7 ms·

How odd is a cluster of plane crashes?

33 points by yitchelle 11y ago

simonh 11y ago
One of the first computer programs I ever wrote just put random dots on a screen and I remember looking at how 'lumpy' the results were. Intuitively most people assume that random distributions will be even distributions, but in fact they are usually full of clusters. Some mathematicians can reliably distinguish 'random' sequences created by people from truly random ones because people produce distributions that are too smoothed out. 'I won't put a dot here because it's too close to that other dot', but a truly random distribution doesn't care where previous dots were and won't avoid the conjunction.
- dmichulke 11y ago
  I don't disagree at all and I remember seeing studies or articles writing about that phenomenon. Just wanted to add that a few years ago (when you wrote your first computer program) a randomly selected random number generator was more likely to be less random than a randomly selected random number generator today.
  - yitchelle 11y ago
    The method described in the article on how the statistician approached the problem is very similar to the approach used in the birthday problem[1]. [1] - https://en.wikipedia.org/wiki/Birthday_problem https://en.wikipedia.org/wiki/Birthday_problem
  - DanBC 11y ago
    Early random number generation was terrible. You wouldn't get clumps / no clumps, you'd get fuzzy straight lines.
  - simonh 11y ago
    Approximately 35 years ago :)
- lordnacho 11y ago
  The iPod shuffle algo was an example of this. It just didn't feel "random" enough for people to have a truly random generator.
  - ececconi 11y ago
    I would think about this all the time when I got a shuffle in high school. After listening to it for a while, I could really predict the next track that would come on even in shuffle mode.
  - wodenokoto 11y ago
    Isn't that also why it is shuffle and not random? When shuffling, you don't risk getting the same song twice in a row.
    - eru 11y ago
      People are seldom precise when they talk about randomness. For example, you could mean: picking tracks with `putting back'. You could shuffle tracks like you shuffle a deck of cards, ie picking without putting back. It's still random, but with a different distribution. Even picking the first track with 90% probability, and the rest with some other smaller probability is random. It's just not a uniform distribution.
- benihana 11y ago
  Great point. Here's a visualization of what you're talking about: http://bl.ocks.org/mbostock/fe3f75700e70416e37cd http://bl.ocks.org/mbostock/fe3f75700e70416e37cd >Uniform random is pretty terrible. There is both severe under- and oversampling: many samples are densely-packed, even overlapping, leading to large empty areas. (Uniform random sampling also represents the lower bound of quality for the best-candidate algorithm, as when the number of candidates per sample is set to one.) Taken from: http://bost.ocks.org/mike/algorithms/ http://bost.ocks.org/mike/algorithms/
- martin-adams 11y ago
  I do wonder if that is down to sample size. For example, I wrote a program to simulate the roll of a dice. There is a one in 6 chance of getting any one number. But roll it 6 million times, you'll be very close to have each number 1 million times. So I would anticipate the clumping to even out as the sample size increases. Or maybe the evening out is a large collection of clumps. Anyone know if this is the correct way to think about it?
  - mungoman2 11y ago
    Well depends of what you mean by very close. It's the relative numbers that approaches the true distribution, not the absolute numbers. Do the experiment you decribe and look at the numbers.
    - martin-adams 11y ago
      This is what I mean: var results = [0,0,0,0,0,0]; for (var i = 0; i < 6000000; i += 1) { results[Math.floor((Math.random() * 6))] += 1; } console.log(results); The results are: [1002140, 999355, 1000009, 1000401, 1000014, 998081] Basically, they are all within 1% tolerance of each number getting 1 million occurrences. So when you look at the clumping on random dots on a screen, given a large enough sample size, I would anticipate that the probability of keeping a clumped visual distribution would become quite improbable.
      - ijager 11y ago
