How many times are needed for card shuffling? (Q2785036)

It is stated that the paper was drawn from \textit{D. Bayer} and \textit{P. Diaconis} [Ann. Appl. Probab. 2, No.~2, 294-313 (1992; Zbl 0757.60003)]. A pack of \(n\) cards, initially ordered \(1,\dots,n\), is shuffled in the following random way. It is cut into two subpacks, \(A\), the upper one, and \(B\), the lower one. Here \(|A|\) is binomial \((n, 1/2)\). So \(A\) or \(B\) may be empty. Then a new pack is built by choosing the upper cards from \(A\) and \(B\) one by one with probabilities proportional to the remaining numbers in \(A\) and \(B\). This shuffling is repeated with the new pack. The probability that after \(m\) shuffles the order of the cards is a given permutation \(\pi\) of \(1,\dots,n\) is derived. It depends on the number of maximal subsequences of the form \(i, i+1,\dots,i+h\) in \(\pi\). The convergence to uniformity as \(m\to\infty\) is discussed. For large \(n\) one needs slightly more than \(\frac 32\log_2n\) shuffles for good mixing. A card trick based on these results is explained.