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We want to take a random number from natural numbers how much is the probability that,the number be $1$?

When we want to say the probability we say it is $0$ but we say zero for impossible things but that is possible.I know that every number divided by infinity is zero but maybe we take $1$ what about that?

Taha Akbari
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    This is a classic confusion--there's no "uniform distribution" on the natural numbers. – Funktorality Jul 11 '16 at 06:36
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    It's also a classic confusion that probability $0$ means "impossible". It just means "very improbable". Of course, impossible things also have probability $0$, but there are many possible-but-not-likely things that have probability $0$ of happening. – Arthur Jul 11 '16 at 06:37
  • @Arthur.Is there any liks to this classic confustion? – Taha Akbari Jul 11 '16 at 06:39
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    https://en.wikipedia.org/wiki/Almost_surely is about probability $1$ instead, but it's the same concept. – Arthur Jul 11 '16 at 06:39
  • This MSE thread gets at the same idea. – Funktorality Jul 11 '16 at 06:40
  • Arthur, would you say the for something with zero probability to be possible there must be an infinite number of possible outcomes? – fleablood Jul 11 '16 at 06:40
  • @fleablood Not at all, but there would probably have to be some infinity going on in the background. For instance, flip a coin until you get tails. Let the random variable $X$ be $1$ if you ever finish, and $0$ if you just get heads all the way (i.e. $X$ is the number of tails you get). Then $P(X = 0) = 0$ (but theoretically possible) and $P(X = 1) = 1$, and those are the only possible outcomes. – Arthur Jul 11 '16 at 06:43
  • But in your example there are an infinite number of outcomed. T, ht, hht, etc. But point taken. I was vague in my wording. – fleablood Jul 11 '16 at 06:48
  • @Arthur What does "possible" mean in this context? If I have a Bernoulli RV with $p=1$ then $P(X=0)=0$ and $P(X=1)=1$, and those are the only outcomes. I don't see how I can in any way distinguish my example from yours, even though my example is entirely "finite". – Mario Carneiro Jul 11 '16 at 06:55

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There are two issues here:

Firstly, you haven't specified what probability distribution on the natural numbers we should assume. You probably mean one which is in some sense uniform: $Pr(0)=Pr(1)=Pr(2)=\cdots$ to infinity. However there is no such distribution, as explained here.

The other issue is that "possible" isn't a very well-defined mathematical term. When we talk about probability formally, we usually talk about a space of outcomes $X$ (eg. in this case that would be the natural numbers) and a probability measure $P$ which takes subsets of $X$ and gives the probability of the outcome being in that subset (there's also something else called a $\sigma$-algebra but there's no need to add that extra confusion here). Probabilists then do not distinguish between probability measures that differ on sets of zero probability events. For example, the $P$ which uniformly picks a number in the interval $[0,1]$ and the $P'$ which uniformly picks a number in $[0,1]\setminus\{0.42\}$ are considered the same distribution since in either case the outcome has a zero probability of being in the set $A=\{0.42\}$. However, in the first case you would say $A$ is "possible," in the second case you would not.

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Funktorality
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  • Even more to the point, what does it mean for an event to "occur"? Probability theory doesn't seem to handle this question well. The usual argument that probability zero events are possible is that any result from the uniform distribution $[0,1]$ is probability zero of occuring, yet some value must occur. But it's not clear how a specific value "occuring" is supposed to be modeled. – Mario Carneiro Jul 11 '16 at 07:00
  • Put another way: suppose I select a number uniformly on $[0,1]$. The number $1/2$ is exactly as likely to happen as $2$ is, and there is no consistent mathematical approach that distinguishes them. (The punchline: I was actually secretly selecting numbers uniformly on $[0,2]$ and repeating until I got a number in $[0,1]\cup{2}$. But how can you tell?) – Mario Carneiro Jul 11 '16 at 07:17