        I think the clumping is more about how many consecutive same numbers there are in the array. For example, a naive me would generate: [3,4,5,1,2,4,2,1,2,6] whereas a true random distribution might generate: [3,4,4,4,4,1,2,2,2]. When you generate 6 million samples you would indeed have about 1 million per number. However you would still have subsequences of the same number.
      - gus_massa 11y ago
        The problem is not when you have much more samples than slots, in that case you should get "almost" the same number of hits in each slot as you measured. The problem is when you get less samples than slots. For example if you roll 3 dices, you should get "1/2" hits for each result. So the question is: How probable is that one result is repeated? (i.e. you get two equal dices out of the three) [spoiler alert] All equal: 3% Two equal and one different: 42% Three different: 55% "Intuitively", it's expected that with a uniform distribution you get something like 1% + 9% + 90% instead (this numbers are completely made up). Most people don't expect that the probability of getting a repeated dice is so close to the probability of getting all of them different.
- leni536 11y ago
  An other great example is how random is a terrible algorithm for dithering. https://en.wikipedia.org/wiki/Dither#Algorithms https://en.wikipedia.org/wiki/Dither#Algorithms
- detrino 11y ago
  Perhaps just an artifact of a weak PRNG or not truly uniform distribution?
  - Sharlin 11y ago
    No. A truly uniform random distribution indeed has "clumps".
    - detrino 11y ago
      And a poor PRNG or poor implementation of a uniform distribution (both extremely common, for example: rand() % max) has more clumps, my question is still valid. Edit: I wrote a program[1] to fill random pixels based on rand/modulo and mersenne twister/uniform distribution and I could not tell the difference in the generates images. [1] https://gist.github.com/anonymous/9c726ea636f06900ba3a https://gist.github.com/anonymous/9c726ea636f06900ba3a
smoyer 11y ago
Pleases mark as 2014 ... I also believe this is a repeat but can't seem to find the original.
- smoyer 11y ago
  Here's the previous discussion: https://news.ycombinator.com/item?id=8085254 https://news.ycombinator.com/item?id=8085254 I had tried the Algolia search box from my tablet and got no results. Using Google (constrained to HN) brought up today's discussion and the one from last year. site:news.ycombinator.com plane crash cluster
phantom784 11y ago
"If there is a crash on 1 August, the chance that the next crash occurs one day later on 2 August is 1/365. But the chance the next crash is on 3 August is (364/365) x (1/365), because the next crash occurs on 3 August only if there is no crash on 2 August." Is this not the gambler's fallacy? Shouldn't the odds be unaffected by a previous crash?
- darkxanthos 11y ago
  In this case I would agree with you. I don't understand the reasoning for the last sentence at all.
  - LeUsername 11y ago
    The focus is on being the next immediate crash. If you are asking whether there is a crash on 3 August, it is the unaffected odds of 1/365. But if you are asking whether the (first) next crash after 1 August is on 3 August, both the following must happen: a) there is no crash on 2 August b) there is a crash on 3 August The gambler fallacy would be "I have seen 9 days without crash, on the 10th day there is higher odds of crash". But you are not asking this, you are asking: "What is the chance that in the next 10 days, exactly first 9 days would be no-crash and on the 10th day there is a crash". EDIT: Also, another view: If you have seen 9 blacks in a row, there is an even chance 1/2 (roulette without zero) that the next will be black or red. Because in all sequences of length 10 starting with 9 blacks (this is given, you observed this), there is equal number of those with 9 consecutive blacks and a red and of those with 9 consecutive blacks and a black. This is gambler fallacy, believing that after repeated colour, you have higher chance of the other colour. Here, you are asking: how many of all sequences of length 10 have exactly 9 blacks and 1 red. And there are all kinds of sequences with 1 red in first 9 places, 2 reds, 3 reds, etc., so the odds of observing the 9 blacks in a row in the first place before asking about the 10th spin is lower, but it does not change the odds of the 10th spin itself.
    - darkxanthos 11y ago
      Super helpful! Thanks!

