Born rule is a postulate in quantum mechanics that gives an interpretation of probability amplitudes. Namely, probability amplitude $c\in\mathbb{C}$ associated with some state $\lvert i\rangle$ encodes the probability of finding the quantum system in the state $\lvert i\rangle$. The value of this probability – according to the Born rule – is proportional to the modulus squared of the probability amplitude $c$.

$$P_i = |c|^2$$

A single qubit $\lvert\psi\rangle$ has two probability amplitudes, $a\in\mathbb{C}$ which is associated with the state $\lvert 0\rangle$, and $b\in\mathbb{C}$ which is associated with the state $\lvert 1\rangle$:

$$\lvert\psi\rangle = a\lvert 0\rangle + b\lvert 1\rangle$$

If we attempt to measure the qubit, it will collapse to either state $\lvert 0\rangle$ with probability $P_0$, or state $\lvert 1\rangle$ with probability $P_1$. What is the value of these probabilities? That's exactly the question that the Born rule answers! It's the modulus square of the probability amplitudes.

$$ P_0 = |a|^2,\quad P_1 = |b|^2 $$

These are real probabilities and must satisfy the second axiom of probability i.e. they must add up to 1 (meaning 100%):

$$|a|^2 + |b|^2 = 1$$

Let's use a concrete examples for probability amplitudes $a\in\mathbb{C}$ and $b\in\mathbb{C}$.

$$\lvert\psi\rangle = \left(\frac{1}{2\sqrt{2}} + i\frac{1}{2\sqrt{2}}\right)\lvert 0 \rangle + \left(-\frac{\sqrt{3}}{2\sqrt{2}} + \frac{i\sqrt{3}}{2\sqrt{2}}\right)\lvert 1 \rangle$$

Recall that a modulus of a complex number written in Cartesian form is just the Pythagorean theorem $|z| = |x +iy| = \sqrt{x^2 + y^2}$. But Born rule simplifies this for us even further, because the square eliminates the square root $|z|^2 = |x + iy|^2 = x^2 + y^2$!

$$P_0 = |a|^2 = \left|\frac{1}{2\sqrt{2}} + i\frac{1}{2\sqrt{2}}\right|^2 = \frac{1}{4} = 25\%$$ $$P_1 = |b|^2 = \left|-\frac{\sqrt{3}}{2\sqrt{2}} + \frac{i\sqrt{3}}{2\sqrt{2}}\right|^2 = \frac{3}{4} = 75\%$$

Thus if we measure this qubit it will collapse to state $\lvert 0\rangle$ with 25% probability, or to state $\lvert 1\rangle$ with 75% probability.

What happens if we measure the qubit twice? Let's say measure the previous qubit and – against the odds – it collapses to state $\lvert 0\rangle$, even though it only had 25% chance of doing so. What happens if we measure it again? Would it flip to $\lvert 1\rangle$ with 75% probability? The answer is no. No matter how many times we measure it across the $\lvert 0\rangle, \lvert 1\rangle$ basis, after collapse it's always going to retain the same state.

To see why, start from the collapsed qubit: $$\lvert\psi\rangle = \lvert 0\rangle$$

This is actually just a short hand way of writing the following: $$\lvert\psi\rangle = 1\cdot\lvert 0\rangle + 0\cdot\lvert 1\rangle$$

Which is also a short hand way of writing the following because probability amplitudes are still complex numbers: $$\lvert\psi\rangle = (1 + i0)\lvert 0\rangle + (0 + i0)\lvert 1\rangle$$

Applying the Born rule, we get: $$P_0 = |a|^2 = |1 + i0|^2 = 1 = 100\%$$ $$P_1 = |b|^2 = |0 + i0|^2 = 0 = 0\%$$

Therefore, after measuring the qubit across the $\lvert 0\rangle, \lvert 1\rangle$ basis twice, the probability of remaining in the state $\lvert 0\rangle$ is 100%, while the probability of recollapsing to state $\lvert 1\rangle$ is 0%. Meaning, its state will almost surely not change due to double measurement.

But this does not mean the qubit is permanently destroyed! We can easily re-entangle the qubit by applying the Hadamard gate or by rotating the Stern-Gerlach machine and measure it across a different basis (e.g. $\lvert +\rangle, \lvert -\rangle$ instead of $\lvert 0\rangle, \lvert 1\rangle$).