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Is the number of qubits of the quantum computer held constant or may it vary with input size? --AxelBoldt

I think it is assumed that there's always enough of them, just as we do with Turing tapes. --Seb

What does this 1/4-clause mean in long run ? That chance of failure in many runs is (1/4)^N, 1/2*(1/2)^N or what ? --Taw

The probability that the algorithm fails N times in a row is (1/4)N. Actually I think 1/4 is more or less arbitrary; choosing any other (rational?) number in ]0,1/2[ would not change the class. --Seb