conditioning depends on the presenter's protocol."
Their solution seems wrong to me, but the authors have fancy credentials (Ph.D.s from Oxford) and positions at Oxford and Cambridge, so I'm hesitant to write it off as an error. It is possible I could have misread the problem or completely miscalculated the conditional probabilities. The details do, after all, seem rather nuanced.
Is this problem/solution an error? Am I just nitpicking wording?
 A: 
As far as I can tell, the authors are confusing "the conditional probability that the third door conceals the car" with the conditional probability of winning by switching when the location of the car is unknown.

They are the same thing.   If the remaining door conceals a car, then you will win by switching.

To see what I mean, suppose we label the doors 1, 2, and 3, and we let 
  C(d), G (d)  be the events that the dth door contains a car, contains a goat, or is first selected by the contestant, respectively. 

Simplify it.   You only need to label two doors.   The events we are interesting in are: What the contestants first choice hides, which of the other two doors the host reveals, what the remaining door hides.
(Notice, though, what the contestant's first choice hides determines what the remaining door hides; the car can only be behind one of them.)  
Without loss of generality, the contestant chooses whichever door and it hides either (1/3) a car or (2/3) a goat.   There are two doors remaining; one on the left of the other; now label those doors $1$ and $2$.   After the host opens one of these such that it will reveal a goat, then evaluate the probability the other hides a car.


*

*Case 1: The contestant's choice hides a goat.   The host reveals whichever door also hides a goat.   Protocol is irrelevant; the host has no choice if a goat is to be revealed.   There is $50\%$ chance this will be either door and the other will certainly hide a car.


$$\begin{align}
\mathsf P_\star(F{=}g, H{=}1, R{=}c) &= \tfrac 23\tfrac 12=\tfrac 13
\\ \mathsf P_\star(F{=}g, H{=}2, R{=}c) &= \tfrac 23\tfrac 12 = \tfrac 13 
\\ \mathsf P_\star(F{=}g, H{=}1, R{=}g) &= 0\\ \mathsf P_\star(F{=}g, H{=}2, R{=}g) &= 0 \end{align}$$


*

*Case 2: The contestant's choice hides a car.   The host may therefore reveal either of the two doors, as they both hide a goat.   Which door he then reveals depends on the protocol.


*

*Protocol I: Random Protocol: The host will then open either door with equal probability.
$$\begin{align}\mathsf P_1(F{=}c, H{=}1, R{=}g) &= \tfrac 13\tfrac 12=\tfrac 16
\\ \mathsf P_1(F{=}c, H{=}2, R{=}g) &= \tfrac 13\tfrac 12 = \tfrac 16 
\\ \mathsf P_1(F{=}c, H{=}1, R{=}c) &= 0\\ \mathsf P_1(F{=}c, H{=}2, R{=}c) &= 0 \end{align}$$

*Protocol II: Leftmost Protcol: The host will open door 1.
$$\begin{align}\mathsf P_2(F{=}c, H{=}1, R{=}g) &= \tfrac 13\\ \mathsf P_2(F{=}c, H{=}2, R{=}g) &= 0 \\ \mathsf P_2(F{=}c, H{=}1, R{=}c) &= 0\\ \mathsf P_2(F{=}c, H{=}2, R{=}c) &= 0 \end{align}$$
Now evaluate $\mathsf P_1(R{=}c\mid H{=}1)$, $\mathsf P_1(R{=}c\mid H{=}2)$ , $\mathsf P_2(R{=}c\mid H{=}1)$ , $\mathsf P_2(R{=}c\mid H{=}2)$ . 
$$\begin{align}\mathsf P_1(R{=}c\mid H{=}1) & =\dfrac{\mathsf P_1(R{=}c\cap H{=}1)}{\mathsf P_1(R{=}c\cap H{=}1)+\mathsf P_1(R{=}g\cap H{=}1)} \\ & = \dfrac{\tfrac 13}{\tfrac 13+\tfrac 16} \\ & =\dfrac 23 \\[2ex] \mathsf P_1(R{=}c\mid H{=}2) & = \dfrac{2}{3} \\[2ex] \mathsf P_2(R{=}c\mid H{=}1) & = \dfrac{\tfrac 13}{\tfrac 13+\tfrac 13} \\ & = \dfrac 12 \\[2ex] \mathsf P_2(R{=}c\mid H{=}2) & = \dfrac{\tfrac 13}{\tfrac 13+0} \\ &= 1 \end{align}$$
Conclusion: 
If the host uses Protocol I (never reveal a car and choose randomly when the contestant chooses the car), then the car will be behind the remaining door with probability $2/3$ regardless of which door the host opens.
If the host is known to use Protocol II (opening the leftmost door unless there is a car behind it) then the car will be behind the remaining door $1/2$ of the times when the host opens the leftmost door, and always when the host opens the rightmost door.


This solutions seems completely wrong to me for multiple reasons:
1.The deterministic protocol adds information, and the probability shouldn't be reduced with more information.

The deterministic probability adds information and depending on what information is added (open leftmost or open rightmost), the conditional probability that the car is behind the remaining door either decreases to $1/2$ or increases to $1$.

2.By switching, the contestant is guaranteed the car as long as he does not select the car first, and this outcome is independent of the chosen protocol, since the protocol only applies when the contestant selects the car first.

Yes.   However the contestant does not know whether he selected the car first, except that protocol II may add some clues.   Because a protocol II host never chooses the rightmost door unless there is a car behind the leftmost door.
So if a protocol II host selects the rightmost door you surely know where the car hides. 
Though if a protocol II host chooses the leftmost door it is even odds whether switching will win a car or lose it.
Of course, knowing which protocol the host uses is the trick; that's the business about the host selecting protocol before the show using a biased coin. 
However, the situations where a protocol II host chooses the rightmost door happen $1/3$ of the time while those where the host chooses the leftmost door happen $2/3$ of the time, so before playing the game you have a total probability of $\tfrac 13\cdotp1+\tfrac 23\tfrac 12=\tfrac 23$ for winning if you switch.

3.This question appears to have already been answered on StackExchange, and the answer there is not in agreement with this given solution.

They are in fact in agreement.   Protocol doesn't matter until the host chooses a door.
A: This does not address your particular beef with the book's analysis, but I would like to point out that in this particular description, it is not even clear whether the presenter knows where the prize is, or whether he will always open a door (maybe some shows he does not; indeed, maybe he only opens a door whenever you did initially pick the door with the prize ... And if you know that, then you have a $0$ chance of winning by switching). So the analysis from the book is already assuming all kinds of things that are not at all clear from the book's description of the problem.
A: I apologise for not providing a complete answer, but I do not have enough reputation points to comment yet.
I have also just struggled with understanding the solution to this question, and I think that the authors meant to prompt readers for the conditional probability that the third door is a car given the action of the presenter. So if the presenter always chooses the second door as the left one amongst the undisclosed doors that house a goat, one should be able to verify that the probability that the third door is a car given that the presenter chose the left door as the second door is $1/2$.
If that is really what the authors meant, I think the question is poorly phrased.
