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I am trying to calculate the probability of picking perfect squares out of first $n$ positive integers.

There are $\operatorname{floor}(\sqrt n)$ number of perfect squares less than $n$, if we assume picking each number is equally likely then probability of picking perfect squares less than $n$ is $p(n) =\operatorname {floor}(\sqrt n) /n$

But if we consider all positive integers then ( I think but I am not sure) the probability is $\lim_{n\to \infty} p(n) = 0$. That means the probability of selecting a square is $0$ even though there are infinitely many squares! Does this mean I that I will not get any squares when I pick any positive integers? This appears like a paradox.

I think I have not very much understood what it means by $P(A) = 0$, does this mean the event $A$ is impossible or something else?

Can you give some insights about this?

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    $\begingroup$ There are no uniform distribution on natural numbers: You can't pick a number 'with equal probability'. The limit you discribe is called density in number theory, which is not as relevant to probability. $\endgroup$
    – Trebor
    Sep 26 at 6:35
  • $\begingroup$ What is the probability that if you select a positive integer $\leq n$, the number that you select is $2$, assuming that $n \geq 2.$ The probability is $\frac{1}{n}$. This means that as $n \to \infty$, the probability of selecting the number $2$ goes to $0$. However, if you are asked to select any positive integer, while the probability of selecting the number $2$ is $0$, it is still not impossible to select the number $2$. $\endgroup$ Sep 26 at 6:51
  • $\begingroup$ In the same way, if a Real number $r$ is selected at random from the interval $[0,1]$ the chance that $r = (1/2)$ will be $0$. Despite this, it is not impossible for the number selected to equal $(1/2).$ $\endgroup$ Sep 26 at 6:55
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The problem is that you are assuming that you can define a uniform distribution on the natural numbers by considering the limit of the uniform distribution on $\{1,\ldots,n\}$ as $n\to\infty$.

However, there is no such thing as a uniform distribution on the natural numbers.

If $\Pr(N=n)=p$ for every $n\in\mathbb{N}$ then if $p=0$ we have $\sum_{n\in\mathbb{N}}\Pr(N=n)=0$, whereas if $p>0$ we have $\sum_{n\in\mathbb{N}}\Pr(N=n)=\infty$. However, we need $\sum_{n\in\mathbb{N}}\Pr(N=n)=1$ for a probability distribution, so it doesn't exist.

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One can understand the problem using the notion of natural density. Specifically, $$ d(A)=\lim_{n\to\infty}\frac{|A\cap \{1,\ldots,n\}|}{n} $$ if the limit exists. In your case, $$ d(A)=\lim_{n\to\infty}\frac{\lfloor \sqrt{n} \rfloor}{n}=0. $$

See this question for another (related) interpretation.

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When dealing with a sample set that has infinite possibilities, a probability of zero is not the same as something being impossible.

A classic example of this is picking a random number $x$ between zero and one. It isn't impossible for you to pick $\frac12$, as $0<\frac12<1$. However, we can show that the probability is zero:

Suppose the probability is $\epsilon>0$. We can then define $N=\lceil \frac1\epsilon \rceil$. We can then define the set $M=\{\frac12,\frac1{N+2},\frac1{N+3},...\frac1{2N+2}$. We have that $M\subset (0,1)$ and therefore $P(x=\frac12)=P(x=\frac12\cap x\in M)+ P(x=\frac 12\cap x\not\in M)=P(x=\frac12\cap x\in M)=P(x=\frac12\cap x\in M)= P(x\in M)P(x=\frac12\mid x\in M)\leq P(x=\frac12\mid x\in M)=\frac1{N+1}<\epsilon$. However, this contradicts our assumption that $P(x=\frac12)=\epsilon$, and therefore we must conclude that $P(x=\frac12)=0$

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