9 ms·

To be fair, it appears they fixed the math later, or at least roughly. (2 out of every 1M) x (the US population) [they cite ~311M]. That gives 622. Not far off

To be fair, it appears they fixed the math later, or at least roughly. (2 out of every 1M) x (the US population) [they cite ~311M]. That gives 622. Not far off the 620 they report in your quote.

petrogradphilos 7y ago
This is like having a weighted coin that comes up heads with probability 2⋅10⁻⁶, flipping it 311 million times, and seeing 0 heads. That's astronomically unlikely. To see this, observe that the number of heads follows a binomial distribution with n = 311 million and p = 2⋅10⁻⁶. This can be well approximated¹ by a normal distribution with mean μ = np = 622 and standard deviation σ = Sqrt[np(1 - p)] = 25. 99.7% of the time², when you sample from this distribution, the sampled value will be within 3 standard deviations of the mean, i.e., between μ - 3σ = 547 and μ + 3σ = 697. Results further from the mean are more unlikely. For example, seeing a value more than 7 standard deviations from the mean (i.e., less than 447 or more than 797) is about a 1 in 2 trillion event³. Since 0 is about 25 standard deviations from the mean, the probability of seeing 0 heads is on the order of 10⁻¹³⁸. [1] https://math.stackexchange.com/questions/2021801/conditions-needed-to-approximate-a-binomial-distribution-using-a-normal-distribu https://math.stackexchange.com/questions/2021801/conditions-... [2] https://en.wikipedia.org/wiki/68–95–99.7_rule https://en.wikipedia.org/wiki/68–95–99.7_rule [3] https://www.johndcook.com/blog/table-of-normal-tail-probabilities/ https://www.johndcook.com/blog/table-of-normal-tail-probabil...
- paulmd 7y ago
  > This is like having a weighted coin Tangent: there is no such thing. You can weight a die, you cannot weight a coin. Intuitively this should make sense because even if you made one side of the coin from lead and the other from balsa wood, all you are doing is changing the center of gravity of the coin. The coin spins about its center of gravity, not the geometric center of the coin, so this makes no difference. https://www.stat.berkeley.edu/~nolan/Papers/dice.pdf https://www.stat.berkeley.edu/~nolan/Papers/dice.pdf
  - GuiA 7y ago
    I wonder if you could build a (admittedly thick) coin with some sensing/actuation going on in order to make it land on an arbitrary side.
  - Supermancho 7y ago
    A magnetic coin, might be the "trick coin" that was being posited.
  - klarstrup 7y ago
    There are examples in that paper of scenarios of effectively influenced coin tosses...
    - rtkwe 7y ago
      I feel like it's not super difficult if you're catching the coin, if you're able to position the coin in the same place and apply the same force consistently you'll largely get the same height and speed of rotation. When I was a bored kid I was able to get a pretty consistent coin toss where it would land on the opposite of the side at the beginning. [0] [0] Of course I didn't do any statistical analysis or a huge number of trials for this to really tell if I was able to do it but it felt pretty consistent.
  - kgwgk 7y ago
    The same is true for a die :-) It could make a difference for the coin if you “rolled” it instead of catching it in the air.
  - munk-a 7y ago
    Possibly, though if we want to be pedantic "weighting" has shifted from merely meaning "increased in weight" to meaning "biased through manipulation" - colloquially you can describe a situation as "weighted against someone" if you think that person may be unfairly disadvantaged - like in a multi-talent contest a continued draw of activities that favor one party using a hidden method that is suspected to be directed to the end of favoring that party. You absolutely can manufacture unfair coins either using a two surfaced approach (like a weak magnetic field acting on a magnetized coin) but even a one surfaced approach is possible if you make use of carvings on the surface of the coin to get a favorable result from air resistence - lastly you can even achieve it through density, if one side of a coin is significantly more dense then the other then it will tend to land face down - you can play with this a bit by trying to flip a weighted cylinder and observing the landing pattern - I might suggest taking a coin roll (like you get in a bank) and gluing some coins into one end of it - then try and flip it in the air so it lands coin-side up. It is, however, very hard to bias a coin significantly without skewing the dimensions or having clear alterations visible on the coin.
- throwawayhhakdl 7y ago
  Except we don’t have perfect information. Schizophrenia is misdiagnosed fairly frequently. I couldn’t find stats on undiagnosed schizophrenia, or schizophrenia diagnosed without visual hallucinations (which is probably the more relevant metric)
- drc500free 7y ago
  Another way to get there is that, given N shots at a 1/N event, you expect 0 hits 36% of the time (1/e). You can divide the population up into into 600 groups with 500000 people in them, and each of those has a 36% chance of not ever hitting. So seeing no cases is like flipping a 36% coin 600 times and hitting 600 times.
- TimonKnigge 7y ago
  Your 'approximation' is doing a lot of the work here.. Since you have a binomial distribution why not just use it directly? (1 - 2e-6)^(3e6) ≈ 0.002 So about 0.2%. Still highly unlikely but orders of magnitude more likely than what your normal distribution-detour gave.
  - petrogradphilos 7y ago
    > (1 - 2e-6)^(3e6) ≈ 0.002 Using the binomial directly is a good way to get the probability of 0 heads. Note, though, that the U.S. population is in the neighborhood of 300 million, not 3 million (as you seem to have used). (1 - 2⋅10⁻⁶)^(3⋅10⁸) ≈ 10⁻²⁶¹ https://www.wolframalpha.com/input/?i=%281+-+2+*+10%5E-6%29%5E%283*10%5E8%29 https://www.wolframalpha.com/input/?i=%281+-+2+*+10%5E-6%29%...
- petrogradphilos 7y ago
  > Since 0 is about 25 standard deviations from the mean, the probability of seeing 0 heads is on the order of 10⁻¹³⁸. Note: the above figure comes from the normal approximation to the binomial, which loses accuracy towards the tails. The exact probability of seeing 0 heads is (1 - p)^n = (1 - 2⋅10⁻⁶)^(311⋅10⁶) ≈ 10⁻²⁷⁰ [1]. [1] https://www.wolframalpha.com/input/?i=%281+-+2+*+10%5E-6%29%5E%28311*10%5E6%29 https://www.wolframalpha.com/input/?i=%281+-+2+*+10%5E-6%29%...
- deleted 7y ago
  [deleted]
- davrosthedalek 7y ago
  That's only true if you test all 311 mio for both. While I think that blindness is self-analyzing, that's not true for schizophrenia. So how many blind people were evaluated for schizophrenia? This might actually be a lower percentage than for normal-seeing persons. People are less experienced with the behavior of blind people, so it's harder for surrounding people (and probable even for the blind themselves) to recognize it and push people to go get diagnosed.

