If I have a $0.00048\%$ chance of dying every second, how to numerically calculate the chance I have of dying in a day?

Question

Hypothetically, if I have a 0.00048% chance of dying when I blink, and I blink once a second, what chance do I have of dying in a single day?

I tried $1-0.0000048^{86400}$ but no calculator I could find would support this. How would I work this out manually?

Answer 1

When $n$ is large, $p$ is small and $np<10$, then the Poisson approximation is very good. In that case, the answer is approximately: $$P =1 - e^{-\lambda}=1-0.6605 = 0.3395$$, where $\lambda = np = 0.41472.$

Answer 2

As @Saketh and @dxiv indicate, you want to take a large power: $(1 - p)^{86400}$, where $p$ is tiny. Calculators don't do well at this. But if you use the rule that $$ a^b = \exp(b \log a) $$ then you can compute $$ b \log a \approx 86400 \log .9999952 \approx -0.41472099533 $$ and compute $e$ to that power to get approximately $0.6605...$, and hence your probability of dying is 1 minus that, or about 34%.

The key step is in using the logarithm to compute the exponent, for your calculator's built-in log function (perhaps called "ln") is very accurate near 1, and exponentiation is pretty accurate for numbers like $e$ (a little less than $3$) with exponents between $0$ and about $5$.

Answer 3

Basically the way you do this is use complementary probability.

The chance of you not dying every second is $99.99952\% = 0.9999952$.

$(0.9999952)^{86400}= 0.660524544429 = 66.052\%$ is the chance you don't die.

The chance you do die is $1-66.052\% = \boxed{33.948\%}$.

I want to die :0

Answer 4

Many systems (the online system WolframAlpha, Mathematica, R, etc.) will happily compute the given expression, but you can also use the series $$(1 + p)^n = 1 + \binom{n}{1}p + \binom{n}{2}p^2 + \cdots + p^n.$$ In our case, $p = -0.0000048$ and $n = 86400$. The first few terms are easily computable with a hand-held calculator, and just going to the $p^2$ and $p^4$ terms is good enough for two and three decimal places, respectively.

Answer 5

Why not use a simple spreedshet as in this figure?

Sometime ago all this was done using a ''little magic book'' called Logarithm Table ! (see my answer here) or was calculated wit a slide rule.

Answer 6

$$ 86400_{10}=10101000110000000_2 $$

which means

$$ 0.9999952^{86400}=0.9999952^{2^7} \times 0.9999952^{2^8} \times 0.9999952^{2^{12}} \times 0.9999952^{2^{14}} \times 0.9999952^{2^{16}} $$

Now, $x^{2^y}$ can be calculated by taking $x$, squaring it, then squaring the result, then again squaring the result, etc. until the total of $y$ squarings are done: exponentiation by squaring. Any decent calculator should be able to do it quite easily without losing too much precision (typing in a number, then pressing $\times$ followed by $=$ $y$ times usually does the trick).

So,

\begin{align} 0.9999952^{2^7}&=&((((((0.9999952^2)^2)^2)^2)^2)^2)^2&=0.99938578723137220775212944322376\\ 0.9999952^{2^8}&=&(0.9999952^{2^7})^2&=0.9987719517200695609221118676042\\ 0.9999952^{2^{12}}&=&(((0.9999952^{2^8})^2)^2)^2)^2&=0.98053116682488583016015535720841\\ 0.9999952^{2^{14}}&=&((0.9999952^{2^{12}})^2)^2&=0.92436950624567200131913471410336\\ 0.9999952^{2^{16}}&=&((0.9999952^{2^{14}})^2)^2&=0.73010015546967242085058682162284 \end{align}

And, finally, find the product of the 5 numbers above, which is:

$$ 0.66052454443033066313263272049394 $$

which is the chance of not dying, so, the chance of dying is:

$$ 1-0.66052454443033066313263272049394=0.33947545556966933686736727950606 $$

(used windows calculator in the process)

Answer 7

The best way to compute this kind of quantities on a computer is using the functions expm1(x) and log1p(y), which compute, respectively, $e^x-1$ and $\ln(1+y)$, and are more accurate than the naive formulas for tiny values of their argument. They are part of the IEEE floating point arithmetic standard and are provided in the standard libraries of most programming languages.

Rewrite your probability as $$1-(1-p)^n = -(e^{n \ln (1-p)}-1) = -\operatorname{expm1}(\operatorname{log1p}(-p)*n).$$

So, for instance, in Python you'd use the following

In [1]: from numpy import expm1, log1p
In [2]: -expm1(log1p(-4.8e-6)*86400)
Out[2]: 0.33947545556966929

In this case the number 0.339 is rather large, so the last subtraction is tame and some of these safeguards are not needed, but for better accuracy for all values of $p$ and $n$ you should use these library functions.

Answer 8

Chance of remaining alive for n seconds is $(1-p)^n$.
$\log (1-p)^n = n \log (1-p)$

The Maclaurin series for $\log(1 − x)$ is $\log(1-x) = -x-{\tfrac {1}{2}}x^{2}-{\tfrac {1}{3}}x^{3}-{\tfrac {1}{4}}x^{4}-\cdots \!$

which yields the approximation $\space \log(1-x) \sim -x$ for $0<x<<1$

Hence $(1−p)^n \sim e^{-np}$

The approximation and numeric result for staying alive: 0.6605, is the same as the answer above given by @dezdichado . However it should be noted that the @dezdichado answer derives from the Poisson approximation of the Binomial, in the case where n is large while p and k are small: Poisson Approximations. In our case, the number of deaths $k$ is $0$. When $k=0$ the binomial simplifies exactly to $(1-p)^n$, and the only part of Poisson approximation remaining is due to the truncation of the Maclaurin series.

Answer 9

For those who prefer a more programmatic syntax, using the calc arbitrary precision command-line calculator:

calc '100*(1-(1-.0000048)^86400)'

Output (percentage odds of dying in a single day):

    ~33.94754555696693368674

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For a longer precision, prepend a config("display", some_precision_value); to the calc code. Here's the result up to 1,000,000 decimal places, (about ten seconds to run on an Intel Core i3):

calc 'config("display", 1000000)
      100*(1-(1-.0000048)^86400)' | fold | less

The complete answer is 604,800 digits long, (plus one more char for the leading ~), the last five digits being ...06624. (To count the the number of digits, replace fold | less above with tail -n +2 | tr -d '[:space:]' | wc -c.)