5 ms·

I hate the units which you cannot easily add: 10 micromorts + 10 micromorts = 19,999999 micromorts 500 milimorts + 500 milimorts = 750 milimorts Playing russi

I hate the units which you cannot easily add: 10 micromorts + 10 micromorts = 19,999999 micromorts 500 milimorts + 500 milimorts = 750 milimorts Playing russian roulette using a sixshooter loaded with 3 rounds is 50% chance of death. It's 75% if you do another try. "Adding" another try does not translate to "adding" the metric. It just adds confusion.

anamexis 6y ago
While that is annoying, I would argue this isn't an issue with the units, but just a fact about how probability works. Setting aside micromorts, you can't say a 50% chance of death + a 50% chance of death = 100% chance of death, either.
- kubanczyk 6y ago
  Nah, it is a issue with units for me. Speaking primitively: % and ‰ look weirdly and add weirdly - they are easy to remember. But when some unit looks like a meter or gram (especially using the micro- prefix here), it should behave accordingly.
  - periodontal 6y ago
    Of course, even your example units don't always behave like you want. For velocities, 150 megameters/second + 150 megameters/second ≈ 240 megameters/second.
    - all-fakes 6y ago
      Sorry, can you explain why the velocities should add that way? I've always just seen velocities add like regular old vectors, so this looks very strange.
      - JadeNB 6y ago
        Relativistic effects.
      - Gladdyu 6y ago
        That'd be special relativity for you - nothing ever exceeds the speed of light, hence a corrective factor needs to be applied. https://en.wikipedia.org/wiki/Special_relativity https://en.wikipedia.org/wiki/Special_relativity
      - identity0 6y ago
        Relativity. If they could add that way, you would get 300 megameters/s, which is breaking the speed of light.
- ineedasername 6y ago
  I guess it depends on how you're adding those 50% together: add three bullets to a 6 chamber revolver for a game of Russian roulette and 3 bullets = 50% chance of death. If you take two tries, sure, that's not 100% chance of death. But 3 bullets = 50% chance, so if you add another another 3 bullets then those percentages combine quite well for 100% chance of death :) (assuming your aim is good & lethality on hit is 100%)
perfmode 6y ago
i understand this is possibly intended as hyperbole. yet hating a probability or a unit of measure is like hating a cloud or a tree or a ray of the sun. it just is. and this attachment this rejection of what is engenders a senseless suffering. one might respond, measures aren't real. but then, what's the use in hating what is unreal? on the other hand, the way we are drawn to elegance and beauty can be awakening and motivating in the quest of truth.
- teej 6y ago
  Humans choose measures. They don’t just exist out in the world. Inches vs Centimeters doesn’t exist in nature.
  - recursive 6y ago
    Lengths exist in the world, and regardless which length measure you choose, they can all be added the same way.
im3w1l 6y ago
I think a better way is to say that 500 millimort + 500 millimort = 1000 millimort like normallly. And then say that you can't add millimorts of successive events to get the millimorts of the sequence.
- spenczar5 6y ago
  That doesn't make sense to me: 500 millimorts means "50% chance of death." 1000 millimorts means "100% chance of death." Flipping a coin twice doesn't guarantee that you got heads.
  - keeganpoppen 6y ago
    his point is that the problem is not that you can never add millimorts together, but rather people try to add them together when the two things being added are correlated. i can add the millimorts of playing russian roulette once w/ the millimorts of playing 5 more rounds with the same gun just fine.
    - spenczar5 6y ago
      I think you have it backwards. The addition fails hardest when things are independent - which is true of most millimort risks in these tables. Smoking a cigarette and riding a motorcycle are totally uncorrelated in terms of how they cause death, so adding their millimorts naively won’t work at all.
kurthr 6y ago
I agree... it's like a % rise in the DJI. Why can't they just use dBuM where 60dBuM=Death?
barbegal 6y ago
In almost all cases adding is a valid approximation. Until you get into the 10s or 100s of milimorts range you can add them and the error will be less than 1%.
adwn 6y ago
Well, then you're gonna love electrical engineering: Adding a 10 Ohm resistor in parallel to another 10 Ohm resistor gives you a total resistance of 5 Ohm. Which is why you don't write 10 Ohm + 10 Ohm in this case, but 10 Ohm || 10 Ohm = 5 Ohm. This would solve your problem as well – just overload a different operator: 500 millimorts || 500 millimorts = 750 millimorts.
- jefftk 6y ago
  On the other hand, adding a 10 Ohm resistor in series with another 10 Ohm resistor gives you a total resistance of 20 Ohm, just as one would expect.
  - keeganpoppen 6y ago
    lol i don't know why this is being downvoted it just plainly wasn't a good example to use... there are two ways to combine resistors and they just arbitrarily took the one that doesn't add linearly, while the other one... does. and if you asked me (ie a layman) to "add two resistors together", i sure as hell wouldn't wire them in parallel lol.
