, 2011

Let the two teams be $A$ and $B$, and represent each possible series as a bit string, where a $1$ represents a win by team $A$ and a $0$ a win by team $B$. Let $S$ be the set of all possible bit strings. We need to determine whether there are more bit strings of length $6$ or length $7$ in $S$. For each bit string $s \in S$, the bit string obtained from $s$ by interchanging $0$ and $1$ is also in $S$, so we may restrict our attention to bit strings ending in a $1$.

The number of bit strings of length $6$ is equal to $\binom{5}{3}$ (exactly $3$ of the first $5$ bits must be a $1$ for $A$ to win the tournament in the sixth game), while the number of bit strings of length $7$ is equal to $\binom{6}{3}$. Since $\binom{6}{3} > \binom{5}{3}$, we conclude that the tournament is more likely to end in exactly $7$ games.