The issue can be characterized as a confusion of prior and posterior probability
or maybe as the dissatisfaction of not knowing the joint distribution of certain random variables.
Conditioning
As an introductory example,
we consider a model for the experiment of drawing, without replacement,
two balls from an urn with $n$ balls numbered from $1$ to $n$.
The typical way to model this experiment is with two random variables $X$ and $Y$,
where $X$ is the number of the first ball and $Y$ is the number of the second ball,
and with the joint distribution $P(X=x \land Y=y) = 1/(n(n-1))$
for all $x,y \in N := \{1,\dots,n\}$ with $x \neq y$.
This way, all possible outcomes have the same, positive probability,
and the impossible outcomes (e.g., drawing the same ball twice) have zero probability.
It follows $P(X=x)=1/n$ and $P(Y=x)=1/n$ for all $x \in N$.
Let the experiment be conducted and the second ball revealed to us,
while the first ball is kept secret.
Denote $t$ the number of the second ball.
Then, still, $P(X=x)=1/n$ for all $x \in N$.
However, for each $x \in N$, our degree of belief that the event $X=x$ happened,
should now be $P(X=x \vert Y=t) = P(X=x \land Y=t) / P(Y=t)$,
which in case of $x \neq t$ is $1/(n-1)$,
and in case of $x = t$, it is $0$.
This is the probability of $X=x$ conditioned on the information that $Y=t$ happened,
also called the posterior probability of $X=x$,
meaning, the updated probability of $X=x$ after we obtained the evidence that $Y=t$ happened.
It is still $P(X=x)=P(Y=x)=1/n$ for all $x \in N$,
those are the prior probabilities.
Not conditioning on evidence means ignoring evidence.
However, we can only condition on what is expressible in the probabilistic model.
In our example with the two balls from the urn,
we cannot condition on the weather or on how we feel today.
In case that we have reason to believe that such is evidence relevant to the experiment,
we must change our model first in order to allow us to express this evidence as formal events.
Let $C$ be the indicator random variable that says if the first ball
has a lower number than the second ball, that is, $C = 1 \Longleftrightarrow X < Y$.
Then $P(C=1) = 1/2$.
Let again $t$ be the number of the second ball,
which is revealed to us, but the number of the first ball is secret.
Then it is easy to see that $P(C=1 \vert Y=t) = (t-1) / (n-1)$.
In particular $P(C=1 \vert Y=1) = 0$,
which in our model means that $C=1$ has certainly not happened.
Moreover, $P(C=1 \vert Y=n) = 1$,
which in our model means that $C=1$ has certainly happened.
It is still $P(C=1) = 1/2$.
Confidence Interval
Let $X = (X_1, \dots, X_n)$ be a vector of $n$ i.i.d random variables.
Let $(l,u)$ be a confidence interval estimator (CIE) with confidence level $\gamma$
for a real parameter of the distribution of the random variables in $X$,
that is, $l$ and $u$ are real-valued functions with domain $\mathbb{R}^n$,
such that if $\theta \in \mathbb{R}$ is the true value of the parameter,
then $P(l(X) \leq \theta \leq u(X)) \geq \gamma$.
Let $C$ be the indicator random variable that says if $(l,u)$ determined the correct parameter,
that is, $C = 1 \Longleftrightarrow l(X) \leq \theta \leq u(X)$.
Then $P(C=1) \geq \gamma$.
Let us collect data so that we have values $x = (x_1,\dots,x_n) \in \mathbb{R}^n$,
where $x_i$ is the realization of $X_i$ for all $i$.
Then our degree of belief that the event $C=1$ happened should be $\delta := P(C=1 \vert X = x)$.
In general, we cannot compute this conditional probability, but we know that it is either $0$ or $1$,
since $(C = 1 \land X = x) \Longleftrightarrow ((l(x) \leq \theta \leq u(x)) \land X = x)$.
If $l(x) \leq \theta \leq u(x)$ is false, then the latter statement is false, and thus $\delta=0$.
If $l(x) \leq \theta \leq u(x)$ is true, then the latter statement is equivalent to $X=x$, and thus $\delta=1$.
If we only know the values $l(x)$ and $u(x)$ and not the data $x$,
we can still argue in a similar way that $\delta \in \{0,1\}$.
It is still $P(C=1) \geq \gamma$.
If, for our degree of belief that $C=1$ happened,
we like this prior probability more,
then we must ignore $x$, and this also means ignoring the confidence interval $[l(x),u(x)]$.
Saying that $[l(x),u(x)]$ contained $\theta$ with probability at least $\gamma$,
would mean acknowledging this evidence and at the same time ignoring it.
Learning More, Knowing Less
What makes this situation so difficult to grasp may be the fact
that we cannot compute the conditional probability $\delta$.
But this is not particular to the CIE situation,
rather it may occur whenever we have insufficient information about the joint distribution of random variables.
As an example, let $X$ and $Y$ be discrete random variables and let their marginal distributions be given,
that is, for each $x \in \mathbb{R}$, we know $P(X=x)$ and $P(Y=x)$.
This does not give us their joint distribution, that is,
we do not know $P(X=x \land Y=y)$ for any $x,y \in \mathbb{R}$.
Assume that a result of this experiment should be reported as the value of the random vector $(X,Y)$,
that is, results should be reported as pairs of real numbers.
Let the underlying experiment be conducted, and assume that we learn that $Y=7$ happened,
while the value for $X$ is still unknown to us.
This does not change $P(X=x)$ for any $x$.
However, it would be problematic to say that the result of the experiment was of the form $(x,7)$,
where $x \in \mathbb{R}$,
and that the probability for each particular real number $x$ for being the first component of this pair was $P(X=x)$.
It is problematic since in this way, we would acknowledge the evidence $Y=7$
and at the same time ignore it.
We acknowledge the evidence $Y=7$ by reporting the second component of the pair as being $7$.
We ignore it by using the prior probability $P(X=x)$, where in fact
our degree of belief for $X=x$ should now be
$P(X=x \vert Y=7) = P(X=x \land Y=7) / P(Y=7)$, which unfortunately we cannot compute.
It may be unsatisfactory that in a sense,
knowing more about $Y$ forces us to say less about $X$.
But to the best of my knowledge this is how things are.
