No, one doesn't need to measure the material for years - or even millions or billions of years. It's enough to watch it for a few minutes (for time $t$) and count the number of atoms $\Delta N$ (convention: a positive number) that have decayed. The lifetime $T$ is calculated from
$$ \exp(-t/T) = \frac{N - \Delta N}{N}$$
where $N$ is the total number of atoms in the sample. This $N$ can be calculated as
$$N={\rm mass}_{\rm total} / {\rm mass}_{\rm atom}.$$
If we know that the lifetime is much longer than the time of the measurement, it's legitimate to Taylor-expand the exponential above and only keep the first uncancelled term:
$$ \frac{t}{T} = \frac{\Delta N}{N}.$$
The decay of the material proceeds atom-by-atom and the chances for individual atoms to decay are independent and equal.
To get some idea about the number of decays, consider 1 kilogram of uranium 238. Its atomic mass is $3.95\times 10^{-25}$ kilograms and its lifetime is $T=6.45$ billion years. By inverting the atomic mass, one sees that there are $2.53\times 10^{24}$ atoms in one kilogram. So if you take one kilogram of uranium 238, it will take $2.53\times 10^{24}$ times shorter a time for an average decay, e.g. the typical separation between two decays is
$$t_{\rm average} = \frac{6.45\times 10^9\times 365.2422\times 86400}{2.53\times 10^{24}}{\rm seconds} = 8.05\times 10^{-8} {\rm seconds}. $$
So one gets about 12.4 million decays during one second. (Thanks for the factor of 1000 fix.) These decays may be observed on an individual basis. Just to be sure, $T$ was always a lifetime in the text above. The half-life is simply $\ln(2) T$, about 69 percent of the lifetime, because of some simple maths (switching from the base $e$ to the base $2$ and vice versa).
If we observe $\Delta N$ decays, the typical relative statistical error of the number of decays is proportional to $1/(\Delta N)^{1/2}$. So if you want the accuracy "1 part in 1 thousand", you need to observe at least 1 million decays, and so on.
Do rates of nuclear decay depend on environmental factors?
There are two known environmental effects that can matter:
(1) The first one has been scientifically well established for a long time. In the process of electron capture, a proton in the nucleus combines with an inner-shell electron to produce a neutron and a neutrino. This effect does depend on the electronic environment, and in particular, the process cannot happen if the atom is completely ionized.
(2) In some exceptional examples, such as 187Re, there are beta decays with extremely low energies (in the keV range, rather than the usual MeV range). In these cases, there are significant effects due to the Pauli exclusion principle and the surrounding electron cloud. See Ionizing a beta decay nucleus causes faster decay?
Other claims of environmental effects on decay rates are crank science, often quoted by creationists in their attempts to discredit evolutionary and geological time scales.
He et al. (He 2007) claim to have detected a change in rates of beta decay of as much as 11% when samples are rotated in a centrifuge, and say that the effect varies asymmetrically with clockwise and counterclockwise rotation. He believes that there is a mysterious energy field that has both biological and nuclear effects, and that it relates to circadian rhythms. The nuclear effects were not observed when the experimental conditions were reproduced by Ding et al. [Ding 2009]
Jenkins and Fischbach (2008) claim to have observed effects on alpha decay rates at the 10^-3 level, correlated with an influence from the sun. They proposed that their results could be tested more dramatically by looking for changes in the rate of alpha decay in radioisotope thermoelectric generators aboard space probes. Such an effect turned out not to exist (Cooper 2009). Undeterred by their theory's failure to pass their own proposed test, they have gone on to publish even kookier ideas, such as a neutrino-mediated effect from solar flares, even though solar flares are a surface phenomenon, whereas neutrinos come from the sun's core. An independent study found no such link between flares and decay rates (Parkhomov 2010a). Laboratory experiments[Lindstrom 2010] have also placed limits on the sensitivity of radioactive decay to neutrino flux that rule out a neutrino-mediated effect at a level orders of magnitude less than what would be required in order to explain the variations claimed in [Jenkins 2008]. Despite this, Jenkins and Fischbach continue to speculate about a neutrino effect in [Sturrock 2012]; refusal to deal with contrary evidence is a hallmark of kook science. They admit that variations shown in their 2012 work "may be due in part to environmental influences," but don't seem to want to acknowledge that if the strength of these influences in unknown, they may explain the entire claimed effect, not just part of it.
Jenkins and Fischbach made further claims in 2010 based on experiments done decades ago by other people, so that Jenkins and Fischbach have no first-hand way of investigating possible sources of systematic error. Other attempts to reproduce the result are also plagued by systematic errors of the same size as the claimed effect. For example, an experiment by Parkhomov (2010b) shows a Fourier power spectrum in which a dozen other peaks are nearly as prominent as the claimed yearly variation.
Cardone et al. claim to have observed variations in the rate of alpha decay of thorium induced by 20 kHz ultrasound, and claim that this alpha decay occurs without the emission of gamma rays. Ericsson et al. have pointed out multiple severe problems with Cardone's experiments.
In agreement with theory, high-precision experimental tests show no detectable temperature-dependence in the rates of electron capture[Goodwin 2009] and alpha decay.[Gurevich 2008] Goodwin's results debunk a series of results from a group led by Rolfs, e.g., [Limata 2006], which used an inferior technique.
He YuJian et al., Science China 50 (2007) 170.
YouQian Ding et al., Science China 52 (2009) 690.
Jenkins and Fischbach (2008), https://arxiv.org/abs/0808.3283v1, Astropart.Phys.32:42-46,2009
Jenkins and Fischbach (2009), https://arxiv.org/abs/0808.3156, Astropart.Phys.31:407-411,2009
Jenkins and Fischbach (2010), https://arxiv.org/abs/1007.3318
Parkhomov 2010a, https://arxiv.org/abs/1006.2295
Parkhomov 2010b, https://arxiv.org/abs/1012.4174
Cooper (2009), https://arxiv.org/abs/0809.4248, Astropart.Phys.31:267-269,2009
Lindstrom et al. (2010), http://arxiv.org/abs/1006.5071 , Nuclear Instruments and Methods in Physics Research A, 622 (2010) 93-96
Sturrock 2012, https://arxiv.org/abs/1205.0205
F. Cardone, R. Mignani, A. Petrucci, Phys. Lett. A 373 (2009) 1956
Ericsson et al., Comment on "Piezonuclear decay of thorium," Phys. Lett. A 373 (2009) 1956, https://arxiv.org/abs/0907.0623
Ericsson et al., https://arxiv.org/abs/0909.2141
Goodwin, Golovko, Iacob and Hardy, "Half-life of the electron-capture decay of 97Ru: Precision measurement shows no temperature dependence" in Physical Review C (2009), 80, 045501, https://arxiv.org/abs/0910.4338
Gurevich et al., "The effect of metallic environment and low temperature on the 253Es α decay rate," Bull. Russ. Acad. Sci. 72 (2008) 315.
Limata et al., "First hints on a change of the 22Na βdecay half-life in the metal Pd," European Physical Journal A - Hadrons and Nuclei
May 2006, Volume 28, Issue 2, pp 251, https://link.springer.com/article/10.1140/epja/i2006-10057-1
Best 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, 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.