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?
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.