3 ms·

The probability calculation is incorrect, you can't just multiply the numbers. This should be obvious from the fact that if we wait long enough the probability

by lutorm 17y ago

The probability calculation is incorrect, you can't just multiply the numbers. This should be obvious from the fact that if we wait long enough the probability according to that expression will exceed 1. If I have a probability of throwing a 6 on a dice of 1/6, throwing 6 dice doesn't give me a probability of 1 of getting (at least) a six. The correct calculation is: If the probability of one meteor hitting an airplane is 5.7e-13, then the probability of it missing is 1-5.7e-13. For no plane to be hit, all of the 750e6 hours * 125 meteors/hour must miss. That probability is (1-5.7e-13)^(750e6*125)=0.948, ie the probability of at least one hit is 5.2%. The answer is close to the same, but that's just luck. If the probability was higher, it would be increasingly off.

tomsaffell 17y ago
Thanks for pointing out the error in his math. But I think you got something wrong too - thrown off big error on his part (~4000x). The correct calculation is: p(one or meteor strikes) = 1 - ( (1 - p(strike of any meteor))^ (total # meteors) ) (assuming statistical independence, etc). p(strike of any meteor) = 5.82e-13 (total # meteors) = 3000 * 365 * 20 (for testing over 20 years) p(one or meteor strikes) = 1.3e-5 = 0.0013% The error of multiplying 720e6 hrs (the total # of flight hours) by 125 meteors/hr is that it does not give you the total number of meteors in the time period (which is what you need), it gives you a number ~4000 times larger (because at any one time, there are ~4000 planes in the sky...)
- lutorm 17y ago
  Except that if there are 4000 planes in the sky, that does make (assuming the planes are not overlapping, which seems largely safe) the probability 4000 times larger than for one plane. As long as the planes don't overlap, the flight hours is the correct number to use, because more planes in the sky gives higher probability. It's not the number of meteors that matter, but the number of "potential meteor strikes". Equivalently, you can say that the probability of a meteor hitting any one plane at a given time is the fractional area taken up by planes, which if there are 4000 planes in the air is 4000*5.82e-13. (Again assuming the planes don't overlap, which since 5.82e-13<<1 is a safe assumption.)
  - tomsaffell 17y ago
    Ah, yes - I mis calculated p(strike of any meteor) by a factor of 3500 (the # of planes in the sky, according to the article). I think this should be applied to the probability, not the count of the occasions - but the numbers are small enough that it doesn't make much difference. p(one or meteor strikes) = 1 - ( (1 - p(strike of any meteor))^ (total # meteors) ) (assuming statistical independence, etc). p(strike of any meteor) = 5.82e-13 * 3500 = 2.04E-9 (total # meteors) = 3000 * 365 * 20 (for testing over 20 years) p(one or meteor strikes) = 0.0436 = 4.36%
    - lutorm 17y ago
      Yeah, it doesn't make a difference until the planes cover a significant area of the Earth, in which case just multiplying by the numbers of planes will be wrong for the same case that the initial calculation was wrong. Then it would be 1-(1-5.82e-13)^3500 in this case, too.
- lutorm 17y ago
  Just to play this game a little further, the probability is probably (!) smaller, because those meteors are not distributed uniformly across the Earth. Due to the direction of motion of the Earth through space and its sense of rotation, most meteors hit the area that is facing forward in space. That's the area on Earth where it's between midnight and noon, with the maximally forward one at 6am, ie sunrise. Since I suspect, due to the way civilization works, there are more planes flying between noon and midnight than between midnight and noon, and the ones that are flying during that time are mostly intercontinental flights that go preferentially across the pole, there are probably less planes than average where those meteors hit. But now I'm just speculating. ;-)
gus_massa 17y ago
It is not luck. The probability of the article is almost correct, because the multiplication of the numbers is a first order approximation to the exact result. Using Taylor series, if A<1 (1-A)^B = 1 – B A + B(B-1)/2 A^2 - B(B-1)(B-2)/6 A^3 + B(B-1)(B-2)/24 A^4 - ... so 1-(1-A)^B ~= B A - B(B-1)/2 A^2 + ... And if A is small enough (A<<1) 1-(1-A)^B ~= B A - B(B-1)/2 A^2 So, AB is a good approximation and if B is big (B>>1) the second order correction is -B(B-1)/2 A^2 ~= -(AB)^2 /2 Then, with A=5.7e-13 and B=750e6*125=9.4E10, then Exact: 1-(1-A)^B = 0.0520 Approximation: AB = 0.0534 Second Order: -(AB)^2 /2 = -0.0014
- lutorm 17y ago
  Yeah, I meant luck in the sense that, if the author consciously used the first-order expansion over the true expression, the article doesn't state that only going to first order was reasonable. Hence one has to assume no such check was made. Just stating the calculation like he did might induce others to think that the expression applies regardless of the magnitude of the result. And then it will be pure luck whether it works or not.

