\problem  \textbf{The World Series}

New York and Atlanta have both made it to the World Series again in 2000.
Assume that New York wins each game with probability 2/3 independently of
the outcomes of other games\footnote{---they wish \texttt{:-)}}.  Let $R$ be the
random variable denoting the number of the 4th game won by New York, e.g.,
if NY wins the 1st, 2nd, 4th, and 6th games, then $R=6$; let $R \eqdef 0$ if
NY doesn't win four games.

Let's generalize slightly and assume the teams keep playing,
possibly forever, until NY has indeed won 4 games.  This makes $R$ a
random variable taking on all natural number values.

\bparts

\ppart For $n>0$, give a simple closed form expression for $\prob{R = n}$.

\solution{

For $0<n<3$, we have $\prob{R=n} = 0$. For $4\leq n$, we have
\[
\prob{R=n} = \binom{n-1}{3} \cdot (2/3)^4  \cdot \frac{1}{3^{n-4}}=
\frac{8(n-1)(n-2)(n-3)}{3^{n+1}}.
\]
}


\ppart What is the most likely value of $R$?

\solution{5}

\ppart What is the probability that NY never wins 4 games, i.e., that $R=0$?

\solution{0}

\ppart What is the probability that New York really wins the 7 game
Series?

\solution{
\[
\prob{4 \leq R\leq 7}= \prob{R=4}+\prob{R=5}+\prob{R=6}+\prob{R=7} = 0.8267
%\frac{?}{?}.
\]
}

\eparts