# How would you figure the odds of nobody on the planet dying on a given day?

What would the odds be of having a day on this planet with no one dying?
I know it’s a fairly easy task of figuring out the odds of a single person dying on a given day by using the data we already have, but how would you figure out the opposite?

Well, what exactly do you mean by “the odds of nobody on the planet dying on a given day”?

For example, I wager that in the entire recorded history of humanity (and well beyond), there has never been a 24 hour period in which nobody on the planet died. Does that mean the answer is 0? I’m guessing not, but this is perhaps illustrative of the ambiguity in the question (ambiguity which may ultimately resolve in the realization that there is no particular coherent concept being referred to).

For the sake of argument, let’s say the global population is 1 million people and five people die every day on average.

The odds any individual surviving is 999995/1000000.

The odds of any two individuals surviving is (999995/1000000)^2

The odds of everyone surviving is (999995/1000000)^1000000

In this example, that would be about 0.6% of a chance or one day in two hundred would be death free.

According to Wikipedia, the gross mortality rate is currently 8.24 people per thousand per year. There are approximately 6.7 billion people, so an expected mortality rate of about 55 million per year, or about 151 thousand per day. Modeling this as a Poisson process, we can say that the chance of no one dying in a given day is approximately 151,000[sup]0[/sup] * e[sup]-151,000[/sup]/0!, which is about 10[sup]-65,600[/sup]. That’s a pretty small chance.

You are assuming a probability distribution such that, on any given day, everyone in the world has independent and equiprobable chances of dying. Why use such a distribution?

(I’m also not sure why you make the global population 1 million with 5 deaths a day on average, but ok…)

Because any other distribution would require a great deal more assumptions, most likely. If we assume that deaths are clumped, for instance (which is undoubtedly true of the real world; see for instance Haiti), then you have to estimate or find some sort of parameter characterizing how clumped the deaths are, which would require a lot more research.

But you can’t just ignore these factors, and claim to have an answer to the question. A large factor you ignore is that old people have higher probability of dying the next year. Come to think of it, I’m with Indistinguishable. The question needs to be stated differently, and possibly isn’t referring to a coherent concept.

Good grief.

Somebody gives a decent simple answer with understandable assumptions that somebody with basic math and reasoning skills could understand and the nitpickers jump out of the wood work. But its not even nitpickers with the “real” answer, because you know, that would probably take way too much work and much typing to give.

I for one await the correct answer now.

Regarding the assumptions:

Clumpiness just means that the Poisson mean is not “number of deaths” but instead “number of death events”. Since any death event with more than a dozen people killed would usually make the news, we have a rough sense of the rate of those, and it is small. (That is, the probability of a significant clump-death on any given day is small enough to ignore, as it would affect the answer on the order of (1-p[sub]event[/sub]), with p[sub]event[/sub] being small, maybe a few percent.)

For smaller clumps: I reckon (without cite) that the deaths are primarily independent anyway (with the old and infirm contributing the most deaths), and that the mean deaths per death-event is something much smaller than 2 and likely very close to 1.

For the distribution of death rates: imagine that only the oldest 20% of the population ever died. That is, if you are among the youngest 80%, you cannot die. That means that the pool of possible dying people goes from 6 billion to a little over 1 billion. The number of deaths in a day is still 0.00015 billion, so the per-person-per-day probability-of-death goes up to 0.00015, which is still much less than unity, so the Poisson assumption is still excellent.

So, letting [symbol]m[/symbol] = 151,000 = number of deaths in a day, and letting 1+b = mean number of people in each death event, with b very small (I figure), then the answer is pretty much what was given before:

probability of no death = e[sup]-m/(1+b)[/sup] ~= 1/(10[sup]10[sup]4.8[/sup][/sup])

or so.