No-feedback card guessing for dovetail shuffles
No-feedback card guessing for dovetail shuffles
复制标题
燕尾洗牌的无反馈猜牌
DOI：
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发表时间：
1998
期刊：
影响因子：
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通讯作者：
M. Ciucu
中科院分区：
文献类型：
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作者：
M. Ciucu
我们考虑以下问题。一副2n张牌，从上到下依次标有1到2n，面朝下放在桌子上。甲板上有k个燕尾牌，面朝下放回桌子上。猜的人试着从上面开始，一次猜一张牌。被猜测的牌的身份不会被透露，也不会被告知猜测者某个特定的猜测是否正确。目标是使正确猜测的次数最大化。我们证明了，对于k2log2 (2n) + 1，最好的策略是猜牌1是前半部分，牌2n是后半部分。这个结果可以解释为，需要执行log 2 （2n）次的级数才能得到一个混合良好的牌组，Bayer和Diaconis已经证明了这一点[3]。我们还证明，如果k = c log 2 （2n）且1 < c < 2，则上述猜测策略不是最好的。
We consider the following problem. A deck of 2n cards labeled consecutively from 1 on top to 2n on bottom is face down on the table. The deck is given k dovetail shuues and placed back on the table, face down. A guesser tries to guess at the cards one at a time, starting from top. The identity of the card guessed at is not revealed, nor is the guesser told whether a particular guess was correct or not. The goal is to maximize the number of correct guesses. We show that for k 2 log 2 (2n) + 1 the best strategy is to guess card 1 for the rst half of the deck and card 2n for the second half. This result can be interpreted as indicating that it suuces to perform the order of log 2 (2n) shuues to obtain a well mixed deck, a fact proved by Bayer and Diaconis 3]. We also show that if k = c log 2 (2n) with 1 < c < 2 then the above guessing strategy is not the best.