Quantum random generation implemented with IBM Q's AER results with non-even distribution even without noise model. Opened issue is: https://github.com/Qiskit/qiskit-aer/issues/437

res_circuit.png

res_counts.png

res_output.txt

Qiskit local environment
{   'qiskit': '0.13.0',
    'qiskit-aer': '0.3.2',
    'qiskit-aqua': '0.6.1',
    'qiskit-ibmq-provider': '0.3.3',
    'qiskit-ignis': '0.2.0',
    'qiskit-terra': '0.10.0'}

Is the new noise model ideal in terms of readout errors?
True

res_qutip_counts.png

res_readout_counts.png

run.py

import numpy as np

import qiskit

from qiskit.tools.visualization import plot_histogram
from qiskit import QuantumCircuit, ClassicalRegister, QuantumRegister, Aer

from qiskit.providers.aer.noise import NoiseModel
from qiskit.providers.aer.noise.errors import ReadoutError

from qutip import basis, snot

backend_aer = Aer.get_backend("qasm_simulator")

import pprint
pp = pprint.PrettyPrinter(indent=4)

print("Qiskit local environment")
pp.pprint(qiskit.__qiskit_version__)

# setup simplest circuit that reproduces the issue

N = 1
shots = 1024
q = QuantumRegister(N)
c = ClassicalRegister(N)
qc = QuantumCircuit(q, c)
qc.h(q[0])
qc.measure(q, c)
diagram = qc.draw(output="mpl")
diagram.savefig("res_circuit.png", format="png")

# set the noise model to be None

nm = None

# setup the backend

job = qiskit.execute(qc, backend_aer, shots=shots, noise_model=nm)
result = job.result()
counts = result.get_counts()
title = "Noise model set to None"
figure = plot_histogram(counts, title=title)
figure.savefig("res_counts.png", format="png")

# create a new noise model that ensures no readout errors

nm = NoiseModel()
probabilities = []
for i in range(2**N):
    lst = np.zeros(2**N)
    lst[i] = 1.0
    probabilities.append(lst)
error = ReadoutError(probabilities)
nm.add_all_qubit_readout_error(error)
print("\nIs the new noise model ideal in terms of readout errors?")
pp.pprint(error.ideal())

# run the backend again with no-readout-error noise model

job = qiskit.execute(qc, backend_aer, shots=shots, noise_model=nm)
result = job.result()
counts = result.get_counts()
title = "Ensured lack of readout errors"
figure = plot_histogram(counts, title=title)
figure.savefig("res_readout_counts.png", format="png")

# not let us perform equivalent calculation using QuTip

psi0 = basis(2, 0)
psif = snot()*psi0

P0 = np.power(np.abs(psif.overlap(basis(2, 0))), 2.0)
P1 = np.power(np.abs(psif.overlap(basis(2, 1))), 2.0)

qutip_counts = {
    "0": int(P0*shots),
    "1": int(P1*shots)
}

title = "Expected counts (QuTip)"
figure = plot_histogram(qutip_counts, title=title)
figure.savefig("res_qutip_counts.png", format="png")

resolved in Qiskit/qiskit-aer#437 (comment), quoting the answer from @nonhermitian

Measurements are random samples of the probability distribution that is 50/50. You should not expect to get exactly 50/50, just like you should not expect that when flipping a coin a finite number of times.

QuTiP is doing a statevector simulation and computing expectation values. Expectation values are what you would get having performed the experiment infinity many times.