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Large numbers tend to have a large number of digits. Given a number with N digits, the probability that it doesn't have any 9s is (9/10)^N because each digit i
by radq 15y ago
Large numbers tend to have a large number of digits.
Given a number with N digits, the probability that it doesn't have any 9s is (9/10)^N because each digit is equally likely at a position.
(9/10)^N becomes very small as N becomes large, so the probability of a large number not having 9s is 0.
- HSO 15y agoHow does that make terms with the 9 special?
- radq 15y agoIt doesn't, this is true for any digit.
- HSO 15y agoWhich was my point ;-)
- Dylan16807 15y agoCan you elaborate on that? I can't find any point or contribution made by that post if it wasn't an actual question.
- ComputerGuru 15y agoHSO's original comment was sarcasm posed as a question. Poorly executed, but that's where his original point was made ;)
- Dylan16807 15y agoWell yes, I understand that it was sarcasm. But while looking at it as sarcasm I don't see any point actually being made.
- politician 15y agoSo, this is just a parlor trick, right? It sounds like this is just a complicated way of saying that this is what the harmonic series looks like when you've removed all terms with more than N digits.
- andrewcooke 15y agowell, almost all, yes.
- Confusion 15y agoNot really, because you've removed 'almost all' terms, but there are many ways to remove 'almost all' terms that do not result in a convergent series. For instance, you could remove 999 out of every 1000 terms. I'd say that counts as removing 'almost all' terms. Nevertheless, the resulting series does not converge, for the exact same reason that removing every other term doesn't work. The amount of terms that you remove has to increase faster than that.
- SamReidHughes 15y agoAnd in particular, removing 100% of the terms doesn't mean much either, since, for example, the series 1/p for all primes does not converge, or 1/floor(n log n). You need to remove them harder than that.
- redthrowaway 15y agoIt's only 0 as far as statisticians and cheap calculators are concerned. The number of N-digit numbers without 9s in them is much higher than the number of similar (N-1)-digit numbers, it's just the proportion that falls. The probability isn't 0, it just asymptotically approaches 0, as the number of non-9-containing numbers approaches infinity. In essence, the cardinality of the set of all numbers that do not contain 9 is infinity; the cardinality of the set of all numbers is infinitier. This despite the fact that both sets are countably infinite.
- neutronicus 15y agoCardinality is not the term you want to be using.
- redthrowaway 15y agoI can see how it's imprecise, but I'm struggling to find a better descriptor. "Size" doesn't really work.
- Dylan16807 15y agoThe way an infinitely unlikely but theoretically possible event is described is in fact 'probability 0'. Sure, in a specific finite range the chance is a specific positive number, but the overall chance is 0.