# Quick probability question

I was wondering whether how safe air travel really, i.e. even for people who flew a huge amount. Hence, this question:

If the probability of any person (assume that it is the same for every person) dying on any flight due to a crash (assume all crashes lead to deaths, and that all “flights” have an equal probability of resulting in a crash/death) is 1/10,000,000, and the probability of crashing/dying on any given flight is independent of crashing/dying on any other flight:

What is the number of such flights that each of N persons, assuming that each of the N persons takes exactly the same number of flights, must have taken for the expected value of dead people among these N to have reached 0.5N? In other words, what is the expected number of flights such that the initial group of N people is cut in half due to the crashes/deaths?

I really have no idea how to proceed.

There is not really enough information to answer this question. Since multiple people die in one plane crash deaths are not independent even if crashes are. You said you are assuming independence, though so that has to mean that our N people never go on the same flight (or go on flights that might crash into each other. If the chance of dying on a flight is 1 in a million, then going on half a million flights will give you an expected number of deaths of 1/2. (Note this is not the same as saying you have a 50% chance of dying.) If each person goes on 1/2 million flights (remember none of our N ever fly together) will have an expected number of deaths of N/2.

But this is all really a bit silly because each person can’t go on the same number of flights. Some of them will die on early flights.

The “better” question you can ask though is if N people take a total of N/2 million flights, the expected number of deaths is N/2. And this doesn’t depend on whether they fly together or not or deaths are otherwise independent.

IMHO - I do not believe this is correct.

If you have a 1 in 10 chance of something happening during an event - and you want to know how many times the event has to occur to have a 50/50 chance of something happening - it is tempting to say the answer is 5, but the same logic can be applied to say that you have a 100% chance of the something happening if the event happens 10 times.

This is clearly not the case.

I know I have seen the math on this before. It is simple, but not so simple that I remember how to do it

If your chance of dying in one flight is one in a million, your chance of not dying is 0.999999.
For not dying in (up to) N flights, it is 0.999999 to the Nth power.

Your change of dying is 1-(0.999999 to the Nth power)

(9/10)[sup]n[/sup] = 0.5

Take logs of both sides and solve for n.

Old joke that worked better before the 9/11 incidents and subsequent security measures:

A professional statistician read an item in Statistics Monthly magazine that said the odds of a given airplane flight having a bomb on it was one in a thousand, whereas the odds of a flight having two bombs on it was one in a million. So every time he flew he carried a bomb…

Independence doesn’t matter because expectation is always additive. The expected number of deaths from k flights is k*N/10,000,000. If this is equal to N/2, then k must be 5,000,000.

If you want to google for the answer, look up “survivor function” in reliability mathematics. Take L to be your chance of failure (death) on any given flight. It will be the average number of passengers multiplied by the total fatality hull-loss rate you give. (One in ten million is actually about right for this).

A conservative approximation for independent events is e^(-Lt). In this case t is the number of flights. So solve 0.5 = e^(-0.00000001 t) to find your answer.

Note that it is an approximation, but for very small failure rates like this the approximation is far less than the error from all the assumptions you have to make.

This isn’t right. For sufficiently large k, this gives the expected number of deaths is greater than the original number of passengers.

The OP asked for an expected number of deaths equal to half the population not a 50% chance of dying. I tired to point out the difference in my original answer along with the notion that it can’t really work this way. Once you’re dead, you can’t fly again. (Well OK you can fly again, I guess, but you can’t die again in another play crash.

But if there is a constant probability p of dying in a plan crash, then the expected number of deaths is p*N where N is the number of passengers flying. This is true regardless of independence, etc. But it is very hard to work this into everyone “takes exactly the same number of flights” language.

This assumes people can die multiple times.

Ah, the Baldrick gambit.

The OP’s stated problem includes assumptions that are not consistent with reality, therefore any solution will include results that are inconsistent with reality. This frequently happens when you simplify statistics.

p[sub]crash[/sub] = 1/10[sup]7[/sup]
p[sub]no crash[/sub] = 1 - p[sub]crash[/sub] = 1 - 1/10[sup]7[/sup]
p[sub]death[/sub] = 0.5
p[sub]not dead yet[/sub] = 1 - p[sub]death[/sub] = 1 - 0.5 = 0.5
p[sub]no crash after N flights[/sub] = p[sub]no crash[/sub][sup]N[/sup]
p[sub]no crash after N flights[/sub] = p[sub]not dead yet[/sub]
(1 - 1/10[sup]7[/sup])[sup]N[/sup] = 0.5
log((1 - 1/10[sup]7[/sup])[sup]N[/sup]) = log(0.5)
log((1 - 1/10[sup]7[/sup])[sup]N[/sup]) = N * log(1 - 1/10[sup]7[/sup])
N * log(1 - 1/10[sup]7[/sup]) = log(0.5)
N = log(0.5)/log(1 - 1/10[sup]7[/sup])
N = 6931471.45902588014280599285

If p is the probability of an event, then the probability of getting 0 hits in k runs is given by
(1-p)^k

As a short cut, if p is very small and k is very large (1-p)^k is approximately equal to exp(-k*p)

in your case your case p=10^-7 and you want the value to equal 0.5 so

exp(-k/10,000,000) = 0.5, So
k = -ln(0.5)*10,000,000 = 6,931,472.2