    - adwn 6y ago
      > [...] it just plainly wasn't a good example to use... there are two ways to combine resistors and they just arbitrarily took the one that doesn't add linearly [...] Nothing arbitrary about it. But if you don't like putting resistors in parallel, go ahead and put them in a series, and then compute the electrical conductance of the pair: 10 S "+" 10 S = 5 S ("S" is Siemens) > and if you asked me (ie a layman) to "add two resistors together", i sure as hell wouldn't wire them in parallel lol Maybe that's the reason why people don't ask you to fix their wiring or to design their schematics for them? Parallel impedances are just as ubiquitous and important as serial ones.
cortesoft 6y ago
I think you are confusing a few things. First, yes, if you want to figure out your chance of dying from doing two things, you would have to do the math as you describe. You’re Russian roulette example isn’t a problem with the units. The chance of the bullet firing is not constant in each round, which is why the odds work as you describe. Each round has a different micromort number. The point of a micromort is to compare activities for risk, not to calculate the total chance of dying after some series of activities. That would be impossible anyway; there are infinite activities you go through, each with their own micromort level.
impendia 6y ago
To expand on barbegal's comment, in my opinion the genius of the "micromort" unit is the observation is that you can add them. In fact, 10 micromorts + 10 micromorts is 19.9999 micromorts (you have too many nines). But within the relevant standard of accuracy, 19.9999 = 20. The same is true for more complicated estimates. For example, 10 micromorts every day for a year is 3643.3 micromorts, which is practically 10 times 365. It's not completely accurate, but it's certainly good enough for the intended flavor of computation. Also, as you observed, the system breaks down with large probabilities. I think a good analogy here is Newtonian mechanics. If you drop a ball of a building, and want to know how long it will take to fall, then for all practical purposes you can ignore the effects of relativity. But if you were to throw the ball at half the speed of light, then you would need to be more careful. (In more ways than one...)
- bo1024 6y ago
  Here's a fun way to see this approximation: 1 + x ~= e^x when x is close to zero (either positive or negative). So let m be 10^-6 . Probability of survival of one event = 1 - m ~= e^-m . Probability of survival of two events = (1 - m)^2 ~= (e^-m)^2 = e^-2m ~= 1 - 2m.
  - klodolph 6y ago
    To add to this, this is why we have those funny functions log1p() and expm1() in the standard library. To add micromorts: double add_micromorts(double x, double y) { return -1e6 * expm1( -log1p(x * 1e-6) - log1p(y * 1e-6) ); } This is not the most accurate way of doing things, it is much simpler to return x + y - x * y * 1e-6; But the formula with expm1 and log1p lets you easily do things like multiply 10 micromorts/day by 10.5 days.
    - smabie 6y ago
      They're mostly there for converting from log returns and back, but yeah they're good for any kind of probabilities as well.
quicklime 6y ago
So in your particular example you can't add micromorts - you can only die once, so if you die in the first round, you don't get to play the second round. And if you could, the number of known-to-be-empty-chambers and/or bullets has changed, which changes the probability. But you can add micromorts in the case where n people are playing separate games of Russian Roulette. As another example, the Wikipedia page says that this is equivalent to 1 micromort: > Traveling 6000 miles (10,000 km) by jet (cancer due to increased background radiation) If you're just trying to calculate your own risk as a passenger, you can't just multiply 1 micromort by the number of 10,000 km trips you take. But if you're operating a flight, then you can multiply it by the number of passengers on this 10,000 km flight. It has more to do with the nature of the thing you're trying to model, than it does with the nature of the unit.
staplor 6y ago
This is pedantic, but going twice in Russian roulette would have an 80% mortality rate. This is because the second shot is a 3/5ths chance, not 1/2.
- deleted 6y ago
  [deleted]
- erik_seaberg 6y ago
  Don't they spin again before each round? That's the difference between (3/6) * (3/6) and (3/6) * (2/5) survival.
unholiness 6y ago
I'd argue adding is the right calculation in many situations, since in most cases you wouldn't receive the benefits of the next risk if you die the first time. So for example, if working in a coal mine is 10 micromorts per work day, and you choose to do it each day, it seems pretty reasonable to call that 10 * 5 days * 50 work weeks = 2500 micromorts per year. You're "paying" that cost each time you decide to go to work, and if you decide to work next year it won't somehow "cost" less micromorts – the possible yous who died last year won't be around to make the decision.

