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.