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I've read the odds of dying as around 1 : 250 000 each day, which presumably is based on the global death rate. I use this when people buy lottery tickets to point out they have more chance of dying than winning the money. This forum thread is one example how the odds are worked out, but there are also no doubt plenty of others.

Is this a good way to base it? Is it actually measurable at all? And does this chance increase when you drive a car, fly, cycle etc.?

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  • I would think the vastly different lifestyles and genetics of different people could make this hard to figure out. Interesting thought though.
    – TheEnigmaMachine
    Commented Jun 30, 2011 at 18:20
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    @AgentKC Age, too.
    – Carson Myers
    Commented Jun 30, 2011 at 18:51
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    Assuming all people are equally likely to die (an admittedly unreasonable assumption), then we can just divide the world annual death rate by the world population, which results in a probability of dying in a given year of 1 in 120. Further assuming that one is equally likely to die on any given day in a year, the probability of dying on a given day is roughly 1 in 44000. Those probabilities are likely lower bounds, given that there is a small minority of people that lead particularly dangerous lives and those statistics also include infant mortality.
    – ESultanik
    Commented Jun 30, 2011 at 19:18
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    @ESultanik Perhap I'm being fooled by statistics, but doesn't that imply an average life expectancy of 120 years per person? If we apply the "world average" to each individual then, given that the world population is expanding, can we each expect to be born more often than we die? The way I'd estimate it (estimate "the average chance per day of dying during an average lifetime") would be to use the average life expectancy: e.g. 69.2 years/person => 25,258 days/person ... almost exactly 10 times number given in the question title.
    – ChrisW
    Commented Jul 1, 2011 at 3:09
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    The chance of dying is 1:1. The chance of winning a lottery with many participants is closer to 1:10000000. That'll totally convince them.
    – Mateen Ulhaq
    Commented Apr 20, 2012 at 22:10

3 Answers 3

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This is based on numerous flawed assumptions. It should not be calculated in such a simplistic way, because:

  • chance of dying in given time is not uniformly distributed among people, there are great variety of factors (genetic, behavioral and environmental); For example there are numerous wars going on right now, greatly increasing chances of death in these zones.
  • chance of dying of one single person is not uniformly distributed in time, Gompertz–Makeham law of mortality applies;
  • total death rate takes in account infant death, children death etc. For calculating the life expectancy for person who has already lived X years, you should only take in account deaths of people X or older.

Graph of Gompertz-Makeham law [Image Source]

Note, that vertical scale in above graph is logarithmic.

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  • 5
    I agree that averaging out over every living person is a particularly naive way of modelling mortality, but that doesn't mean it "cannot be calculated".
    – Oddthinking ♦
    Commented Jul 1, 2011 at 1:28
  • @Odd: better now?
    – vartec
    Commented Jul 1, 2011 at 9:56
  • Sneaky solution, but yes!
    – Oddthinking ♦
    Commented Jul 1, 2011 at 10:13
  • The whole idea is wrong. If you add more and more attributes, which you think of, that influence your individual rest of life expectation, you get more and more accuracy for a smaller and smaller group of people, with a smaller and smaller certainity. In the end you have a subset of the earth population of size 1, and a statement, which is totally wrong, because it does not take into account a sickness, which you already have, but don't know of.
    – user unknown
    Commented Jul 1, 2011 at 15:46
  • @user: more specific you go, lesser the standard deviation is, thus confidence is higher.
    – vartec
    Commented Jul 1, 2011 at 16:09
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In Survival Analysis, hazard is defined as the instantaneous probability of the event occurring (such as death) per unit time.

If the hazard of death were constant over our lifetimes and were the same for everybody, and if the mean life expectancy were 80 years, say, that comes out to a hazard rate of 1/(80*365.25) = 1/29220 per day. That's about 10 times larger than 1/250000.

Of course, the hazard of death is not constant or the same for everybody. It starts out high at birth, then drops to near zero, then gradually rises, in what's called a "bathtub curve" [Ref], and depends on all kinds of risk factors. For someone in their 20s or 30s it might well be as small as 1/250000. Risky behavior, like drinking and driving, will have the opposite effect.

Also, see Actuarial Table.

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  • Nice answer. What the single "chances of dying" number does is to approximate a really complicated probability distribution function by only its mean value. If the original distribution is far from constant, as is the case here, this approximation is terrible.
    – Lagerbaer
    Commented Jun 30, 2011 at 21:06
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    @Lagerbaer: Thanks. Of course, for the purpose of the OP's original point, that the probability of winning the lottery is even smaller, it does the job. It saddens me how many people consider the lottery an investment.
    – Mike Dunlavey
    Commented Jun 30, 2011 at 21:19
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    Mike, you make a claim that mortality follows a bath-tub curve without reference (particularly notable because it contradicts @vartec's claim about the shape of Gompertz–Makeham's curve). Could you please fix?
    – Oddthinking ♦
    Commented Jul 1, 2011 at 1:26
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    The Wikipedia link does an excellent job of explaining what is meant by bathtub curve. It never claims that humans follow one. (You argument about the semi-log distorting it beyond recognition would be fair if the X axis had been the log axis. Looking at the different cumulative distribution functions (cdf) equations in their definitions shows they really are different.)
    – Oddthinking ♦
    Commented Jul 1, 2011 at 2:15
  • @Oddthinking: No, the link does not say that humans follow one. I just know that from my business, and @vartec's curve does show it in a small way, but it's there. I don't agree with you about X should be the log. If you make the Y axis linear, you'll see that after basically age 0, the hazard is an exponential curve with the "knee" about age 60, and it's very low before that.
    – Mike Dunlavey
    Commented Jul 1, 2011 at 2:20
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The math in the title of the question apears to be incorrect.

This chance should be 1 in 25,000, assuming a life expectancy of 68.5 years. That is, the "average" person with such an expectancy will live 25,000 days. It also assumes that a person will have an equal chance of dying on the first day or the 25000th.

But as early as the 19th century, Benjamin Gompertz worked out that mortality increases (exponentially) with age. That is, you are MUCH more likely to die on the 25000th day than on the first.

http://science-of-aging.healthaliciousness.com/timelines/gompertz-aging-human-mortality.php

So your ACTUAL chances of dying on a given day depends on how old you are. If your are 60, your chances of dying today are greater than 1 in 25,000. And if you are 20, less.

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  • many babies die on the first day, definitely more than people who die on the 25000th day: "More than a million babies die the same day they are born. In the U.S., 11,000 die on their birthdays" nbcnews.com/healthmain/…
    – DavePhD
    Commented Sep 30, 2023 at 2:28

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