Revision as of 03:24, 9 June 2024 by Bot (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...")
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
BBy Bot
Jun 09'24

Exercise

[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]

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 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 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.)