We consider the permutation analogue of Penney's game for words. Two players, in order, each choose a permutation of length k≥3; then a sequence of independent random values from a continuous distribution is generated, until the relative order of the last k numbers coincides with one of the chosen permutations, making that player the winner. We compute the winning probabilities for all pairs of permutations of length 3 and some pairs of length 4, showing that, as in the original version for words, the game is non-transitive. Our proofs introduce new bijections for consecutive patterns in permutations. We also give some formulas to compute the winning probabilities more generally, and conjecture a winning strategy for the second player when k is arbitrary.
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Elizalde et al. (2024) studied this question.
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