exercise:06a04b2704: Difference between revisions
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(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 previous problem, assume that <math>p = 1 - \bar{p}</math>. <ul><li> Show that under either service convention, the first player will win more often than the second player if and only if <math>p > .5</math>. </li> <li> In volleyball, a t...") |
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In the previous problem, assume that <math>p = 1 - \bar{p}</math>. | In the previous problem, assume that <math>p = 1 - \bar{p}</math>. | ||
<ul><li> Show that under either service convention, the first player will win more often than the second | <ul style="list-style-type:lower-alpha"><li> Show that under either service convention, the first player will win more often than the second | ||
player if and only if <math>p > .5</math>. | player if and only if <math>p > .5</math>. | ||
</li> | </li> | ||
Latest revision as of 00:13, 13 June 2024
In the previous problem, assume that [math]p = 1 - \bar{p}[/math].
- Show that under either service convention, the first player will win more often than the second player if and only if [math]p \gt .5[/math].
-
In volleyball, a team can only win a point while it is serving. Thus, any
individual “play” either ends with a point being awarded to the serving team or with the service
changing to the other team. The first team to win [math]N[/math] points wins the game. (We ignore here the
additional restriction that the winning team must be ahead by at least two points at the end of the
game.) Assume that each team has the same probability of winning the play when it is serving, i.e.,
that [math]p = 1 - \bar{p}[/math]. Show that in this case, the team that serves first will win more than half the
time, as long as [math]p \gt 0[/math]. (If [math]p = 0[/math], then the game never ends.) Hint: Define [math]p'[/math] to be
the probability that a team wins the next point, given that it is serving. If we write [math]q = 1 -
p[/math], then one can show that
[[math]] p' = \frac p{1-q^2}\ . [[/math]]If one now considers this game in a slightly different way, one can see that the second service convention in the preceding problem can be used, with [math]p[/math] replaced by [math]p'[/math].