Why is naive Bayes overconfident? In the fourth edition of "Artificial Intelligence: a modern approach" by Russel and Norvig, they write in section 12.6, regarding the Naive Bayes Model for text classification, the following:

The naive Bayes model assumes that words occur independently in documents, with frequencies determined by the document category. This independence assumption is clearly
violated in practice. For example, the phrase “first quarter” occurs more frequently in business (or sports) articles than would be suggested by multiplying the probabilities of “first” and
“quarter.” The violation of independence usually means that the final posterior probabilities
will be much closer to 1 or 0 than they should be; in other words, the model is overconfident in its predictions. On the other hand, even with these errors, the ranking of the possible
categories is often quite accurate.

(Emphasis mine)
I do not see why the assumption of conditional independence would lead the naive Bayes model to be overly confident in its predictions.
Just to make sure I understand their statement correctly, I assume that they mean that that naive Bayes is overly confident compared to non-naive Bayes. As an example, assuming we wanted to determine whether an article is a sport article or not  and that $s = \text{sport article}, f=\text{"first" occured in article}, q=\text{"quarter" occured in article}$, we get the non-naive Bayes model as:
$$
P(s | f, q) = \dfrac{P(s)P(f, q|s)}{P(f, q|s)P(s)+P(f, q|\neg s)P(\neg s)} \quad (1.)
$$
The conditional independence assumption then gives the naive Bayes model
\begin{align}
P(s | f, q) = \dfrac{P(s)P(f, q|s)}{P(f, q|s)P(s)+P(f, q|\neg s)P(\neg s)}  = \\\\ 
\dfrac{P(s)P(f|s)P(q|s)}{P(f|s)P( q|s)P(s)+P(f|\neg s)P(q|\neg s)P(\neg s)} \quad (2.)
\end{align}
As far as I can see then, the statement amounts to saying that the numerator and denominator in $(2.)$ usually takes the value further away from $0.5$ than in $(1.)$. Is there a theoretical explanation for this or is it more of an empirical fact that just happens to be true?

For sports articles, the given bigram is much more frequent and will have much higher probability than the one is computed with naive assumption as the product of the probabilities for the corresponding unigrams, when compared to non-sports articles, as per assumption, so that we have,
$\frac{P(f, q|s)}{P(f|s)P(q|s)} \gg \frac{P(f, q|\neg s)}{P(f|\neg s)P(q|\neg s)} \quad \quad (3)$
here all probability values are $\in [0,1]$.
Now, we have, in (1), (ignoring the trivial cases and assuming non-zero probabilities)
$P(s | f, q) = \dfrac{P(s)P(f, q|s)}{P(f, q|s)P(s)+P(f, q|\neg s)P(\neg s)} = \dfrac{1}{1+\dfrac{P(f, q|\neg s)P(\neg s)}{P(f, q|s)P(s)}}$
Similarly from (2), we have,
$P_{naive}(s | f, q) = \dfrac{P(s)P(f|s)P(q|s)}{P(f|s)P( q|s)p(s)+P(f|\neg s)P(q|\neg s)P(\neg s)}=\dfrac{1}{1+\dfrac{P(f|\neg s)P(q|\neg s)P(\neg s)}{P(f|s)P(q|s)P(s)}}$
$\Rightarrow\frac{P_{naive}(s|f,q)}{P(s|f,q)}$
$=\dfrac{1+\dfrac{P(f, q|\neg s)P(\neg s)}{P(f, q|s)P(s)}}{1+\dfrac{P(f|\neg s)P(q|\neg s)P(\neg s)}{P(f|s)P(q|s)P(s)}}$
$=\dfrac{1+\dfrac{P(f, q|\neg s)}{P(f, q|s)}}{1+\dfrac{P(f|\neg s)P(q|\neg s)}{P(f|s)P(q|s)}}$, (let's assume a uniform prior for simplicity, i.e., $P(s)=P(\neg s)$)
$\gg 1$, from (3)
Hence, we have,
$P_{naive}(s|f,q) \gg P(s|f,q)$ (overconfidence)
