Let [math]D[/math] be a deck of [math]n[/math] cards. We have seen that there are [math]a^n[/math] [math]a[/math]-shuffles of [math]D[/math]. A coding of the set of [math]a[/math]-unshuffles was given in the proof of Theorem. We will now give a coding of the [math]a[/math]-shuffles which corresponds to the coding of the [math]a[/math]-unshuffles. Let [math]S[/math] be the set of all [math]n[/math]-tuples of integers, each between 0 and [math]a-1[/math]. Let [math]M = (m_1, m_2, \ldots, m_n)[/math] be any element of [math]S[/math]. Let [math]n_i[/math] be the number of [math]i[/math]'s in [math]M[/math], for [math]0 \le i \le a-1[/math]. Suppose that we start with the deck in increasing order (i.e., the cards are numbered from 1 to [math]n[/math]). We label the first [math]n_0[/math] cards with a 0, the next [math]n_1[/math] cards with a 1, etc. Then the [math]a[/math]-shuffle corresponding to [math]M[/math] is the shuffle which results in the ordering in which the cards labelled [math]i[/math] are placed in the positions in [math]M[/math] containing the label [math]i[/math]. The cards with the same label are placed in these positions in increasing order of their numbers. For example, if [math]n = 6[/math] and [math]a = 3[/math], let [math]M = (1, 0, 2, 2, 0, 2)[/math]. Then [math]n_0 = 2,\ n_1 = 1,[/math] and [math]n_2 = 3[/math]. So we label cards 1 and 2 with a 0, card 3 with a 1, and cards 4, 5, and 6 with a 2. Then cards 1 and 2 are placed in positions 2 and 5, card 3 is placed in position 1, and cards 4, 5, and 6 are placed in positions 3, 4, and 6, resulting in the ordering [math](3, 1, 4, 5, 2, 6)[/math].

• Using this coding, show that the probability that in an [math]a[/math]-shuffle, the first card (i.e., card number 1) moves to the [math]i[/math]th position, is given by the following expression:
  [[math]] {{(a-1)^{i-1}a^{n-i} + (a-2)^{i-1}(a-1)^{n-i} + \cdots + 1^{i-1}2^{n-i}}\over{a^n}}\ . [[/math]]
• Give an accurate estimate for the probability that in three riffle shuffles of a 52-card deck, the first card ends up in one of the first 26 positions. Using a computer, accurately estimate the probability of the same event after seven riffle shuffles.