exercise:D9d70350c9: Difference between revisions

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<div class="d-none"><math>
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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
Section \ref{sec 3.3}.  Using [[guide:21bfd24860#eq 3.3.1 |Equation]] in that section, together with the formula
[[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">
E(T) = \sum_{s = 0}^\infty P(T  >  s)
E(T) = \sum_{s = 0}^\infty P(T  >  s)
</math>
</math>
that was proved in Exercise [[exercise:D9107332f5 |Exercise]], show that
that was proved in [[exercise:D9107332f5 |Exercise]], show that


<math display="block">
<math display="block">
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.)
Chapter~\ref{chp 3}, 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.)

Latest revision as of 19:32, 24 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, 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.)