Let $X_i$ be a discrete random variable which is team $i$'s number of wins.
$X_i$ follows a binomial distribution with $n=162$ trials and probability $p=\frac{1}{2}$.
Let $Z=\max\left\{X_1, X_2, X_3, X_4, X_5\right\}$ be the number of wins of the team with the most wins. We are interested in $\mathbb{E}[Z]$.
If the cdf of $X_i$ is $F(x)$, the cdf of
$Z$ is $[F(x)]^5$ since each of the $X_i$ are independent. To see this:
\begin{align*}
F_Z\left( z \right)= &\Pr\left( Z\leq z \right) \\
=& \Pr\left( \max\left\{ X_1, X_2, X_3, X_4, X_5 \right\}\leq z \right) \\
=& \Pr\left( X_1 \leq z, X_2 \leq z, X_3 \leq z, X_4 \leq z, X_5 \leq
z\right)\\
=& \Pr\left( X_1 \leq z\right)\Pr\left( X_2 \leq z\right)\Pr\left( X_3 \leq z\right)\Pr\left( X_4 \leq z\right)\Pr\left( X_5 \leq z\right)
\end{align*}
This means that the pmf of $Z$ is $[F(z)]^5 - [F(z-1)]^5$.
Let's write out the pmf and cdf of $X_i$ explicitly. The pmf is:
\[ f(x) = {n \choose x} p^x (1- p)^{n-x} = {n \choose x}\left(\frac{1}{2}\right)^n \]
The cdf is:
\[ F(x) = \sum_{k=0}^x {n \choose k} \left(\frac{1}{2}\right)^n \]
Therefore the pmf of $Z$ is:
\[ f_Z(z) =\left[\sum_{k=0}^z {n \choose k} \left(\frac{1}{2}\right)^n \right]^5 -
\left[\sum_{k=0}^{z-1} {n \choose k} \left(\frac{1}{2}\right)^n \right]^5
\]
Therefore the expected value of $Z$ is given by:
\[
\mathbb{E}[Z]=\sum_{j=0}^{n=162}
j\left(
\left[\sum_{k=0}^j {n \choose k} \left(\frac{1}{2}\right)^n \right]^5 -
\left[\sum_{k=0}^{j-1} {n \choose k} \left(\frac{1}{2}\right)^n \right]^5
\right)
\]
This turns out to be about 88.394.
To check the answer, we can simulate the result. See here.
Special thanks to Nick Saponara for helping out with the solution.