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Assuming risk is evenly distributed between birth and death, you would expect a 1 in 80 chance of dying in any given year. But since the risk is loaded towards
by brilee 5y ago
Assuming risk is evenly distributed between birth and death, you would expect a 1 in 80 chance of dying in any given year. But since the risk is loaded towards the later years of life, 1 in 200 during the earlier years sounds about right. Perhaps it's 2x higher than I would have guessed otherwise, but it's the right order of magnitude.
- brianwawok 5y agoNo that is bad math around 1/80. If the average person had a 1/80 chance to die each year, the average life expectancy would be 40 years. Think about this another way. You have a gun with 80 chambers and 1 bullet. How many times on average can you point it at your head and pull the trigger before it goes off? Would you still argue 80 times? On average it is the last chamber?
- leni536 5y agoThat's not how it works. For a geometric distribution with p=1/80 the mean or expected value is indeed 80. https://en.wikipedia.org/wiki/Geometric_distribution https://en.wikipedia.org/wiki/Geometric_distribution Your gun example has a uniform distribution between 1 and 80 with an expected value of roughly 40 if you don't spin the chamber each time between pulling the trigger. If you spin it each time, then it's again the geometric distribution and the expected value is 80. If you don't spin the chamber between each time, then each time you pull the trigger the probability of dying at that round is not 1/80, the probability goes up and up at each round, it's only 1/80 on the first round.
- capitalsigma 5y ago"evenly distributed" implies uniform distribution
- notfbi 5y agoTo be fair, they said the risks are evenly distributed, not that deaths are.
- adverbly 5y agoI had the same initial thought, but this isn't quite a perfect model... that distribution averaging assumes that the only factor at play is covid... but there are other chances of death with highly irregular probability distributions as a function of age... with other nonlinear weightings as a function of age you could get pretty different numbers in the end...
- deleted 5y ago[deleted]
- adverbly 5y agoAre you sure? Is it not the same as this: https://math.stackexchange.com/questions/1009490/average-number-of-heads-in-a-row https://math.stackexchange.com/questions/1009490/average-num... E = p/(1-p) So pretty close to 80 years, no? I guess it gets more complicated because you'd die of other reasons as you age so there's no point including the eventualities where you reached 120 for example... I guess it gets pretty complicated in the end...
- i_cannot_hack 5y agoYour last example is not applicable. The question is comparable to a gun with infinitely many chambers, each with a 1/80 chance of containing a bullet: How many times on average can you point it at your head and pull the trigger? The differences are that you have no guarantee that the first 80 chambers will contain exactly 1 bullet, that more than one chamber can contain a bullet, and that you can pull the trigger more than 80 times.
- minitoar 5y agoI think motor vehicle related mortality really shifts the risk forward.