33 points by yitchelle 11y ago

One of the first computer programs I ever wrote just put random dots on a screen and I remember looking at how 'lumpy' the results were. Intuitively most people assume that random distributions will be even distributions, but in fact they are usually full of clusters. Some mathematicians can reliably distinguish 'random' sequences created by people from truly random ones because people produce distributions that are too smoothed out. 'I won't put a dot here because it's too close to that other dot', but a truly random distribution doesn't care where previous dots were and won't avoid the conjunction.

dmichulke 11y ago
I don't disagree at all and I remember seeing studies or articles writing about that phenomenon. Just wanted to add that a few years ago (when you wrote your first computer program) a randomly selected random number generator was more likely to be less random than a randomly selected random number generator today.
- yitchelle 11y ago
  The method described in the article on how the statistician approached the problem is very similar to the approach used in the birthday problem[1]. [1] - https://en.wikipedia.org/wiki/Birthday_problem https://en.wikipedia.org/wiki/Birthday_problem
- DanBC 11y ago
  Early random number generation was terrible. You wouldn't get clumps / no clumps, you'd get fuzzy straight lines.
- simonh 11y ago
  Approximately 35 years ago :)
lordnacho 11y ago
The iPod shuffle algo was an example of this. It just didn't feel "random" enough for people to have a truly random generator.
- ececconi 11y ago
  I would think about this all the time when I got a shuffle in high school. After listening to it for a while, I could really predict the next track that would come on even in shuffle mode.
- wodenokoto 11y ago
  Isn't that also why it is shuffle and not random? When shuffling, you don't risk getting the same song twice in a row.
  - eru 11y ago
    People are seldom precise when they talk about randomness. For example, you could mean: picking tracks with `putting back'. You could shuffle tracks like you shuffle a deck of cards, ie picking without putting back. It's still random, but with a different distribution. Even picking the first track with 90% probability, and the rest with some other smaller probability is random. It's just not a uniform distribution.
benihana 11y ago
Great point. Here's a visualization of what you're talking about: http://bl.ocks.org/mbostock/fe3f75700e70416e37cd http://bl.ocks.org/mbostock/fe3f75700e70416e37cd >Uniform random is pretty terrible. There is both severe under- and oversampling: many samples are densely-packed, even overlapping, leading to large empty areas. (Uniform random sampling also represents the lower bound of quality for the best-candidate algorithm, as when the number of candidates per sample is set to one.) Taken from: http://bost.ocks.org/mike/algorithms/ http://bost.ocks.org/mike/algorithms/
martin-adams 11y ago
I do wonder if that is down to sample size. For example, I wrote a program to simulate the roll of a dice. There is a one in 6 chance of getting any one number. But roll it 6 million times, you'll be very close to have each number 1 million times. So I would anticipate the clumping to even out as the sample size increases. Or maybe the evening out is a large collection of clumps. Anyone know if this is the correct way to think about it?
- mungoman2 11y ago
  Well depends of what you mean by very close. It's the relative numbers that approaches the true distribution, not the absolute numbers. Do the experiment you decribe and look at the numbers.
  - martin-adams 11y ago
    This is what I mean: var results = [0,0,0,0,0,0]; for (var i = 0; i < 6000000; i += 1) { results[Math.floor((Math.random() * 6))] += 1; } console.log(results); The results are: [1002140, 999355, 1000009, 1000401, 1000014, 998081] Basically, they are all within 1% tolerance of each number getting 1 million occurrences. So when you look at the clumping on random dots on a screen, given a large enough sample size, I would anticipate that the probability of keeping a clumped visual distribution would become quite improbable.
    - ijager 11y ago
      I think the clumping is more about how many consecutive same numbers there are in the array. For example, a naive me would generate: [3,4,5,1,2,4,2,1,2,6] whereas a true random distribution might generate: [3,4,4,4,4,1,2,2,2]. When you generate 6 million samples you would indeed have about 1 million per number. However you would still have subsequences of the same number.
    - gus_massa 11y ago
      The problem is not when you have much more samples than slots, in that case you should get "almost" the same number of hits in each slot as you measured. The problem is when you get less samples than slots. For example if you roll 3 dices, you should get "1/2" hits for each result. So the question is: How probable is that one result is repeated? (i.e. you get two equal dices out of the three) [spoiler alert] All equal: 3% Two equal and one different: 42% Three different: 55% "Intuitively", it's expected that with a uniform distribution you get something like 1% + 9% + 90% instead (this numbers are completely made up). Most people don't expect that the probability of getting a repeated dice is so close to the probability of getting all of them different.
leni536 11y ago
An other great example is how random is a terrible algorithm for dithering. https://en.wikipedia.org/wiki/Dither#Algorithms https://en.wikipedia.org/wiki/Dither#Algorithms
detrino 11y ago
Perhaps just an artifact of a weak PRNG or not truly uniform distribution?
- Sharlin 11y ago
  No. A truly uniform random distribution indeed has "clumps".
  - detrino 11y ago
    And a poor PRNG or poor implementation of a uniform distribution (both extremely common, for example: rand() % max) has more clumps, my question is still valid. Edit: I wrote a program[1] to fill random pixels based on rand/modulo and mersenne twister/uniform distribution and I could not tell the difference in the generates images. [1] https://gist.github.com/anonymous/9c726ea636f06900ba3a https://gist.github.com/anonymous/9c726ea636f06900ba3a

smoyer 11y ago

Pleases mark as 2014 ... I also believe this is a repeat but can't seem to find the original.

smoyer 11y ago
Here's the previous discussion: https://news.ycombinator.com/item?id=8085254 https://news.ycombinator.com/item?id=8085254 I had tried the Algolia search box from my tablet and got no results. Using Google (constrained to HN) brought up today's discussion and the one from last year. site:news.ycombinator.com plane crash cluster

phantom784 11y ago

"If there is a crash on 1 August, the chance that the next crash occurs one day later on 2 August is 1/365. But the chance the next crash is on 3 August is (364/365) x (1/365), because the next crash occurs on 3 August only if there is no crash on 2 August." Is this not the gambler's fallacy? Shouldn't the odds be unaffected by a previous crash?

darkxanthos 11y ago
In this case I would agree with you. I don't understand the reasoning for the last sentence at all.
- LeUsername 11y ago
  The focus is on being the next immediate crash. If you are asking whether there is a crash on 3 August, it is the unaffected odds of 1/365. But if you are asking whether the (first) next crash after 1 August is on 3 August, both the following must happen: a) there is no crash on 2 August b) there is a crash on 3 August The gambler fallacy would be "I have seen 9 days without crash, on the 10th day there is higher odds of crash". But you are not asking this, you are asking: "What is the chance that in the next 10 days, exactly first 9 days would be no-crash and on the 10th day there is a crash". EDIT: Also, another view: If you have seen 9 blacks in a row, there is an even chance 1/2 (roulette without zero) that the next will be black or red. Because in all sequences of length 10 starting with 9 blacks (this is given, you observed this), there is equal number of those with 9 consecutive blacks and a red and of those with 9 consecutive blacks and a black. This is gambler fallacy, believing that after repeated colour, you have higher chance of the other colour. Here, you are asking: how many of all sequences of length 10 have exactly 9 blacks and 1 red. And there are all kinds of sequences with 1 red in first 9 places, 2 reds, 3 reds, etc., so the odds of observing the 9 blacks in a row in the first place before asking about the 10th spin is lower, but it does not change the odds of the 10th spin itself.
  - darkxanthos 11y ago
    Super helpful! Thanks!