exercise:D9d70350c9: 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> Let <math>T</math> be the random variable that counts the number of 2-unshuffles performed on an <math>n</math>-card deck until all of the labels on the cards are distinct. This random variable was discussed in Section \ref{sec 3.3}. Using gui...") |
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Let <math>T</math> be the random variable that counts the number of 2-unshuffles performed on an <math>n</math>-card deck | Let <math>T</math> be the random variable that counts the number of 2-unshuffles performed on an <math>n</math>-card deck | ||
until all of the labels on the cards are distinct. This random variable was discussed in | until all of the labels on the cards are distinct. This random variable was discussed in | ||
[[guide:21bfd24860|Card Shuffling]]. Using [[guide:21bfd24860#eq 3.3.1 |Equation]] in that section, together with the formula | |||
<math display="block"> | <math display="block"> | ||
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E(T) = \sum_{s = 0}^\infty \left(1 - {{2^s}\choose n}\frac {n!}{2^{sn}}\right)\ . | E(T) = \sum_{s = 0}^\infty \left(1 - {{2^s}\choose n}\frac {n!}{2^{sn}}\right)\ . | ||
</math> | </math> | ||
Show that for <math>n = 52</math>, this expression is approximately equal to 11.7. (As was stated in | Show that for <math>n = 52</math>, this expression is approximately equal to 11.7. (As was stated in [[guide:1cf65e65b3|Combinatorics]], this means that on the average, almost 12 riffle shuffles of a 52-card deck are required in order for the process to be considered random.) | ||
deck are required in order for the process to be considered random.) | |||
Revision as of 17:23, 14 June 2024
Let [math]T[/math] be the random variable that counts the number of 2-unshuffles performed on an [math]n[/math]-card deck until all of the labels on the cards are distinct. This random variable was discussed in Card Shuffling. Using Equation in that section, together with the formula
[[math]]
E(T) = \sum_{s = 0}^\infty P(T \gt s)
[[/math]]
that was proved in Exercise Exercise, show that
[[math]]
E(T) = \sum_{s = 0}^\infty \left(1 - {{2^s}\choose n}\frac {n!}{2^{sn}}\right)\ .
[[/math]]
Show that for [math]n = 52[/math], this expression is approximately equal to 11.7. (As was stated in Combinatorics, this means that on the average, almost 12 riffle shuffles of a 52-card deck are required in order for the process to be considered random.)