Anybody understand the details of how they compute the F_XEB fidelity estimator? I'm very perplexed that the main article states that "𝑃(𝑥_𝑖) is the probabi

Anybody understand the details of how they compute the F_XEB fidelity estimator? I'm very perplexed that the main article states that "𝑃(𝑥_𝑖) is the probability of bitstring 𝑥_𝑖 computed for the ideal quantum circuit", but Supplementary Information section IV.C seems to argue that if the "qubits are in the maximally mixed state" (i.e. the quantum computer doesn't work) then "the estimator [𝐹_𝑋𝐸𝐵] yields zero fidelity" since 𝑃(𝑥_𝑖)=1/2^𝑛 for every 𝑖 in this case. To me this doesn't make sense, since P(x_i)=1/2^n is clearly the probability mass function of the empirical non-ideal quantum circuit output distribution (and not the ideal one). I've posted a question on the QC stack exchange: https://quantumcomputing.stackexchange.com/questions/8427/quantum-supremacy-some-questions-on-cross-entropy-benchmarking https://quantumcomputing.stackexchange.com/questions/8427/qu...