To be fair, it appears they fixed the math later, or at least roughly. (2 out of every 1M) x (the US population) [they cite ~311M]. That gives 622. Not far off the 620 they report in your quote.

petrogradphilos 7y ago

This is like having a weighted coin that comes up heads with probability 2⋅10⁻⁶, flipping it 311 million times, and seeing 0 heads. That's astronomically unlikely. To see this, observe that the number of heads follows a binomial distribution with n = 311 million and p = 2⋅10⁻⁶. This can be well approximated¹ by a normal distribution with mean μ = np = 622 and standard deviation σ = Sqrt[np(1 - p)] = 25. 99.7% of the time², when you sample from this distribution, the sampled value will be within 3 standard deviations of the mean, i.e., between μ - 3σ = 547 and μ + 3σ = 697. Results further from the mean are more unlikely. For example, seeing a value more than 7 standard deviations from the mean (i.e., less than 447 or more than 797) is about a 1 in 2 trillion event³. Since 0 is about 25 standard deviations from the mean, the probability of seeing 0 heads is on the order of 10⁻¹³⁸. [1] https://math.stackexchange.com/questions/2021801/conditions-needed-to-approximate-a-binomial-distribution-using-a-normal-distribu https://math.stackexchange.com/questions/2021801/conditions-... [2] https://en.wikipedia.org/wiki/68–95–99.7_rule https://en.wikipedia.org/wiki/68–95–99.7_rule [3] https://www.johndcook.com/blog/table-of-normal-tail-probabilities/ https://www.johndcook.com/blog/table-of-normal-tail-probabil...

