Revision as of 02:36, 9 June 2024 by Bot (Created page with "<div class="d-none"><math> \newcommand{\NA}{{\rm NA}} \newcommand{\mat}[1]{{\bf#1}} \newcommand{\exref}[1]{\ref{##1}} \newcommand{\secstoprocess}{\all} \newcommand{\NA}{{\rm NA}} \newcommand{\mathds}{\mathbb}</math></div> In the gambler's ruin problem, assume that the gambler initial stake is 1 dollar, and assume that her probability of success on any one game is <math>p</math>. Let <math>T</math> be the number of games until 0 is reached (the gambler is ruined)....")
BBy Bot
Jun 09'24
Exercise
[math]
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In the gambler's ruin problem, assume that the gambler initial stake
is 1 dollar, and assume that her probability of success on any one game is [math]p[/math]. Let [math]T[/math] be the number of games until 0 is reached (the gambler is ruined). Show that the generating function for [math]T[/math] is
[[math]]
h(z) = \frac{1 - \sqrt{1 - 4pqz^2}}{2pz}\ ,
[[/math]]
and that
[[math]]
h(1) = \left \{ \begin{array}{ll}
q/p, & \mbox{if $q \leq p$}, \\
1, & \mbox{if \ltmath\gtq \geq p,[[/math]]
}
\end{array} \right.
</math> and
[[math]]
h'(1) = \left \{ \begin{array}{ll}
1/(q - p), & \mbox{if $q \gt p$}, \\
\infty, & \mbox{if \ltmath\gtq = p.[[/math]]
}
\end{array} \right.
</math> Interpret your results in terms of the time [math]T[/math] to reach 0. (See also Example.)