Q:

Probability of sweeping in a best-of-N

What is the probability that in a best-of-N ($N$ is some odd number, say 3) series of games between two equally skilled players (probability of winning a single game = $1/2$) one player sweeps (i.e. wins 2-0)? My first thought was just to enumerate the sample space ($aa$, $aba$, $abb$, $bb$, $baa$, $bab$), so we expect to observe a sweep $\frac{2}{6}$ of the time
Similarly for larger $N$ it's $\frac{2}{M}$ (2 because there's only one sequence of outcomes per team that leads to a sweep, not sure what M is but it feels like that's straightforward to prove by induction).
OTOH one could look at it from the perspective of a single player and argue that the
probability of sweeping is $\left(\frac{1}{2}\right)^{(n+1)}$, $N=2n+1$ (so the overall probability of observing a sweep is 2 times that).
Hope that makes sense (I just got up and was watching some StarCraft with my coffee :))

A:

In this example, the probability is not uniformly distributed across the sample space thus you can't just enumerate to get the probability. (e.g. $P(aa)\ne P(aba)$), so we have to take another approach.
First we define that in a best of $n$ games, if a single player wins the first $\frac{n+1}{2}$ games then that player has swept the match (for example best of 7 $\rightarrow$ has to win 4)
Since for sweep to occur, it doesn't really matter which of the two players wins the first game as long as they continue winning, so we can model the question as follows:
Regardless of who wins the first match, what's the probability that match $2$ up to match $\frac{n+1}{2}$ is won by the same person. And we know that the probability of either person winning is $\frac{1}{2}$. Thus, the answer is $\frac{1}{2}^{(\frac{n+1}{2}-1)}=\frac{1}{2}^{(\frac{n-1}{2})}$