by lutorm 17y ago

tomsaffell 17y ago

Thanks for pointing out the error in his math. But I think you got something wrong too - thrown off big error on his part (~4000x). The correct calculation is: p(one or meteor strikes) = 1 - ( (1 - p(strike of any meteor))^ (total # meteors) ) (assuming statistical independence, etc). p(strike of any meteor) = 5.82e-13 (total # meteors) = 3000 * 365 * 20 (for testing over 20 years) p(one or meteor strikes) = 1.3e-5 = 0.0013% The error of multiplying 720e6 hrs (the total # of flight hours) by 125 meteors/hr is that it does not give you the total number of meteors in the time period (which is what you need), it gives you a number ~4000 times larger (because at any one time, there are ~4000 planes in the sky...)

lutorm 17y ago
Except that if there are 4000 planes in the sky, that does make (assuming the planes are not overlapping, which seems largely safe) the probability 4000 times larger than for one plane. As long as the planes don't overlap, the flight hours is the correct number to use, because more planes in the sky gives higher probability. It's not the number of meteors that matter, but the number of "potential meteor strikes". Equivalently, you can say that the probability of a meteor hitting any one plane at a given time is the fractional area taken up by planes, which if there are 4000 planes in the air is 4000*5.82e-13. (Again assuming the planes don't overlap, which since 5.82e-13<<1 is a safe assumption.)
- tomsaffell 17y ago
  Ah, yes - I mis calculated p(strike of any meteor) by a factor of 3500 (the # of planes in the sky, according to the article). I think this should be applied to the probability, not the count of the occasions - but the numbers are small enough that it doesn't make much difference. p(one or meteor strikes) = 1 - ( (1 - p(strike of any meteor))^ (total # meteors) ) (assuming statistical independence, etc). p(strike of any meteor) = 5.82e-13 * 3500 = 2.04E-9 (total # meteors) = 3000 * 365 * 20 (for testing over 20 years) p(one or meteor strikes) = 0.0436 = 4.36%
  - lutorm 17y ago
    Yeah, it doesn't make a difference until the planes cover a significant area of the Earth, in which case just multiplying by the numbers of planes will be wrong for the same case that the initial calculation was wrong. Then it would be 1-(1-5.82e-13)^3500 in this case, too.
lutorm 17y ago
Just to play this game a little further, the probability is probably (!) smaller, because those meteors are not distributed uniformly across the Earth. Due to the direction of motion of the Earth through space and its sense of rotation, most meteors hit the area that is facing forward in space. That's the area on Earth where it's between midnight and noon, with the maximally forward one at 6am, ie sunrise. Since I suspect, due to the way civilization works, there are more planes flying between noon and midnight than between midnight and noon, and the ones that are flying during that time are mostly intercontinental flights that go preferentially across the pole, there are probably less planes than average where those meteors hit. But now I'm just speculating. ;-)

gus_massa 17y ago

It is not luck. The probability of the article is almost correct, because the multiplication of the numbers is a first order approximation to the exact result. Using Taylor series, if A<1 (1-A)^B = 1 – B A + B(B-1)/2 A^2 - B(B-1)(B-2)/6 A^3 + B(B-1)(B-2)/24 A^4 - ... so 1-(1-A)^B ~= B A - B(B-1)/2 A^2 + ... And if A is small enough (A<<1) 1-(1-A)^B ~= B A - B(B-1)/2 A^2 So, AB is a good approximation and if B is big (B>>1) the second order correction is -B(B-1)/2 A^2 ~= -(AB)^2 /2 Then, with A=5.7e-13 and B=750e6*125=9.4E10, then Exact: 1-(1-A)^B = 0.0520 Approximation: AB = 0.0534 Second Order: -(AB)^2 /2 = -0.0014

lutorm 17y ago
Yeah, I meant luck in the sense that, if the author consciously used the first-order expansion over the true expression, the article doesn't state that only going to first order was reasonable. Hence one has to assume no such check was made. Just stating the calculation like he did might induce others to think that the expression applies regardless of the magnitude of the result. And then it will be pure luck whether it works or not.