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?