1

I've been thinking about this and it seems to confuse me more and more. Sure, at first site it looks to be half, but is it really half?

One can argue that for each even number there corresponds an odd number, but this argument can be challenged by saying that for each even number there can correspond two odd numbers which will mean the probability is a third...

Am I thinking wrong somewhere??

B2VSi
  • 1,017
  • 4
    The answer depends on the probability distribution on $\mathbb{N}$. Note that there is no uniform distribution on $\mathbb{N}$. – Daniel Fischer Sep 30 '16 at 13:02
  • Thanks, didn't think of that! – B2VSi Sep 30 '16 at 13:05
  • If you consider a (*FINITELY ADDITIVE*) translation-invatiant probability measure on $\mathbb N,$ it's clear that the probability of picking an even number is $\frac12.$ The fact that you can make each even number correspond to two odd numbers, or that you can make a one-one correspondence between the odd numbers and the numbers divisible by $100,$ is irrelevant. Thee is a uniform probability distribution on the interval $(0,1),$ and the fact that there is a bijection between $(0,\frac13) $ and $(\frac13,1)$ does not mean that the those two intervals have the same probability. – bof Sep 30 '16 at 13:14
  • @bof: There isn't a finitely additive translation invariant probability measure on $\mathbb N$ though, is there? – Deusovi Sep 30 '16 at 13:18
  • @Deusovi You may be interested in my answer to the duplicate question that I linked above. – aduh Sep 30 '16 at 13:23
  • @Deusovi Such measures do exist (assuming a weak form of the axiom of choice) which are defined for every subset of $\mathbb N.$ Moreover such a measure can be chosen so that the measure of any set is no less than its lower density and no greater than its upper density. – bof Sep 30 '16 at 13:26

1 Answers1

2

It depends on what probability distribution one defines on $\mathbb N$ (as observed, this cannot be a uniform distribution, but that doesn't mean no distribution exists). Consider the following probability assignments on $\mathbb N$:

Prob(1) = 1/2, Prob(2) = 1/4, ..., Prob($n$) = 1/$2^n$ [this is a legitimate probability distribution since the sum of all terms is 1]

Here obviously the probability of picking an even number is 1/4 + 1/16 + ..., which is 1/3, not 1/2. On the other hand, we could 'split the distribution in two' thus:

Prob(1) = prob(2) = 1/4, Prob(3) = Prob(4) = 1/8, ..., Prob($2k-1$) = Prob($2k$) = 1/$4^k$

Here obviously the probability of picking an even number is 1/2 since the odd and even terms must separatly have the same sum.

PMar
  • 21