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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.?

Sklivvz
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Chris S
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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 Jun 30 '11 at 18:20
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    @AgentKC Age, too. – Carson Myers Jun 30 '11 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](http://bit.ly/iluO3O). 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 Jun 30 '11 at 19:18
  • @Chris S: Actuaries are able to accurately predict the rate of death of sufficiently large numbers of people because, while the odds of dying for any given person is unpredictable, the odds across a large number is astonishingly predictable. This is how life-insurance companies establish their reserves for payouts each year. So while any given person's likelihood of dying a given day is not measurable, one can determine the likelihood of a very large number of people dying on a given day with great accuracy (the larger the number, the greater the accuracy - see law of large numbers). – Brian M. Hunt Jun 30 '11 at 19:29
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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 Jul 01 '11 at 03:09
  • @ChrisW: With those numbers, [you are correct](http://bit.ly/kvyoRz). I suspect, however, that the annual death rate as reported by Wolfram Alpha is low. Also, in reality, the probability of dying in a given year is actually a function of one's age, so that probability of 1/120 isn't really a constant, which invalidates the calculation of the expected value for life expectancy. – ESultanik Jul 01 '11 at 12:18
  • However, the chance of dying is perhaps 1 : 25 000, not 25 000 : 1. Corrected the question, but kept the 250 000 as number. – user unknown Jul 01 '11 at 15:38
  • You may survive the other day wihtout winning, you may win and die, you may loose and die, or loose the money but survive. And you might not win the Jackpot, but make a smaller win with higher probability (but not higher than 250 000 : 1 I guess:) ). – user unknown Jul 01 '11 at 15:50
  • I think the odds are as simple as you do or you don't. If its your time to go its your time to go. There's too many variables to accuratly determine the odds. Technicaly I could sneeze and die, or I could get shot on the way home, or slip in the shower.....who knows. Its life –  Sep 20 '11 at 20:16
  • If you take the forum numbers, 151/6602224 (chopping off the last three digits from both numerator and denominator), the ratio s/b be about 1:26000, not 1:250000 (one extra zero, plus some rounding). A life of 26000 days would correspond to a life expectancy of 71 years. – Tom Au Sep 21 '11 at 14:54
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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 Apr 20 '12 at 22:10

3 Answers3

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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.

vartec
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    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 Jul 01 '11 at 01:28
  • @Odd: better now? – vartec Jul 01 '11 at 09:56
  • Sneaky solution, but yes! – Oddthinking Jul 01 '11 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 Jul 01 '11 at 15:46
  • @user: more specific you go, lesser the standard deviation is, thus confidence is higher. – vartec Jul 01 '11 at 16:09
  • Not necessarily. First you may look at the whole earth population. Later you focus on a single street. Maybe your specific user lives in a street with many old men, but is himself young? If you say, you only concentrate in a direction, where you focus more and more on relevant data, you would imply to know what is relevant beforehand. You wouldn't need to analyze a distribution, if you knew before, what the likelihood to die tomorrow is. – user unknown Jul 01 '11 at 20:22
  • @user: there is reliable data, as for example effect of age on probability of dying within a year (which you can see in the chart above), there are many more factors, that are highly reliable. It makes for example no sense at all to use global data, which includes 3rd world countries if you're interested in probabilities for developed countries etc. – vartec Jul 02 '11 at 14:04
  • If you say 'effekt of age', you already assume that the age is of importance here. It might be so, if there is no other information, which is much stronger correlated to the life expectation, like a desease, for example or high risk hobbies. Since the age was not given in the question, nor the nation of the asker, the question is perfectly valid for the whole earth population. Why do you start with a look at the age, and not at the sex? Why don't you look for the country, he lives in? This would be legitimate attemps, to refine the question towards the person who is asking, too. But ... – user unknown Jul 02 '11 at 18:12
  • ... But that's not what the question asked for. You could ask for smoking, for driving a car, driving drunk, for drinking, for illnesses in the family and genetic risks, body-mass-index, for living in a neighborhood with high rates of violence or suicides. The reason to pick the table of age is only the availability of these tables, isn't it? – user unknown Jul 02 '11 at 18:18
  • @user: my point exactly, there are so many variables, that you cannot say that for any given person chances are 1:250000. – vartec Jul 02 '11 at 22:35
  • It would be nice to draw some concrete numbers from that wonderful graph. A 20-to-40-year-old in the US has a per-day chance of roughly 1:250,000 (confirming the claim), while a 60-year-old is closer to 1:35,000. – Aleksandr Dubinsky Jan 13 '16 at 03:59
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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.

