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

Solution 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 $\lambda = np = 0.41472.$

Solution 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 $5$.

Solution 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

Solution 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.