Bdtyktl

3 $begingroup$ The following problem I can only seem to solve by simulation. Suppose we take a deck and just label the cards from 1-52 in order, with 1 being the card on top. Now suppose we cut the deck at approximately the middle and complete the cut. We could assume that there's an equal probability that we cut at each of 3 cards near the exact middle; that is, we either cut at exactly the middle (26 cards in hand), or we cut up to 29 cards or as few as 23 cards, all with equal probability. Then we could ask, what's the probability that the $n$ th card is now on top? The answer is simply $0$ for most of the cards, and $frac{1}{7}$ that cards 24, 25, 26, 27, 28, 29, or 30 are on top. But suppose we perform this cut twice, what then? I think the simplest answer unfortunately is just to sum up all ...