SHUFFLING CARDS AND STOPPING-TIMES

EXAMPLE1. Top in at random shuffle. Consider the following method of mixing a deck of cards: the top card is removed and inserted into the deck at a random position. This procedure is repeated a number of times. The following argument should convince the reader that about n log n shuffles suffice to mix up n cards. The argument depends on following the bottom card of the deck. This card stays at the bottom until the first time (TI) a card is inserted below it. Standard calculations, reviewed below, imply this takes about n shuffles. As the shuffles continue, eventually a second card is inserted below the original bottom card (this takes about n/2 further shuffles). Consider the instant (T,) that a second card is inserted below the original bottom card. The two cards under the original bottom card are equally likely to be in relative order low-high or high-low. Similarly, the first time a h r d card is inserted below the original bottom card, each of the 6 possible orders of the 3 bottom cards is equally likely. Now consider the first time T,-, that the original bottom card comes up to the top. By an inductive argument, all (n l ) ! arrangements of the lower cards are equally likely. When the original bottom card is inserted at random, at time T = q,-, + 1, then all n! possible arrangements of the deck are equally likely.

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