Oddthinking
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Mike Dunlavey
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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 Jun 30 '11 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 Jun 30 '11 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 Jul 01 '11 at 01:26
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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 Jul 01 '11 at 02:15
  • @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 Jul 01 '11 at 02:20
  • @Lagerbaer: I've heard it called less charitable things. On the other hand, by wife's uncle ran the RI lottery for 18 years. His take is nuanced. Yes there are compulsive gamblers. On the other hand, it put the RI numbers racket out of business. Go figure. – Mike Dunlavey Jul 01 '11 at 02:24
  • "I just know that from my business" - Ah, that's the root issue. You know that, but we don't, and we need a reliable reference to empirical data and statistical analysis to convince ourselves it is true. If you did that, you could blow-off the rest of my wittering about whether the curves match. – Oddthinking Jul 01 '11 at 02:29
  • @Oddthinking: Hey, we all witter sometimes. @vartec's curve does show it. The hazard at birth is about .01, and it drops by a couple orders of magnitude by age 5, after which it starts the roughly exponential climb, staying under the radar until about age 60, when it starts shooting up. – Mike Dunlavey Jul 01 '11 at 02:35
  • @Lagerbaer: Pretty much. See also [Popular support for increased inequality?](http://worthwhile.typepad.com/worthwhile_canadian_initi/2011/02/popular-support-for-increase-inequality.html) – Borror0 Jul 01 '11 at 03:10
  • @Mike: I understood the question as being very unspecific, for whom it is asked, so I would assume for the widest possible audience: worldwide, and every age - not only 33 years old Britains. – user unknown Jul 01 '11 at 16:01
  • @user: Right. Or even 67 year old Uh-mericuns :) – Mike Dunlavey Jul 01 '11 at 16:56
  • @Mike, you still haven't given any references for your claim that human mortality follows the definition of the bathtub curve that you have linked to. The Wikipedia page even says "While the bathtub curve is useful, not every product or system follows a bathtub curve hazard function" You've argued in comments that they are superficially similar, but I see no sign of a "constant failure rate" zone in Vartec's graph, and it is certainly doesn't follow the suggested model from Wikipedia. [This only counts as "wittering" once you've proven me wrong. :-) ] – Oddthinking Jul 03 '11 at 02:27
  • @Oddthinking: It's not a superficial similarity. Just take @vartec's curve. That's human, right? and it's roughly Gompertz-Markham, right? On a linear scale, the values at every 10 years are .01 .0002 .001 .001 .002 .0025 ... roughly. Plot it on graph paper with 1 square = .01 You'll see the shape. The description in the bathtub article isn't supposed to be an accurate picture, but to describe how it goes down, stays down, then goes up. Gompertz-Markham is one of the formulas considered to make that shape. – Mike Dunlavey Jul 03 '11 at 12:52
  • @Mike, we differed in our opinions of whether the term bathtub curve is broad enough to cover this, but the critical point here is our opinions don't count if we can cite the literature. I went ahead and did the search I was inviting you to do, and added a cite to support your position (Interestingly, from a paper that critiques Gompertz–Makeham.) According to Google Scholar, there's an even better quote to support your (well, now, our) position in [here](http://onlinelibrary.wiley.com/doi/10.1002/sim.4780070808/abstract) but behind a paywall, so I can't prove it. – Oddthinking Jul 03 '11 at 22:25
  • @Oddthinking: Thanks. I don't put quite as much faith in citations as some do. On detailed technical subjects (performance tuning of software is one example) the highest-rated published work is actually inadequate. Fortunately there's simple reason, math, and example you can fall back on. – Mike Dunlavey Jul 04 '11 at 02:50
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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.

Tom Au
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