Say we have two teams $A$, $B$ that will play in a best of $n$ series, which to win you must win $w$ games, which is the ceiling of $\frac{n}{2}$. Say that when the two teams play each other, $A$ has a probability $p$ of winning a single game, and $B$ has a probability $q$ of winning a single game.
Say that in our (for example) best of $5$ match, $B$ wins $3-2$ over $A$, what is the probability that $A$ was actually the better team, or that $p>q$?
Rephrased: what is $$ P(p>q\mid B\text{ won 3-2}) $$
My attempt: Since $p+q=1$, we know $q = 1-p$ so in order for $p>q \implies p>\frac{1}{2}$ via substitution. The probability will then be
$$ P(p>q\mid B\text{ won 3-2}) = \int_{.5}^{1} P(p=x\mid B\text{ won 3-2})dx $$
which we can expand with Bayes Theorem to get
$$ \int_{.5}^{1} P(p=x\mid B\text{ won 3-2})dx = \int_{.5}^{1} \dfrac{P(B\text{ won 3-2}\mid p=x)P(B\text{ won 3-2})}{P(p=x)}dx $$
Which is where I get stuck. Any help would be appreciated!
I think this is possible in a Bayesian context. Suppose $A$ wins $m$ out of $n$ matches.
We place an "uninformative" uniform $\textrm{Beta}(1,1)$ priors on $p$, and get the posterior distribution:
\begin{equation} p| D \sim \textrm{Beta}(1+m,1+n-m) \end{equation}
Where $D$ denotes the data. Then we want $P[p>q] = P[p > 1-p] = P[p > .5]$.
This becomes evaluating the Beta cdf, but to check it note that $1-p$, when $p \sim \textrm{Beta}(a,b)$, has distribution:
\begin{equation} q \sim \textrm{Beta}(b,a) \end{equation}
This gives us that $q \sim \textrm{Beta}(1+n-m,1+m)$.
In your example this gives us that $P[p>.5] = .34375$ and $P[q>.5] = .65625$.