Question: I wondered this since my teacher told us about half life of radioactive materials back in school. It seems intuitive to me to think this way, but I wonder if there's a deeper explanation which proves me wrong.

When many atoms are involved, half life can statistically hold, but since decaying of an individual atom is completely random and stateless, can't all the atoms in a 1 kg of matter just decide to decay in the next minute, even if the probability of this event occurring is extremely small?

Answer: 
The short answer is *yes*. No matter how many atoms there are, there is always a (sometimes vanishingly small) chance that *all* of them decay in the next minute. The *fun* answer is actually seeing how small this probability gets for large numbers of atoms.

Let's take [iodine-131](https://en.wikipedia.org/wiki/Iodine-131), which I chose because it has the reasonable half-life of around $8$ days = $\text{691,200}$ seconds. Now $1$ kg of iodine-131 will have around $7.63 \times N\_A$ atoms in it, where $N\_A$ is Avogadro's constant. Using the formula for probability for the decay of an atom in time $t$:

$$
P(t) = 1-\exp(-\lambda t),
$$

and assuming that all decays are statistically independent$^\dagger$, the probability that *all* the atoms will have decayed in one minute is:

$$
(1-\exp(-\lambda \times 60\,\text{s}))^{7.63\times N\_A}
$$

where $\lambda$ is the decay constant, equal to $\frac{\ln 2}{\text{half-life}}$, in this case, almost exactly $10^{-6}\,\text{s}^{–1}$. So
$$
P = (1-\exp(-6\times10^{-5}))^{7.63\times N\_A} \\
\approx(6\times10^{-5})^{7.63\times N\_A} \\
\approx (10^{-4.22})^{7.63\times N\_A} \\
= 10^{-4.22\times7.63\times N\_A} \\
\approx 10^{-1.94\times10^{25}}
$$

(I chose iodine-131 as a concrete example, but pretty much any radioactive atom will result in a similar probability, no matter what the mass or the half-life is.) So if you played out this experiment on $10^{1.94\times10^{25}}$ such setups, you would expect all the atoms to decay in *one* of the setups, on average.

To give you an idea of how incomprehensibly large this number is, there are "only" $10^{78}$ atoms in the universe - that's $1$ followed by $78$ zeroes. $10^{1.94\times10^{25}}$ is $1$ followed by over a million billion billion zeroes. I'd much rather bet on horses.

---

$^\dagger$ This Poisson distribution model is a simplifying, but perhaps crude approximation in this scenario, since even small deviations from statistical independence can add up to large suppressing factors given the number of atoms, and so $10^{1.94\times10^{25}}$ is certainly an upper bound (of course, the approximation is fully justified if the atoms are separated to infinity at $0 \text{ K}$, or their decay products do not have sufficient energy to make more than a $1/N\_A$-order change in the decay probability of other atoms). A more detailed analysis would have to be tailored specifically to the isotope under consideration - or a next-order approximation could be made by making the decay constant $\lambda$ a strictly increasing *function* of time. Rest assured that the true probability, while *much* more difficult to calculate than this back-of-the-envelope estimation, will still run into the mind-bogglingly large territory of $1$ in $1$ followed by several trillions of zeroes.
In order for that to *happen in the real world* you need to start with about 3.8 million kilograms of that material.

Here is how you come up with that number. You start from the the formula connecting the half-life to the number of particles over time

$$ N(t) = N\_0 \left(\frac{1}{2}\right)^\frac{t}{t\_{1/2}} $$

Now you replace $N(t)$ with what you would like to have
$$ N\_0 - 1~\text{kg} = N\_0 \left(\frac{1}{2}\right)^\frac{t}{t\_{1/2}} $$
And you solve for $N\_0$
$$ N\_0 = \frac{1~\text{kg}}{1-\left(\frac{1}{2}\right)^\frac{t}{t\_{1/2}}}$$
At this point is just a matter of plugging in $t=60~\text{s}$ and $t\_{1/2}=5~\text{y}$.