The Fiddler Basketball Association’s All-Star Game consists of two teams: “East” and “West.” Every year these two teams play a game, each with a 50 percent chance of winning that’s independent of the outcomes of previous years.
Many, many years into the future, you look at the most recent results of the All-Star Game. On average, what is the longest current winning streak that one of the teams is on?
Let's assume that we are looking $N$ years into the future. If the current winning streak is $S = 1$, then the last two games would have to either be EW or WE, where the first $N-2$ games could have had any possible outcome. This means there are $2 \cdot 2^{N-2} = 2^{N-1}$ possible configurations with current streak $S=1$ out of a total of $2^N$ possible configurations, so $$\mathbb{P} \{ S = 1 \} = \frac{2^{N-1}}{2^N} = \frac{1}{2}.$$ Similarly, for some generic $k \in \{ 2, \dots, N \},$ there are only two possible configurations of the last $k+1$ games that will yield $S = k$, that is, $W\overbrace{EE\cdots E}^k$ or $E\overbrace{WW \cdots W}^k.$ Again, the first $N-k-1$ games could have had any possible outcome. So we see that $$\mathbb{P} \{ S_N = k \} = \frac{2 \cdot 2^{N-k-1}}{2^N} = 2^{-k},$$
So we see that the for $N$ years into the future we get $$\mathbb{E} [ S_N ] = \sum_{k=1}^N k \mathbb{P} \{ S_N = k \} = \sum_{k=1}^N k 2^{-k}.$$ We will resort to the use of the following friendly partial sum formulae for $$f(t) = \sum_{k=0}^n t^k = \frac{1-t^{n+1}}{1-t}$$ and $$g(t) = f^\prime (t) = \sum_{k=1}^n kt^{k-1} = \frac{1 - (n+1)t^k +nt^{n+1}}{(1-x)^2}.$$ Here we see that the average current streak after $N$ years is $$\mathbb{E} [S_N] = \frac{1}{2} g\left( \frac{1}{2} \right) = \frac{1}{2} \frac{1 - (N+1) 2^{-N} + N 2^{-N-1} }{\left(1 - \frac{1}{2}\right)^2} = 2 \left( 1 - \frac{N+2}{2^{N+1}} \right),$$ which leads naturally when $N \to \infty$ to the long term average of $\mathbb{E}[S_\infty] = 2.$
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