exercise:032fee81db: Difference between revisions
(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> Fermat solved the ''problem of points'' (see Exercise \ref{exer 5.1.11}) as follows: He realized that the problem was difficult because the possible ways the play might go are not equally likely. For example, when the first player needs two more...") |
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Fermat solved the ''problem of points'' (see [[exercise:D1d037085a|Exercise]]) as follows: He realized that the problem was difficult because the possible ways the | |||
play might go are not equally likely. For example, when the first player needs two more games | play might go are not equally likely. For example, when the first player needs two more games | ||
and the second needs three to win, two possible ways the series might go for the first player | and the second needs three to win, two possible ways the series might go for the first player | ||
Latest revision as of 00:17, 13 June 2024
Fermat solved the problem of points (see Exercise) as follows: He realized that the problem was difficult because the possible ways the play might go are not equally likely. For example, when the first player needs two more games and the second needs three to win, two possible ways the series might go for the first player are WLW and LWLW. These sequences are not equally likely. To avoid this difficulty, Fermat extended the play, adding fictitious plays so that the series went the maximum number of games needed (four in this case). He obtained equally likely outcomes and used, in effect, the Pascal triangle to calculate [math]P(r,s)[/math]. Show that this leads to a formula for [math]P(r,s)[/math] even for the case [math]p \ne 1/2[/math].