exercise:6d2d2af699: Difference between revisions

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<div class="d-none"><math>
The letters between Pascal and Fermat, which are often credited with having started probability theory, dealt mostly with the ''problem of points'' described in Exercise [[exercise:D1d037085a |Exercise]].  Pascal and Fermat considered the problem of finding a fair division of stakes if the game must be called off when the first player has won <math>r</math> games and the second player has won <math>s</math> games, with <math>r  <  N</math> and <math>s  <  N</math>.  Let <math>P(r,s)</math> be the probability that player A wins the game if he has already won <math>r</math> points and player B has won
\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> The letters between Pascal and Fermat, which
are often credited with having started probability theory, dealt mostly with the ''problem of
points'' described in Exercise [[exercise:D1d037085a |Exercise]].  Pascal and Fermat considered the problem of
finding a fair division of stakes if the game must be called off when the first player has won
<math>r</math> games and the second player has won <math>s</math> games, with <math>r  <  N</math> and <math>s  <  N</math>.  Let <math>P(r,s)</math> be the
probability that player A wins the game if he has already won <math>r</math> points and player B has won
<math>s</math> points.  Then
<math>s</math> points.  Then
<ul><li> <math>P(r,N) = 0</math> if <math>r  <  N</math>,
 
<ul style="list-style-type:lower-alpha"><li> <math>P(r,N) = 0</math> if <math>r  <  N</math>,
</li>
</li>
<li> <math>P(N,s) = 1</math> if <math>s  <  N</math>,
<li> <math>P(N,s) = 1</math> if <math>s  <  N</math>,
Line 18: Line 8:
<li> <math>P(r,s) = pP(r + 1,s) + qP(r,s + 1)</math> if <math>r  <  N</math> and <math>s  <  N</math>;
<li> <math>P(r,s) = pP(r + 1,s) + qP(r,s + 1)</math> if <math>r  <  N</math> and <math>s  <  N</math>;
</li>
</li>
</ul> and (1), (2), and (3) determine <math>P(r,s)</math> for <math>r \leq N</math> and <math>s \leq N</math>.  Pascal
</ul> and (1), (2), and (3) determine <math>P(r,s)</math> for <math>r \leq N</math> and <math>s \leq N</math>.  Pascal used these facts to find <math>P(r,s)</math> by working backward: He first obtained <math>P(N - 1,j)</math> for <math>j =
used these facts to find <math>P(r,s)</math> by working backward: He first obtained <math>P(N - 1,j)</math> for <math>j =
N - 1,N - 2, \ldots, 0</math>; then, from these values, he obtained
N - 1</math>, <math>N - 2</math>, \dots, 0; then, from these values, he obtained
<math>P(N - 2,j)</math> for <math>j = N - 1, N - 2, \ldots, 0</math> and, continuing backward, obtained all the
<math>P(N - 2,j)</math> for <math>j = N - 1</math>, <math>N - 2</math>, \dots, 0 and, continuing backward, obtained all the
values <math>P(r,s)</math>.  Write a program to compute <math>P(r,s)</math> for given <math>N</math>, <math>a</math>, <math>b</math>, and <math>p</math>.   
values <math>P(r,s)</math>.  Write a program to compute <math>P(r,s)</math> for given <math>N</math>, <math>a</math>, <math>b</math>, and <math>p</math>.  ''
 
Warning'': Follow Pascal and you will be able to run <math>N = 100</math>; use recursion and you will
{{alert-warning| Follow Pascal and you will be able to run <math>N = 100</math>; use recursion and you will
not be able to run <math>N = 20</math>.
not be able to run <math>N = 20</math>.}}

Revision as of 01:10, 13 June 2024

The letters between Pascal and Fermat, which are often credited with having started probability theory, dealt mostly with the problem of points described in Exercise Exercise. Pascal and Fermat considered the problem of finding a fair division of stakes if the game must be called off when the first player has won [math]r[/math] games and the second player has won [math]s[/math] games, with [math]r \lt N[/math] and [math]s \lt N[/math]. Let [math]P(r,s)[/math] be the probability that player A wins the game if he has already won [math]r[/math] points and player B has won [math]s[/math] points. Then

  • [math]P(r,N) = 0[/math] if [math]r \lt N[/math],
  • [math]P(N,s) = 1[/math] if [math]s \lt N[/math],
  • [math]P(r,s) = pP(r + 1,s) + qP(r,s + 1)[/math] if [math]r \lt N[/math] and [math]s \lt N[/math];

and (1), (2), and (3) determine [math]P(r,s)[/math] for [math]r \leq N[/math] and [math]s \leq N[/math]. Pascal used these facts to find [math]P(r,s)[/math] by working backward: He first obtained [math]P(N - 1,j)[/math] for [math]j = N - 1,N - 2, \ldots, 0[/math]; then, from these values, he obtained

[math]P(N - 2,j)[/math] for [math]j = N - 1, N - 2, \ldots, 0[/math] and, continuing backward, obtained all the values [math]P(r,s)[/math]. Write a program to compute [math]P(r,s)[/math] for given [math]N[/math], [math]a[/math], [math]b[/math], and [math]p[/math].

Follow Pascal and you will be able to run [math]N = 100[/math]; use recursion and you will not be able to run [math]N = 20[/math].