What does it mean when it says "probability of death doubles" every 8 years? Can probability of death exceed 136%? I've seen on a lot of websites that your probability of death "doubles" every 8 years.
However, the way they calculate the probability of death seems to lead to counterintuitive conclusions. e.g on http://www.npr.org/blogs/krulwich/2014/01/08/260463710/am-i-going-to-die-this-year-a-mathematical-puzzle they calculate probability of death like this:
At 25 years old, your probability is 1 in 3000
At 33: 1 in 1500
+8 : 1 in 750
Presumably, the progression goes: 1 in 3000, then 1500, then 750, then 375, then 187.5, then 93.2, then 46.875, then 23.4375, then 11.71875, then ~5.86, then ~2.9, then ~1.5, then ~0.7.
But 1 in 0.7 = 1.36533... = over 136%
How can you have over 136% probability of dying? Surely the maximum probability is 100%, and that will never be reached due to uncertainty? 
It seems unintuitive to me. Could someone please tell me if this calculation is correct, or did I miss something? If not, what's the correct way to calculate probability? 
 A: This formula presented in the article is only a rough empirical approximation to the real death numbers. It does not make sense for high ages.
If you compare your results to the numbers cited you will see the prediction deviating increasingly at ages over 100.
A: A simple model of the probability to be still alive at time $x$ is
$$p(x)={1\over2}(1-\tanh x)={e^{-x}\over e^x+e^{-x}}\ .$$

Here $x=0$ corresponds to the age where this probability is $={1\over2}$, and the time scale has to be adjusted to demographical data. The above figure corresponds to $x={t-70\over15}$, $t$ denoting age in years. At any rate
$$p'(x)=-{1\over 2\cosh^2 x}={-2\over e^{2x}+2+e^{-2x}}\ .$$
When $x\ll0$ one has
$$p'(x)\doteq -2e^{2x}\ ,$$
which shows that the probability to die the following day  (or in the next year) increases exponentially with time. While this regime is in force we see a characteristic "doubling time" of this probability.
On the other hand, when $x\gg0$ we have
$$p(x)\doteq e^{-2x}\ ,$$
wich shows that for $x\to\infty$ the probability of survival decreases exponentially with time, giving rise to a typical "half-life-span".
A: It's important to distinguish between two kinds of probability when we talk about continuous quantities, such as time.
The cumulative probability of dying within some specified length of time, such as one year, is a number between $0$ and $1$, or in other words, between $0\%$ and $100\%$. It would not make sense to say that the probability of dying in a given year is $136\%$.
On the other hand, the instantaneous probability of dying is a probability density function, with units of $1/[\mathrm{time}]$. The value of a pdf is not constrained between $0$ and $1$; indeed the comparison doesn't even make sense, because the units aren't the same. You could, for example, have a probability density of dying equal to $p = 1.36\,\mathrm{yr}^{-1}$. That simply means that for very small time periods $\Delta t$, the probability of dying within $\Delta t$ is approximately $p\Delta t$. We will have $p\Delta t < 1$ after all, although $p \not< 1\,\mathrm{yr}^{-1}$.
A: The correct form of the statement is that "your mortality rate doubles roughly every eight years."  
You can think of the "mortality rate" as, for example, the probability of dying during the next second.  Or, more formally, the next infinitesimally small unit of time.  
As long as the rate is much lower than 1/year, then calling the mortality rate "your probability of dying during the next year" is okay.  But once the mortality rate gets larger than 1/year, you need a more precise form of the statement.
