The half-life of a radioactive substance is constant because the governing equation for radioactive decay is inverse exponential, i.e. the number decaying per second at any instant is proportional to the number of existing radioactive atoms.
So:
$$\frac{dN}{dt} \propto N $$
And hence:
$$ \frac{dN}{dt} = - \lambda t $$
where $\lambda$ is a constant signifying the instantaneous rate of radioactive decay.
This leads us to:
$$ N(t) = N_0 e^{-\lambda t} $$
where $N_0$ is the original number of radioactive atoms at some initial time and $\lambda$ is a constant related to the decay-rate for a given radioactive substance.
Regardless of what size $N_0$ may be, when the left-hand side of this equation is some known fraction of $N_0$ then the value of $t$ depends only on the decay-rate ($\lambda$) and the fraction of $N_0$ that we have on the LHS - both of which are constant for a given substance.
For $N(t) = \frac{N_0}{2}$ we have a half-life given by:
$$ t = \frac{ln2}{\lambda} $$
It clear from the graph of this function that the rate of decay slows down continually through the duration of radioactivity.
So your assertion that the second half of a radioactive substance's mass should need no more than its half-life to fully expire its radioactivity is clearly not applicable here: the second half of the radioactive mass will decay at an ever reducing rate and thus take much longer. In fact, to be exact about it, the remaining mass will take forever to lose full radioactivity as its decay rate slows down to near nothing towards the end.
The situation is analogous to unfilling a unit square by continually taking half of the square's remaining area. In this way after taking $N$ withdrawals we have a remaining area of:
$$ A = 1 - (\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + ... + \frac{1}{2^n}) $$
As $n$ tends to infinity, the above area tends to unity.
So it takes an infinite number of withdrawals to unfill the unit square.
The notion of half-life to measure the persistence of a radioactive source is useful insofar as it gives us a relatable means of describing how long a radioactive substance is significantly present and emitting potentially dangerous radiation to the atmosphere. If we are happy with 0.1 % of the radioactive mass being there and know its half-life to be 3 months, we can find the number of half-lives $N$ required to achieve this reduction via:
$$ (\frac{1}{2})^n = \frac{0.125}{100} $$
$$ \Rightarrow n \approx 10 $$
So after 10 x 3 months $\approx 2.5$ years there will be less than 0.1 % of the radioactive mass still there.