1 in ~10 million seems to be the average risk of an american to die from air plane, so, in your lifetime (100 years, if you live to 100 - average lifespan is 80 it looks like), that goes up to 1-(9,999,999/10,000,000)^100 ~ 1 in 100k
Since that number is about the average risk of an american to die (and not the risk per flight), using it to calculate the risk per flight is erroneous
Actually the 1 in 10 million figure I used is based on number of fatal crash per flight. I thought it was roughly 1 in 10 million flights ending in fatal crash, but that was couple years ago. It's actually more like 1 in 3.7 million flights. so if you take 100 of those flights, I figured the odds would be 1 in 37,000.
No they're using the gambler's fallacy, each flight you take is unaffected by any flight you've taken previously. Flying a load of times isn't reducing the denominator.
It’s not the gambler’s fallacy since they’re treating each flight as an independent event:
Probability of dying on one flight: 1/10,000,000
Probability of surviving one flight: 1-1/10,000,000 = 9,999,999/10,000,000
Probability of surviving 100 flights: (9,999,999/10,000,000)^100
Probability of not surviving 100 flights:
1-(9,999,999/10,000,000)^100 ~ 1/100,000.
I don’t know if the initial assumption of 1/10,000,000 survival rate is accurate, but the math is correct.
And when you have a probability that’s very small, you can skip all that and approximate by multiplying the probability by the number of events: 1/10,000,000 * 100 = 1/100,000. That’s thanks to the binomial approximation.
Isn't gambler's fallacy more about expectation of future result skewing based on past outcome?
If there's a very dangerous airplane ride that results in 10% of all flights ending in crash, if you take that plane 10 times, wouldn't you expect 1 crash on average?
Since that number is about the average risk of an american to die (and not the risk per flight), using it to calculate the risk per flight is erroneous
I think?