How Lucky Can a Baseball Team Get?

Fiddler

The Fiddler Baseball League consists of exactly two teams of equal skill: the Algebraists and the Geometers. Over the course of a season, these two teams play each other 162 times. Each team has an equal chance of winning each game, and the results of games are independent of one another.

At the end of the season, on average, how many games would you expect the team with the better record to have won? (If the teams have the same record, then you should include one of them in your calculation.)

Solution

Looking at a total of $2g$ games between two teams :

• Either player can win all $2g$ games $\rightarrow2 \cdot 2g \cdot {2g \choose 0}$.
• Either player can win $2g-1$ games $\rightarrow2 \cdot (2g-1) \cdot {2g \choose 1}$.
• Either player can win $2g-2$ games $\rightarrow2 \cdot (2g-2) \cdot {2g \choose 2}$.
• ...
• Either player can win $2g-k$ games $\rightarrow2 \cdot (2g-k) \cdot {2g \choose k}$.
• ...
• Either player can win $g+2$ games $\rightarrow2 \cdot (g+2) \cdot {2g \choose {g-2}}$.
• Either player can win $g+1$ games $\rightarrow2 \cdot (g+1) \cdot {2g \choose {g-1}}$.
• Both players can win $g$ games $\rightarrow g \cdot {2g \choose g}$.

The expected number of games won by the team with the better record can be expressed as :

\begin{align*} G &= \left(\left(2\cdot \sum\limits_{k=0}^{g-1} (2g-k) \cdot {2g \choose k}\right) + g \cdot {2g \choose g}\right)\cdot \dfrac{1}{2^{2g}} \\ &= \left(\left(2\cdot \sum\limits_{k=0}^{g-1} \dfrac{(2g-k)\cdot(2g)!}{(2g-k)!\cdot k!}\right) + g \cdot {2g \choose g}\right)\cdot \dfrac{1}{2^{2g}} \\ &= \left(\left(2\cdot2g\cdot \sum\limits_{k=0}^{g-1} \dfrac{(2g-1)!}{(2g-1-k)!\cdot k!}\right) + g \cdot {2g \choose g}\right)\cdot \dfrac{1}{2^{2g}} \\ &= \left(\left(2\cdot2g\cdot \sum\limits_{k=0}^{g-1} {2g-1 \choose k}\right) + g \cdot {2g \choose g}\right)\cdot \dfrac{1}{2^{2g}} \\ &= \dfrac{\left(\left(2g\cdot \sum\limits_{k=0}^{2g-1} {2g-1 \choose k}\right) + g \cdot {2g \choose g}\right)}{2^{2g}} \\ &= \dfrac{2g\cdot2^{2g-1} + g \cdot {2g \choose g}}{2^{2g}} \\ &= g + \dfrac{g \cdot {2g \choose g}}{2^{2g}} \\ \end{align*}

Answer

Substituting $g = 81$ yields $$\boxed{\approx 86.0699}$$

Extra Credit

After some expansion, the Fiddler Baseball League now boasts 30 teams. Over the course of a season, each team plays each other team five times. (Thus, each team plays a total of 145 games.) As before, each team has an equal chance of winning each game, and the results of games are independent of one another.

At the end of the season, on average, how many games would you expect the team with the best record to have won? (If more than one team has the same best record, then you should include one of them in your calculation.)</p

Solution

I ran a simulation 10 million times and achieved...

Answer

$$\boxed{\approx 84.982}$$

Rohan Lewis

2026.08.10

Code can be found here.