anamexis 6y ago

While that is annoying, I would argue this isn't an issue with the units, but just a fact about how probability works. Setting aside micromorts, you can't say a 50% chance of death + a 50% chance of death = 100% chance of death, either.

kubanczyk 6y ago
Nah, it is a issue with units for me. Speaking primitively: % and ‰ look weirdly and add weirdly - they are easy to remember. But when some unit looks like a meter or gram (especially using the micro- prefix here), it should behave accordingly.
- periodontal 6y ago
  Of course, even your example units don't always behave like you want. For velocities, 150 megameters/second + 150 megameters/second ≈ 240 megameters/second.
  - all-fakes 6y ago
    Sorry, can you explain why the velocities should add that way? I've always just seen velocities add like regular old vectors, so this looks very strange.
    - JadeNB 6y ago
      Relativistic effects.
    - Gladdyu 6y ago
      That'd be special relativity for you - nothing ever exceeds the speed of light, hence a corrective factor needs to be applied. https://en.wikipedia.org/wiki/Special_relativity https://en.wikipedia.org/wiki/Special_relativity
    - identity0 6y ago
      Relativity. If they could add that way, you would get 300 megameters/s, which is breaking the speed of light.
ineedasername 6y ago
I guess it depends on how you're adding those 50% together: add three bullets to a 6 chamber revolver for a game of Russian roulette and 3 bullets = 50% chance of death. If you take two tries, sure, that's not 100% chance of death. But 3 bullets = 50% chance, so if you add another another 3 bullets then those percentages combine quite well for 100% chance of death :) (assuming your aim is good & lethality on hit is 100%)

perfmode 6y ago

i understand this is possibly intended as hyperbole. yet hating a probability or a unit of measure is like hating a cloud or a tree or a ray of the sun. it just is. and this attachment this rejection of what is engenders a senseless suffering. one might respond, measures aren't real. but then, what's the use in hating what is unreal? on the other hand, the way we are drawn to elegance and beauty can be awakening and motivating in the quest of truth.

teej 6y ago
Humans choose measures. They don’t just exist out in the world. Inches vs Centimeters doesn’t exist in nature.
- recursive 6y ago
  Lengths exist in the world, and regardless which length measure you choose, they can all be added the same way.

im3w1l 6y ago

I think a better way is to say that 500 millimort + 500 millimort = 1000 millimort like normallly. And then say that you can't add millimorts of successive events to get the millimorts of the sequence.

spenczar5 6y ago
That doesn't make sense to me: 500 millimorts means "50% chance of death." 1000 millimorts means "100% chance of death." Flipping a coin twice doesn't guarantee that you got heads.
- keeganpoppen 6y ago
  his point is that the problem is not that you can never add millimorts together, but rather people try to add them together when the two things being added are correlated. i can add the millimorts of playing russian roulette once w/ the millimorts of playing 5 more rounds with the same gun just fine.
  - spenczar5 6y ago
    I think you have it backwards. The addition fails hardest when things are independent - which is true of most millimort risks in these tables. Smoking a cigarette and riding a motorcycle are totally uncorrelated in terms of how they cause death, so adding their millimorts naively won’t work at all.

kurthr 6y ago

I agree... it's like a % rise in the DJI. Why can't they just use dBuM where 60dBuM=Death?

barbegal 6y ago

In almost all cases adding is a valid approximation. Until you get into the 10s or 100s of milimorts range you can add them and the error will be less than 1%.

adwn 6y ago

Well, then you're gonna love electrical engineering: Adding a 10 Ohm resistor in parallel to another 10 Ohm resistor gives you a total resistance of 5 Ohm. Which is why you don't write 10 Ohm + 10 Ohm in this case, but 10 Ohm || 10 Ohm = 5 Ohm. This would solve your problem as well – just overload a different operator: 500 millimorts || 500 millimorts = 750 millimorts.

