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So, I curious and trying to determine what sort life expectancy a human being would have if they were immortal (as in, no more senescence (aging)). Accidental deaths only. I've googled around and found numbers from a few hundred years to nearly 9000 years.

Europe has some great statistics on accidental deaths - https://ec.europa.eu/eurostat/statistics-explained/index.php/Accidents_and_injuries_statistics#Deaths_from_accidents.2C_injuries_and_assault

The headline is that 3.1% of deaths were accidents in 2015.

However, it occurs to me that the best representative sample to use is people aged 15-25 who generally don't really die from illnesses and are in better health so less susceptible to things like falls (and more to traffic accidents). There's a graph on that page that shows accidental deaths are about 35% of all deaths for those age groups.

Unfortunately I lack the mathematical chops to be able to take those numbers and merge them together meaningfully. So.... given the information above (and in the link if I've missed anything useful out), what's the approximate average European life expectancy if we "solved" senescence tomorrow?

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    $\begingroup$ Welcome to stackexchange. This is an interesting question, but more of a small research project than a mathematics problem you need help with. It's likely to be closed as too broad, You would have a better chance at getting help if you tried the problem and showed us the attempt. Finally, perhaps better asked at stats.stackexchange.com. $\endgroup$
    – Ethan Bolker
    Dec 17, 2018 at 17:19
  • $\begingroup$ The rate of accidental death is not constant in time, mostly because technology gives us both the medical treatments to survive accidents and exciting new ways to cause them. $\endgroup$
    – eyeballfrog
    Dec 17, 2018 at 17:32
  • $\begingroup$ @EthanBolker - Thanks but I wouldn't know where to begin on the maths, so if you want my attempt - the chances of death any given year are 3%, so... nope, really have no idea how to approach this. $\endgroup$
    – linux_confusion
    Dec 17, 2018 at 17:35
  • $\begingroup$ @eyeballfrog - Yep, I'm aware of those things, it's more of a thought experiment. $\endgroup$
    – linux_confusion
    Dec 17, 2018 at 17:36
  • $\begingroup$ Also older people tend to be more careful to avoid accidents (although if they still had the bodies of 20-year-olds, they might not be). Then again, someone with the prospect of living for millenia (and who wanted to do so) would be extremely careful... $\endgroup$
    – Robert Israel
    Dec 17, 2018 at 18:29

2 Answers 2

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Suppose we only consider people aged 15-25, as you suggested. Let $d$ be the annual death rate for this age group, i.e. the percentage of people that die every year. Let $a$ be the percentage of said deaths that are accidental. Then $p = d \cdot a$ is the likelihood that someone aged 15-25 will die by accident in a given year.

We wonder about the number of years someone could survive without dying, if the only possible causes of death are by accident. We can model this scenario by a geometric distribution with parameter $p$. That is, if $p$ is the likelihood of death (by accident), then the number of years that pass before a death occurs follows a geometric distribution. But we know that the expectation of a geometric distribution is $1/p$, so the expected number of years lived is $1/p$.

To recap: if all illness was cured and the only cause of death was by accident, then the life expectancy would be $1/p = 1/(d \cdot a)$. You mention that the parameter $a$ should be $a = 0.35$. This source suggests that the mortality rate for people aged 15-25 is rougly $d = 0.001$. Therefore the life expectancy would be:

$$ \frac{1}{0.001 \cdot 0.35} \approx 2850 $$

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I wrote a blog post on the topic where I model this question with an exponential distribution. The answer I get (under some reasonable assumptions) is about 2,200 years.

You can find the post (with full R code so that you can modify the assumptions) here:
https://blog.ephorie.de/how-long-would-you-live-if-you-were-immortal

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