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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 a popular computer game the computer picks an integer from 1 to <math>n</math> at random. The player is given <math>k</math> chances to guess the number. After each guess the computer responds “correct,” “too small,” or “too big.”...")
 
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<div class="d-none"><math>
In a popular computer game the computer picks an integerfrom 1 to <math>n</math> at random.  The player is given <math>k</math> chances to guess the number.  After each guess the computer responds “correct,” “too small,” or “too big.”
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<ul style="list-style-type:lower-alpha"><li> Show that if <math>n \leq 2^k - 1</math>, then there is a strategy that guarantees you
\newcommand{\mat}[1]{{\bf#1}}
\newcommand{\exref}[1]{\ref{##1}}
\newcommand{\secstoprocess}{\all}
\newcommand{\NA}{{\rm NA}}
\newcommand{\mathds}{\mathbb}</math></div> In a popular computer game the computer picks an integer
from 1 to <math>n</math> at random.  The player is given <math>k</math> chances to guess the number.  After
each guess the computer responds “correct,” “too small,” or “too big.”
<ul><li> Show that if <math>n \leq 2^k - 1</math>, then there is a strategy that guarantees you
will correctly guess the number in <math>k</math> tries.
will correctly guess the number in <math>k</math> tries.
</li>
</li>

Latest revision as of 17:02, 14 June 2024

In a popular computer game the computer picks an integerfrom 1 to [math]n[/math] at random. The player is given [math]k[/math] chances to guess the number. After each guess the computer responds “correct,” “too small,” or “too big.”

  • Show that if [math]n \leq 2^k - 1[/math], then there is a strategy that guarantees you will correctly guess the number in [math]k[/math] tries.
  • Show that if [math]n \geq 2^k - 1[/math], there is a strategy that assures you of identifying one of [math]2^k - 1[/math] numbers and hence gives a probability of [math](2^k - 1)/n[/math] of winning. Why is this an optimal strategy? Illustrate your result in terms of the case [math]n = 9[/math] and [math]k = 3[/math].