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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!

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  • $\begingroup$ haven't checked the logic, but if you have that right it will be much easier if you replace the integral with a summation, since these are discrete events. $\endgroup$ – eSurfsnake Aug 26 '18 at 5:53
  • $\begingroup$ I think in this case it would be an integral because we are adding all possible probabilities but I don't know, I did get stuck $\endgroup$ – wjmccann Aug 26 '18 at 5:55
  • $\begingroup$ This may not help you much... but this problem reminds me of one I saw in the book Sheldon Ross in chapter 6 section 5, example 5d (at least in my edition). Punch line is that the probability p that team A wins follows a Beta distribution. $\endgroup$ – HJ_beginner Aug 26 '18 at 5:56
  • $\begingroup$ To follow up with this possibly wrong approach... let $X \sim uniform(0,1)$ and $N \sim Binomial(n+m,p)$ Then $f_{X|N}(x|n) = cx^2(1-x)^3$ then you integrate from $0$ to $1$ to figure out what $c$ is, then with the conditional pdf you can get the cum cdf and see what the probability that $X|N > 0.5$ $\endgroup$ – HJ_beginner Aug 26 '18 at 6:02
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    $\begingroup$ There's not enough information to answer the question; you need a prior for $p$. $\endgroup$ – joriki Aug 27 '18 at 5:18
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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 $P[q>.5] = .65625$.

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