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\noindent{\bf Math 496/889 \hfill Quiz 3 \hfill Name:\rule{5.cm}{.25mm}} \\
\noindent{\bf Friday, October 3, 2003 \hfill SSN:\rule{5.cm}{.25mm}}\\
\bigskip
\begin{tabular}{||l|c|c|c|c|c||}
\hline
Problem & 1 & 2 & 3 & 4 & Total\\
\hline
Possible& 5 & 5 & 10 & 5 & 25 \\
\hline
Points & & & & & \\
\hline
\end{tabular}
\begin{enumerate}
\item (5 points)
Consider a coin-tossing game where a gambler starts with \$10,
tosses a fair coin with heads having probability $p =1/2$ and tails
having probability $q = 1/2$. The gambler wins \$1 for every
``head'' and loses \$1 for every ``tail''. The gambler is ruined
if his fortune reaches \$0 and the gambler is victorious if his
fortune reaches \$20. What is the probability of the gambler being
victorious?
\emph{The probability of ruin is $q_10 = 1 - S_0/a = 1 - 10/20 = 1/2$. The
probability of victory is $p_10 = S_0/a = 1/2$ }
%\vspace{1in}
\item (5 points) What is the expected duration (until either ruin or
victory) for this game? What is the minimum possible length of a
game? What is the maximum possible length of a game?
%\vspace{1in}
\emph{
The expected duration of the game is $D_10 = S_0(a - S_0) = 10(20-10)
= 100$. The minimm possible length of a game is 10 flips, the
maximum possible length of a game is infinity (unlimited duration).
}
\item (10 points)
Use the expected duration of the game from the previous problem
as the length of a specific game. What is an estimate of
the probability of an excess of 10 or more heads over tails in this
game?
%\vspace{2in}
\emph{ We need $\Pr[S_100 > 10] \approx \Pr[ Z >
(10+1/2)/\sqrt{100}] = \Pr[Z > 1.05] = 0.1469$ }
\item (5 points) Since an excess of 10 heads over tails would ensure victory,
does the estimate from the third question
contradict, support, or have no bearing on the answer from the
first question? Explain! (Hint:
Paths with an excess of $10$ heads over
tails in $n$ flips
are paths which from $S_0 = 10$ at some time cross the level $20$ and have
$S_n \ge 20$ but we don't care what else happens.
Paths which have victory are those paths which
touch or cross the level $20$ before touching or crossing
$0$ but we don't care about
what happens afterward.)
\emph{ I think the easiest way to see this is to consider the set of
all paths on $0$ to $100$. Paths with an excess of $10$ heads over
tails in $100$ flips are paths which from $S_0 = 10$ at some time
cross the level $20$ and have $S_100 \ge 20$. Paths which have
victory are those paths which touch or cross the level $20$ before
touching or crossing $0$ but we don't care about what happens
afterword. Some paths in the first set may not be in the second
set, say if they first cross $0$ before crossing $20$. Some paths
in the second set may not be in the first set, since they may not
have $S_100 \ge 20$. The two events are not comparable, and so the
probability calculations neither support nor contradict each other.}
\end{enumerate}
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