Why a qubit is not a probabilistic bit
A qubit is not a bit with randomness attached. A random bit, re-scrambled, stays random — but a qubit sent through two Hadamard gates in a row returns 0 on every single shot. That deterministic un-randomising is interference, and no probabilistic bit can do it.
The most common shortcut for explaining qubits — "it's 0 and 1 at the same time", "it's a coin that hasn't landed yet" — actually describes a probabilistic bit: a bit whose value you don't know. Probabilistic bits are useful and thoroughly classical; nothing quantum about them.
A qubit is a different object, and the difference is testable in one short experiment. The plan: apply the "randomising" gate H once and look at the statistics, then apply it twice in a row and look again. A coin-flip model makes a firm prediction for the second run — and it is wrong.
What does one Hadamard look like?
What would a coin-flip model predict for two flips?
Suppose H really meant "replace the bit with a fresh 50/50 random value". Then applying it twice is just re-scrambling a scrambled bit: still 50/50. Ten times: still 50/50. In any probabilistic model, once the distribution is uniform it stays uniform — probabilities can be diluted, but randomness never cancels.
Now run H twice on the same qubit and check that prediction.
What actually happens with H then H?
Where did the randomness go?
The qubit's state is not a probability — it is a pair of amplitudes, numbers that can be negative. Starting from |0⟩, the first H produces the state (|0⟩ + |1⟩)/√2: amplitude 1/√2 on each outcome, which is why one H alone measures 50/50.
The second H acts on each part. It sends |0⟩ → (|0⟩ + |1⟩)/√2 and |1⟩ → (|0⟩ − |1⟩)/√2. Add the pieces: the |0⟩ contributions are ½ + ½ = 1, and the |1⟩ contributions are ½ − ½ = 0. The outcome 1 is not merely unlikely — its amplitude has been cancelled exactly. That cancellation is interference.
This is the capability a probabilistic bit lacks. Probabilities are never negative, so a classical stochastic process has nothing that can cancel. Amplitudes carry signs, and signs can erase each other. Every quantum algorithm that beats a classical one is, at bottom, an exercise in arranging cancellations so that wrong answers erase and right answers add.
What does this look like on real hardware?
The two-H experiment is a real, calibration-grade test. On the noise-free simulator you get 0 on every shot. On a physical device a small fraction of shots read 1, because gates and readout are slightly imperfect: current superconducting and trapped-ion machines run a single-qubit gate with error around 10⁻³ to 10⁻⁴, while reading the qubit out misassigns the bit on the order of 1% on many devices — usually the largest error in a circuit this short.
So the honest statement has three layers. In theory, H then H returns |0⟩ with certainty. In simulation, you will observe exactly that. On hardware today, you would typically see zeros on roughly 97–99% of shots, limited mostly by readout. Per-device gate and readout fidelities are listed on the QPU spec pages, and you can line devices up side by side on the comparison page.