paulmd 7y ago
> This is like having a weighted coin Tangent: there is no such thing. You can weight a die, you cannot weight a coin. Intuitively this should make sense because even if you made one side of the coin from lead and the other from balsa wood, all you are doing is changing the center of gravity of the coin. The coin spins about its center of gravity, not the geometric center of the coin, so this makes no difference. https://www.stat.berkeley.edu/~nolan/Papers/dice.pdf https://www.stat.berkeley.edu/~nolan/Papers/dice.pdf
- GuiA 7y ago
  I wonder if you could build a (admittedly thick) coin with some sensing/actuation going on in order to make it land on an arbitrary side.
- Supermancho 7y ago
  A magnetic coin, might be the "trick coin" that was being posited.
- klarstrup 7y ago
  There are examples in that paper of scenarios of effectively influenced coin tosses...
  - rtkwe 7y ago
    I feel like it's not super difficult if you're catching the coin, if you're able to position the coin in the same place and apply the same force consistently you'll largely get the same height and speed of rotation. When I was a bored kid I was able to get a pretty consistent coin toss where it would land on the opposite of the side at the beginning. [0] [0] Of course I didn't do any statistical analysis or a huge number of trials for this to really tell if I was able to do it but it felt pretty consistent.
- kgwgk 7y ago
  The same is true for a die :-) It could make a difference for the coin if you “rolled” it instead of catching it in the air.
- munk-a 7y ago
  Possibly, though if we want to be pedantic "weighting" has shifted from merely meaning "increased in weight" to meaning "biased through manipulation" - colloquially you can describe a situation as "weighted against someone" if you think that person may be unfairly disadvantaged - like in a multi-talent contest a continued draw of activities that favor one party using a hidden method that is suspected to be directed to the end of favoring that party. You absolutely can manufacture unfair coins either using a two surfaced approach (like a weak magnetic field acting on a magnetized coin) but even a one surfaced approach is possible if you make use of carvings on the surface of the coin to get a favorable result from air resistence - lastly you can even achieve it through density, if one side of a coin is significantly more dense then the other then it will tend to land face down - you can play with this a bit by trying to flip a weighted cylinder and observing the landing pattern - I might suggest taking a coin roll (like you get in a bank) and gluing some coins into one end of it - then try and flip it in the air so it lands coin-side up. It is, however, very hard to bias a coin significantly without skewing the dimensions or having clear alterations visible on the coin.
throwawayhhakdl 7y ago
Except we don’t have perfect information. Schizophrenia is misdiagnosed fairly frequently. I couldn’t find stats on undiagnosed schizophrenia, or schizophrenia diagnosed without visual hallucinations (which is probably the more relevant metric)
drc500free 7y ago
Another way to get there is that, given N shots at a 1/N event, you expect 0 hits 36% of the time (1/e). You can divide the population up into into 600 groups with 500000 people in them, and each of those has a 36% chance of not ever hitting. So seeing no cases is like flipping a 36% coin 600 times and hitting 600 times.
TimonKnigge 7y ago
Your 'approximation' is doing a lot of the work here.. Since you have a binomial distribution why not just use it directly? (1 - 2e-6)^(3e6) ≈ 0.002 So about 0.2%. Still highly unlikely but orders of magnitude more likely than what your normal distribution-detour gave.
- petrogradphilos 7y ago
  > (1 - 2e-6)^(3e6) ≈ 0.002 Using the binomial directly is a good way to get the probability of 0 heads. Note, though, that the U.S. population is in the neighborhood of 300 million, not 3 million (as you seem to have used). (1 - 2⋅10⁻⁶)^(3⋅10⁸) ≈ 10⁻²⁶¹ https://www.wolframalpha.com/input/?i=%281+-+2+*+10%5E-6%29%5E%283*10%5E8%29 https://www.wolframalpha.com/input/?i=%281+-+2+*+10%5E-6%29%...
petrogradphilos 7y ago
> Since 0 is about 25 standard deviations from the mean, the probability of seeing 0 heads is on the order of 10⁻¹³⁸. Note: the above figure comes from the normal approximation to the binomial, which loses accuracy towards the tails. The exact probability of seeing 0 heads is (1 - p)^n = (1 - 2⋅10⁻⁶)^(311⋅10⁶) ≈ 10⁻²⁷⁰ [1]. [1] https://www.wolframalpha.com/input/?i=%281+-+2+*+10%5E-6%29%5E%28311*10%5E6%29 https://www.wolframalpha.com/input/?i=%281+-+2+*+10%5E-6%29%...
deleted 7y ago
[deleted]
davrosthedalek 7y ago
That's only true if you test all 311 mio for both. While I think that blindness is self-analyzing, that's not true for schizophrenia. So how many blind people were evaluated for schizophrenia? This might actually be a lower percentage than for normal-seeing persons. People are less experienced with the behavior of blind people, so it's harder for surrounding people (and probable even for the blind themselves) to recognize it and push people to go get diagnosed.