So in 5 years, you will be throwing 1825 times.
Say, probability of one of those problematic events is P(E), where E is the event.
Probability of E happening n (Natural) number of times would be either 0 or P(E), with P(E) for n = 1 and 0 for n > 1, because you don’t get more than 1 life to make it happen more than once.
Probability of it happening at least once in the whole 5 years of time, is P(E) × 1825.(check silasmariner’s reply)
So, what is the value you set to P(E) to convince yourself that probability is the correct thing to go by?
I checked wiki for a couple numbers, and it looks like Earth’s ocean surface area is 361 million km^2, and the shoreline is 356,000 km. Even assuming every bit of the shoreline is made of sharp deadly rocks, there’s an extremely low probability you’ll end up near the shore at all, even with almost 1900 chances at it.
Going by masquenox’s comment, you might want to consider the whole volume of the oceans and the instant depressurisation that would occur if you were to survive the pressurisation.
It’s not P(E) x1825 that makes no sense because the probabilities are independent. It’s 1 - (1 - P(E))^1825 – your maths would imply that a 1/1000 chance had a 100% probability of happening after 1000 attempts, which is not how independent probabilities work.
Anyway I’d probably go for it if it were, like, 1/100,000, giving a probability of the bad thing happening once over 5 years at just under 2%
I just remembered I hadn’t tried to write anything in J for like a year so I had a go at seeing how far the approximation was from the real values and got:
So it looks like the ratio probably converges as you increase n (these results are for 1, 1/10, 1/100 etc) but that x1825 is always higher and starts off wildly off. But these numbers are probably mad squiffy because floating point yadda yadda.
Edit again: oh yeah. Binomial expansion. Once the negation of your probability gets very close to 1 the initial coefficient will absolutely dominate. Proof left for reader.
So in
5years, you will be throwing1825times.Say, probability of one of those problematic events is
P(E), whereEis the event.Probability of
Ehappeningn(Natural) number of times would be either0orP(E), withP(E)forn = 1and0forn > 1, because you don’t get more than1life to make it happen more than once.Probability of it happening at least once in the whole
5years of time, isP(E) × 1825.So, what is the value you set to
P(E)to convince yourself that probability is the correct thing to go by?I checked wiki for a couple numbers, and it looks like Earth’s ocean surface area is 361 million km^2, and the shoreline is 356,000 km. Even assuming every bit of the shoreline is made of sharp deadly rocks, there’s an extremely low probability you’ll end up near the shore at all, even with almost 1900 chances at it.
Going by masquenox’s comment, you might want to consider the whole volume of the oceans and the instant depressurisation that would occur if you were to survive the pressurisation.
Yeah the volume is one thing but I assumed we were talking about the surface. If you could end up miles underwater it wouldn’t be worth it
It’s not
P(E) x 1825that makes no sense because the probabilities are independent. It’s1 - (1 - P(E))^1825– your maths would imply that a 1/1000 chance had a 100% probability of happening after 1000 attempts, which is not how independent probabilities work.Anyway I’d probably go for it if it were, like, 1/100,000, giving a probability of the bad thing happening once over 5 years at just under 2%
I just remembered I hadn’t tried to write anything in J for like a year so I had a go at seeing how far the approximation was from the real values and got:
So it looks like the ratio probably converges as you increase n (these results are for 1, 1/10, 1/100 etc) but that
x 1825is always higher and starts off wildly off. But these numbers are probably mad squiffy because floating point yadda yadda.Edit again: oh yeah. Binomial expansion. Once the negation of your probability gets very close to 1 the initial coefficient will absolutely dominate. Proof left for reader.