Jay Gambetta mentions a unitary matrix that is used to represent a Hadamard gate. It corresponds to rotating a coin by $$90^\circ$$, so a coin that's initially heads up becomes vertically oriented with, say, heads facing away from the wall. If the wall is magnetic and you release the coin, it will stick to it with heads up. If, instead, you started with a coin that's tails up and applied the same rotation, it would be vertical with tails facing away from the wall. If you release it (and the wall is still magnetic), you get tails. On the other hand, if the wall is not magnetic and you drop it, it lands heads or tails with equal probability. Using a "floor" measurement doesn't distinguish between the two vertical orientations, but using a "wall" measurement does. It's not so much whether things are predictable or not, it's the type of measurement you do that distinguishes one quantum state (or coin orientation) from another.

This is the whole of it. The only remaining mystery is that the outcome of the coin measurement is considered to be, in theory, completely deterministic, while that of the qubit is considered to be, except in special cases, "intrinsically random." But that's another discussion...

The analogy between qubits and coin flips is popular but can be misleading. (See, for example, this video: https://www.youtube.com/watch?v=lypnkNm0B4A) A coin spinning in the air and landing on the ground is not truly random, though we may describe it as such. The key point is how you measure it.

At any point in time the coin has a definite orientation, though it may be unknown to us. Likewise, qubits have a definite state at any time, which we can describe by a point on the surface of a sphere (the so-called Bloch sphere). Mathematically, a coin's orientation and a qubit's state are equivalent. While in the air, the coin may undergo deterministic and reversible motion (e.g., spinning and falling). Likewise, prior to measurement a qubit may undergo deterministic and reversible transformations (e.g., unitary gate operations on a quantum computer).

Measurement represents an irreversible process. For a coin, it is a series of inelastic collisions with the ground, bouncing and spinning until it comes to rest. If we are completely ignorant of the initial conditions of the coin, the two final orientations (heads or tails) will appear equally likely, but this is not always the case. If I drop it oriented "heads up" from a short height, it will land flat with "heads up" with near certainty. But suppose I was standing next to a large magnetic wall and did this. The coin would hit edge-on and would likely land with either heads or tails showing, with equal probability. One could imagine doing this experiment with various initial orientations of the coin and orientations of the magnetic wall (upright, flat, slanted, etc.). You can imagine that the probability of getting heads or tails will be different, depending on the relative orientations of the coin and wall. (In theory it's all completely deterministic, but in practice we never know the initial conditions that precisely.)

Measurements of qubits are quite similar. I can prepare a qubit in the state $$\frac{1}{\sqrt{2}}[|0\rangle + |1\rangle]$$, measure it in the computational0/1 basis $$\{|0\rangle, |1\rangle\}$$, and get either $$|0\rangle$$ or $$|1\rangle$$ with equal probability. If, however, I measure in the diagonal+/- basis $$\{|+\rangle, |-\rangle\}$$ (analogous to using a slanted magnetic wall), I get $$|+\rangle$$ with near certainty. (I say "near certainty" because, well, nothing in the real world is perfect.) Here, $$|\pm\rangle = \frac{1}{\sqrt{2}}[|0\rangle\pm|1\rangle$$ are the diagonal+/- basis states. For polarized photons, for example, this could be done used polarization filters rotated $$45^\circ$$.

The difference between preparing the state $$\frac{1}{\sqrt{2}}[|0\rangle + |1\rangle]$$ and the state $$\frac{1}{\sqrt{2}}[|0\rangle - |1\rangle]$$ is the difference between preparing a diagonallyvertically oriented coin with either heads up or downtails facing away from the wall. (A good picture would really help here.) We can tell which of the two states is prepared based on the outcome of a suitably chosen measurement, which in this case would be a diagonal+/- basis (or slantedmagnetic wall) measurement.

This is the whole of it. The only remaining mystery is that the outcome of the coin measurement is considered to be, in theory, completely deterministic, while that of the qubit is considered to be, except in special cases, "intrinsically random." But that's another discussion...

The analogy between qubits and coin flips is popular but can be misleading. (See, for example, this video: https://www.youtube.com/watch?v=lypnkNm0B4A) A coin spinning in the air and landing on the ground is not truly random, though we may describe it as such. The key point is how you measure it.

At any point in time the coin has a definite orientation, though it may be unknown to us. Likewise, qubits have a definite state at any time, which we can describe by a point on the surface of a sphere (the so-called Bloch sphere). Mathematically, a coin's orientation and a qubit's state are equivalent. While in the air, the coin may undergo deterministic and reversible motion (e.g., spinning and falling). Likewise, prior to measurement a qubit may undergo deterministic and reversible transformations (e.g., unitary gate operations on a quantum computer).

Measurement represents an irreversible process. For a coin, it is a series of inelastic collisions with the ground, bouncing and spinning until it comes to rest. If we are completely ignorant of the initial conditions of the coin, the two final orientations (heads or tails) will appear equally likely, but this is not always the case. If I drop it oriented "heads up" from a short height, it will land flat with "heads up" with near certainty. But suppose I was standing next to a large magnetic wall and did this. The coin would hit edge-on and would likely land with either heads or tails showing, with equal probability. One could imagine doing this experiment with various initial orientations of the coin and orientations of the magnetic wall (upright, flat, slanted, etc.). You can imagine that the probability of getting heads or tails will be different, depending on the relative orientations of the coin and wall. (In theory it's all completely deterministic, but in practice we never know the initial conditions that precisely.)

Measurements of qubits are quite similar. I can prepare a qubit in the state $$\frac{1}{\sqrt{2}}[|0\rangle + |1\rangle]$$, measure it in the computational basis $$\{|0\rangle, |1\rangle\}$$, and get either $$|0\rangle$$ or $$|1\rangle$$ with equal probability. If, however, I measure in the diagonal basis $$\{|+\rangle, |-\rangle\}$$ (analogous to using a slanted magnetic wall), I get $$|+\rangle$$ with near certainty. (I say "near certainty" because, well, nothing in the real world is perfect.) Here, $$|\pm\rangle = \frac{1}{\sqrt{2}}[|0\rangle\pm|1\rangle$$ are the diagonal basis states. For polarized photons, for example, this could be done used polarization filters rotated $$45^\circ$$.

The difference between preparing the state $$\frac{1}{\sqrt{2}}[|0\rangle + |1\rangle]$$ and the state $$\frac{1}{\sqrt{2}}[|0\rangle - |1\rangle]$$ is the difference between preparing a diagonally oriented coin with either heads up or down. (A good picture would really help here.) We can tell which of the two states is prepared based on the outcome of a suitably chosen measurement, which in this case would be a diagonal basis (or slanted wall) measurement.

This is the whole of it. The only remaining mystery is that the outcome of the coin measurement is considered to be, in theory, completely deterministic, while that of the qubit is considered to be, except in special cases, "intrinsically random." But that's another discussion...