The American League Championship Series of Riddler League Baseball determines one of the teams that will compete in the Riddler World Series. This year’s teams — the Tampa Bay Lines and the Minnesota Twin Primes — are evenly matched. In other words, both teams are equally likely to win each game of the best-of-seven series.
On average, how many games will the series last?
There are obviously two possible winners of the series, but since everything is symmetric let's not bother enumerating the other half of the outcomes as it won't affect the average length. Since the winning team will obviously win the last game, then to determine how many arrangements end in exactly $n$ games, we are really trying to decide how to distribute the $3$ other wins into the $n-1$ first games. In this case, there are $\binom{n-1}{3}$ such ways to do so, therefore the average length of the series is $$A = \frac{\sum_{n=4}^7 n \binom{n-1}{3}}{\sum_{n=4}^7 \binom{n-1}{3}} = \frac{4 \cdot 1 + 5 \cdot 4 + 6 \cdot 10 + 7 \cdot 20}{1 + 4 + 10 + 20} = \frac{224}{35} = 6.4 \,\text{games}.$$
For full clarity, and since we will use it in the Riddler Classic solution, there are a total of $70 = 2 \sum_{n=4}^7 \binom{n-1}{3}$ equally likely outcomes for this best of $7$ series.
Whoops, I guess I should have weighted by the equal likelihood of each game being won, not equal likelihood of the series being won. That would increase the weight of the 4 game series, and skew the average down towards the correct answer of 5.8125 games.
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