Why a qubit is not a probabilistic bit
A qubit is not just a bit with some randomness added. If you scramble a random bit again, it stays random. But send a qubit through two Hadamard gates in a row and it reads 0 on every single shot. That canceling is called interference, and no random bit can do it.
You may have heard that a qubit "is 0 and 1 at the same time," or that it is "a coin that hasn't landed yet." Those slogans really describe a probabilistic bit. That is a normal bit whose value you just don't know yet. Random bits are useful, but there is nothing quantum about them.
A qubit is something different. You can test the difference with one short experiment. First, apply the "randomizing" gate H (the Hadamard gate) once and look at the results. Then apply it twice in a row and look again. A coin-flip model makes a firm guess about the second run. That guess turns out to be wrong.
What does one Hadamard look like?
What would a coin-flip model guess for two flips?
Pretend H really meant "flip a fair coin and replace the bit with the result." Then doing it twice is just flipping the coin again. You still get 50/50. Flip it ten times and you still get 50/50. Think of shuffling a deck of cards. Once it is fully shuffled, more shuffling never puts the cards back in order. In any coin model, randomness can spread out, but it never cancels.
Now run H twice on the same qubit and test that guess.
What actually happens with H then H?
Where did the randomness go?
A qubit's state is not a pair of chances. It is a pair of amplitudes. An amplitude is a number that says how strongly the qubit leans toward an answer. Unlike a chance, an amplitude can be negative. To get the chance of an answer, you square its amplitude.
Let's walk through it. The qubit starts as |0⟩, which means "surely 0." The first H turns it into (|0⟩ + |1⟩)/√2. That is an amplitude of 1/√2 on each answer. Square it: (1/√2)² = ½. So one H alone reads 0 half the time.
The second H acts on each part separately. It turns |0⟩ into (|0⟩ + |1⟩)/√2. It turns |1⟩ into (|0⟩ − |1⟩)/√2. Note the minus sign. Now add up the pieces. For the answer 0 you get ½ + ½ = 1. For the answer 1 you get ½ − ½ = 0. The answer 1 is not just unlikely. Its amplitude has been wiped out completely. That wiping out is called interference.
Here is an everyday picture. Noise-canceling headphones play a sound wave that is the exact opposite of the noise. The two waves add up to silence. Amplitudes can cancel the same way. Unlike sound, though, there is no hidden wave you could ever listen to. You only see the final count of 0s and 1s.
A random bit cannot do this. Chances are never negative, so there is nothing to cancel. Amplitudes have signs, and signs can erase each other. Every quantum algorithm that beats a normal computer works this way. It arranges the cancellations so wrong answers erase and right answers add up.
What does this look like on real hardware?
The two-H test is a real check that labs use to tune their machines. On the simulator, which has no noise, you get 0 on every shot. On a real device, a few shots read 1. That is because the gates and the readout are not perfect. On today's superconducting and trapped-ion machines, a single-qubit gate goes wrong about 1 time in 1,000 to 1 time in 10,000 (an error of 10⁻³ to 10⁻⁴). Reading the qubit out gets the bit wrong about 1% of the time on many devices. In a circuit this short, readout is usually the biggest error.
So the honest answer has three layers. In theory, H then H always gives |0⟩. In simulation, you will see exactly that. On hardware today, you would usually see 0 on about 97–99% of shots, held back mostly by readout. Each device's gate and readout fidelity (how often an operation does the right thing) is listed on the QPU spec pages. You can line devices up side by side on the comparison page.