I was skimming through the [Google quantum supremacy paper](https://www.inverse.com/article/59507-full-quantum-supremacy-paper) but got stuck on this section:

> <img src="https://i.sstatic.net/M86jY.png" width="500"></img>

Questions:

 1. If $\mathcal{F}_{XEB} = 1$ when $x_i$ are sampled from the correct distribution then $\left< P(x_i) \right>_i$ must be $2^{1-n}$ (for any quantum circuit). That seems to restrict the output probability distributions of all quantum circuits to rather high entropy distributions. This is not what I would have expected for any kind of computation. Is this correct?

 2. If $P(x_i)$ is the probability of bitstring $x_i$ computed for the ideal quantum circuit, then how can $\left< P(x_i) \right>_i$ equal $1/2^n$ when the bitstrings $x_i$ are sampled from the uniform distribution? This is the result we would see if $P(x_i)$ had been $1/2^n$ for all $x_i$ (i.e. the probability mass function for the uniform distribution), but it is explicitly stated that $P(x_i)$ is the probability of bitstring $x_i$ computed for the ideal quantum circuit. Intuitively it seems to me more reasonable to assume that $\left< P(x_i) \right>_i$ with $x_i$ sampled from the uniform distribution would be much closer to zero, because if $P(x_i)$ concentrates much of its probability mass to a relatively low number of bitstrings (as I would assume computation to do) then $P(x_i) = 0$ for almost all bitstrings from the uniform distribution.

 3. How can the value of $\mathcal{F}_{XEB}$ correspond to "the probability that no error has occurred while running the circuit"? This sounds so simple/simplistic to me that it's hard to believe.