I’ve been learning a lot of quantum computing. For now, here's rudimentary basics.

What I can tell you is that probabilities are created along a “circuit”. First by applying a Hadamard gate, whereby you would have qbits in superposition, and thus any new spooky action you wish to perform on a target qbit can be done via entanglement after it’s measured, which would then give us information about what the state is of the other entangled qbit during its journey down the rest of its circuit path. I don't mean to imply one necessarily needs to apply Hadamard for every circuit run, but then what's the point of using a Quantum Computer? :)

The math behind how gates are applied is much more complex, but imagine a bunch of matrix operations being performed on each line of a qbits circuit, and sometimes even groupings of circuit lines. These matrices can contain fractions with complex numbers or values such as 1 or 0, mainly because the sum of the rows or columns for any qbit eventually will be the absolute values of each item squared, which will be add up to 1:

\sum_{\alpha \in \Sigma} |{\alpha|^2} = 1

Matrix Operations

Unitary - A matrix is unitary if the matrix U times its conjugate transposed self is equal to 1 (i.e., the identity matrix), or the conjugate transpose of U matrix is the the inverse of U.

UU^\dagger = I \\

U^\dagger = U^{-1}

Hermitian - A matrix U is hermitian where upon applying a conjugate transpose, the matrix U appears to have been unchanged after the operation.

Pauli Operations

All of the below operations are Unitary

Identity Matrix which effectively does nothing as an operation

$I = \begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix}$

NOT GATE - Flips the matrix. So if Bit is 1, it now is 0

$\sigma_X = \begin{bmatrix} 0 & 1 \\ 1& 0 \end{bmatrix}$

Y gate, I assume just turns the qbit along the Y axis. Need more details....

$\sigma_Y = \begin{bmatrix} 0 & -i \\ i & 0 \end{bmatrix}$

This a phase flip

$\sigma_Z = \begin{bmatrix} 1 & 0\\ 0 & -1 \end{bmatrix}$

Hadamard gate which puts a qbit into superposition, effectively giving a 0 qubit a 50/50 chance of being a 1or zero upon measurement.

$H = \begin{bmatrix} \frac{1} {\sqrt{2} } & \frac{1}{\sqrt{2} } \\ \frac{1} {\sqrt{2} } & -\frac{1} {\sqrt{2} } \\ \end{bmatrix}$

To imagine a qbit as a vector, you can think of it as \begin{bmatrix} 1 \\ 0 \end{bmatrix}

This is important because a qbit is something which can represent probabilities between values. In this above case, since we have a 1 in the 0th index, it would denote a Dirac notated value as |0>

When applying the Hadamard gate, we see that the value is now presented as

\frac{1}{\sqrt{2}}|0> + \frac{1}{\sqrt{2}}|1>

Now that we have that, the vector would contain values the equal to 1, but would represent a 50/50 chance of being a 1 or 0 upon measurement.

We can apply any of the above Pauli operations of X, Y, or Z to this value into different states so the qbits probabilities can be altered (e.g., 50%/50% -> %80/20%).

Applying the Hadamard Gate to either of these states gives a new state.

|0> = \begin{bmatrix} 1 \\ 0 \end{bmatrix}, |1> = \begin{bmatrix} 0 \\ 1 \end{bmatrix}, \begin{bmatrix} 1 \\ 0 \end{bmatrix}, |+>=\begin{bmatrix} 1 \\ \frac{1}{\sqrt{2}} \end{bmatrix}, |-> = \begin{bmatrix} 1 \\ -\frac{1}{\sqrt{2}} \end{bmatrix}

Example translations:
H∣0⟩=∣+⟩ 

H∣1⟩=∣−⟩ 

H∣+⟩=∣0⟩ 

H∣−⟩=∣1⟩  

Entanglement

Other things to be mindful of are bell states, which are a representation of maximally entangled (guaranteed change!) qbits.

We have four bell states to be mindful of. Each can be represented as a vector where we denote the position of values. Between 2 qbits there there 4 potential states to represent, so we create a 4x1 vector. In Dirac notation, we can see correlate index 0,1,2,3 to their binary counterparts of 00, 01, 10, 11. Here are the four bell states:

|\Phi^+> = \begin{bmatrix} \frac{1}{\sqrt{2}} \\ 0 \\ 0 \\ \frac{1}{\sqrt{2}} \end{bmatrix}

|\Phi^-> = \begin{bmatrix} \frac{1}{\sqrt{2}} \\ 0 \\ 0 \\ - \frac{1}{\sqrt{2}} \end{bmatrix}

|\Psi^+> = \begin{bmatrix} 0 \\ \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} \\ 0 \end{bmatrix}

|\Psi^-> = \begin{bmatrix} 0 \\ \frac{1}{\sqrt{2}} \\ -\frac{1}{\sqrt{2}} \\ 0 \end{bmatrix}

The + and - signs here denote some level of constructive or destructive interference, but I'm unsure exactly what this means right now.