5 ms·

Aren’t they already the same? You’re just not applying the same metric in both cases. The chance of dying in a car accident isn’t the number of miles driven, i

by setr 3y ago

Aren’t they already the same? You’re just not applying the same metric in both cases. The chance of dying in a car accident isn’t the number of miles driven, it’s the chance of a failure happening and killing you — the distance driven is simply more rolls of the dice. The same with parachute-failure — the number of chutes opened is the number of dice rolls. You should be saying skydiving death rate is X per chute

burnished 3y ago
What is the driving equivalent to one skydive? I think that question genuinely illustrates their point that some risk is best considered as a rate whereas some is as a discrete event but I am very interested if you can elaborate a perspective that reconciles that
- inkcapmushroom 3y ago
  The link gives both of those, actually. Skydiving is 8 micromorts per jump, car travel is one micromort per ~250 miles. So the equivalent chance of death for one skydive is driving 2000 miles.
  - ineedasername 3y ago
    But how to combine them, i.e., micromorts per skydiving jump via a car that travels 250 miles in the air before driving it out of the back of a cargo plane? I think we should just add them together and call it 9 micromorts, assuming the car has an appropriately sized parachute.
  - burnished 3y ago
    That seems to further solidify the original observation. I do appreciate you finding their break even point though!
- setr 3y ago
  I think the answer is that it’s always a rate; it’s just that the skydiving is such a rare event we can implicitly drop the “per chute” because we’re assuming it’s always going to be “1 chute”. If you took the skydiver instructor, doing hundreds of rounds, then suddenly skydiving should no longer be considered a “discrete event” — your determination of how to treat it isn’t based on the event-type itself, but the object undergoing the event.

by setr 3y ago

burnished 3y ago

What is the driving equivalent to one skydive? I think that question genuinely illustrates their point that some risk is best considered as a rate whereas some is as a discrete event but I am very interested if you can elaborate a perspective that reconciles that

inkcapmushroom 3y ago
The link gives both of those, actually. Skydiving is 8 micromorts per jump, car travel is one micromort per ~250 miles. So the equivalent chance of death for one skydive is driving 2000 miles.
- ineedasername 3y ago
  But how to combine them, i.e., micromorts per skydiving jump via a car that travels 250 miles in the air before driving it out of the back of a cargo plane? I think we should just add them together and call it 9 micromorts, assuming the car has an appropriately sized parachute.
- burnished 3y ago
  That seems to further solidify the original observation. I do appreciate you finding their break even point though!
setr 3y ago
I think the answer is that it’s always a rate; it’s just that the skydiving is such a rare event we can implicitly drop the “per chute” because we’re assuming it’s always going to be “1 chute”. If you took the skydiver instructor, doing hundreds of rounds, then suddenly skydiving should no longer be considered a “discrete event” — your determination of how to treat it isn’t based on the event-type itself, but the object undergoing the event.