We consider the permutation analogue of Penney's game for words. Two players, in order, each choose a permutation of length $k\ge3$; 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.