\problemdata      
        {?} % latex-friendly label for the prob.
        {Probability-Variance} % The topic(s) of the problem content.
        {?} % Source (if known)
        {Fall 2001, final} % Usage (list ones you are aware of).
        {Christos Kapoutsis, S02} % Last revision info (author, date).


\begin{problem}
A gambler plays 120 hands of draw poker, 60 hands of black jack,
and 20 hands of stud poker per day.  He wins a hand of draw poker with
probability 1/6, a hand of black jack with probability 1/2, and a hand of
stud poker with probability 1/5.  Assume the outcomes of the card games are
mutually independent.  %He hopes to win 108 hands per day.

\bparts

\ppart What is the expected number of hands the gambler wins in a day?

\solution{
$120(1/6)+60(1/2)+20(1/5)=54$.
}

\ppart What is the variance in the number of hands won per day?

\solution{
The variance can also be calculated using linearity of
variance.  For an individual hand the variance is $p(1-p)$ where $p$ is
the probability of winning.  Therefore the variance is
$120(1/6)(5/6) + 60(1/2)(1/2) + 20(1/5)(4/5) = 50/3 + 15 + 16/5 = 250/15 +
225/15 + 48/15 = 523/15 = 34\ 13/15$.
}

\ppart What would the Markov bound be on the probability that the
gambler will win 108 hands on a given day?

\solution{
The expected number of ganes won is 54, so
by Markov, $\pr{R \geq 108} \leq 54/108 = 1/2$.
}

\ppart What would the Chebyshev bound be on the probability that the
gambler will win 108 hands on a given day?  You may answer with a
numerical expression that is not completely evaluated.

\solution{
\[
\pr{R - 54 \geq 54} \leq \frac{V}{V + 54^2} = \frac{523}{523 +
15(54)^2} \approx 1/80.
\]

}

\ppart Apply the Chernoff bound for the probability that the gambler will
win the 108 hands to find the largest integer, $n$, such that the bound is
less than $e^{-n}$.  \hint $\ln 2 \approx 0.7$.

\solution{
Answer: $n=21$.
\begin{align*}
\pr{R \geq 108} & = \pr{R \geq 2\cdot 54}\\
     & \leq \textup{exp}(- (2\ln 2 - 2 + 1)54)\\
     & = \textup{exp}(-108\ln 2 + 54)\\
     & \approx \textup{exp}(- 108\cdot0.7+54)\\
     & \approx \textup{exp}(- 75.6 -54)\\
     & \approx \textup{exp}(- 21.6)\\
     & \approx \cfrac{1}{2,400,000,000}
\end{align*}
}

\eparts
\end{problem}
