exercise:D9d70350c9: Difference between revisions
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E(T) = \sum_{s = 0}^\infty P(T > s) | E(T) = \sum_{s = 0}^\infty P(T > s) | ||
</math> | </math> | ||
that was proved in | that was proved in [[exercise:D9107332f5 |Exercise]], show that | ||
<math display="block"> | <math display="block"> |
Latest revision as of 18:32, 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.)