jefftk 6y ago
On the other hand, adding a 10 Ohm resistor in series with another 10 Ohm resistor gives you a total resistance of 20 Ohm, just as one would expect.
- keeganpoppen 6y ago
  lol i don't know why this is being downvoted it just plainly wasn't a good example to use... there are two ways to combine resistors and they just arbitrarily took the one that doesn't add linearly, while the other one... does. and if you asked me (ie a layman) to "add two resistors together", i sure as hell wouldn't wire them in parallel lol.
  - adwn 6y ago
    > [...] it just plainly wasn't a good example to use... there are two ways to combine resistors and they just arbitrarily took the one that doesn't add linearly [...] Nothing arbitrary about it. But if you don't like putting resistors in parallel, go ahead and put them in a series, and then compute the electrical conductance of the pair: 10 S "+" 10 S = 5 S ("S" is Siemens) > and if you asked me (ie a layman) to "add two resistors together", i sure as hell wouldn't wire them in parallel lol Maybe that's the reason why people don't ask you to fix their wiring or to design their schematics for them? Parallel impedances are just as ubiquitous and important as serial ones.

cortesoft 6y ago

I think you are confusing a few things. First, yes, if you want to figure out your chance of dying from doing two things, you would have to do the math as you describe. You’re Russian roulette example isn’t a problem with the units. The chance of the bullet firing is not constant in each round, which is why the odds work as you describe. Each round has a different micromort number. The point of a micromort is to compare activities for risk, not to calculate the total chance of dying after some series of activities. That would be impossible anyway; there are infinite activities you go through, each with their own micromort level.

impendia 6y ago

To expand on barbegal's comment, in my opinion the genius of the "micromort" unit is the observation is that you can add them. In fact, 10 micromorts + 10 micromorts is 19.9999 micromorts (you have too many nines). But within the relevant standard of accuracy, 19.9999 = 20. The same is true for more complicated estimates. For example, 10 micromorts every day for a year is 3643.3 micromorts, which is practically 10 times 365. It's not completely accurate, but it's certainly good enough for the intended flavor of computation. Also, as you observed, the system breaks down with large probabilities. I think a good analogy here is Newtonian mechanics. If you drop a ball of a building, and want to know how long it will take to fall, then for all practical purposes you can ignore the effects of relativity. But if you were to throw the ball at half the speed of light, then you would need to be more careful. (In more ways than one...)

bo1024 6y ago
Here's a fun way to see this approximation: 1 + x ~= e^x when x is close to zero (either positive or negative). So let m be 10^-6 . Probability of survival of one event = 1 - m ~= e^-m . Probability of survival of two events = (1 - m)^2 ~= (e^-m)^2 = e^-2m ~= 1 - 2m.
- klodolph 6y ago
  To add to this, this is why we have those funny functions log1p() and expm1() in the standard library. To add micromorts: double add_micromorts(double x, double y) { return -1e6 * expm1( -log1p(x * 1e-6) - log1p(y * 1e-6) ); } This is not the most accurate way of doing things, it is much simpler to return x + y - x * y * 1e-6; But the formula with expm1 and log1p lets you easily do things like multiply 10 micromorts/day by 10.5 days.
  - smabie 6y ago
    They're mostly there for converting from log returns and back, but yeah they're good for any kind of probabilities as well.

quicklime 6y ago

So in your particular example you can't add micromorts - you can only die once, so if you die in the first round, you don't get to play the second round. And if you could, the number of known-to-be-empty-chambers and/or bullets has changed, which changes the probability. But you can add micromorts in the case where n people are playing separate games of Russian Roulette. As another example, the Wikipedia page says that this is equivalent to 1 micromort: > Traveling 6000 miles (10,000 km) by jet (cancer due to increased background radiation) If you're just trying to calculate your own risk as a passenger, you can't just multiply 1 micromort by the number of 10,000 km trips you take. But if you're operating a flight, then you can multiply it by the number of passengers on this 10,000 km flight. It has more to do with the nature of the thing you're trying to model, than it does with the nature of the unit.

staplor 6y ago

This is pedantic, but going twice in Russian roulette would have an 80% mortality rate. This is because the second shot is a 3/5ths chance, not 1/2.

deleted 6y ago
[deleted]
erik_seaberg 6y ago
Don't they spin again before each round? That's the difference between (3/6) * (3/6) and (3/6) * (2/5) survival.

unholiness 6y ago

I'd argue adding is the right calculation in many situations, since in most cases you wouldn't receive the benefits of the next risk if you die the first time. So for example, if working in a coal mine is 10 micromorts per work day, and you choose to do it each day, it seems pretty reasonable to call that 10 * 5 days * 50 work weeks = 2500 micromorts per year. You're "paying" that cost each time you decide to go to work, and if you decide to work next year it won't somehow "cost" less micromorts – the possible yous who died last year won't be around to make the decision.