Random Integers

Probability Level 3

Calvin chooses an integer from 1 to infinity, uniformly at random.

What is the probability that he picks his favourite number, 501?

1 No such probability distribution exists 0 Between 0 and 1

1 solution

Calvin Lin Staff
Sep 21, 2016

There is no uniform distribution on the (positive) integers. Hence, the first step doesn't make sense.

Proof: Suppose there such a distribution existed. Then, let $P ( X = 1 ) = p$ .
Since we have a uniform distribution, we see that $P ( X = n ) = P(X = 1) = p$ .
Thus the total probability is $P ( X \geq 1 ) = P ( X = 1 ) + P ( X = 2 ) + P ( X + 3 ) + \ldots$ .

If $p > 0$ , then this sum would be infinite, contradicting that the total probability is 1.
If $p = 0$ , then we have the countable sum of 0's, which is 0. This contradicts that the total probability is 1.

The wording is critical. "Calvin chooses an integer from $1$ to infinity ". This choice is an impossibility because there is no uniform probability distribution on the positive integers if there are infinitely many. If we qualified the wording by saying, "Calvin chooses a positive integer of finite size at random", then there could be one, and the probability of his having chosen his favorite number "501" (Levi's 501 Blue Jeans?) could range from $0$ to $1$ .

Philosophically speaking (i.e. off the rigorous record) another way of making the same argument would be that if Calvin were to choose a positive integer uniformly at random, then he will never finish specifying it, as it takes an infinite time or information to do so, as the probability of having picked a number infinitely long approaches $1$ . The question becomes, did he ever plan on making or specifying a choice in finite time? The wording says no.

Otherwise, with the wording as it is now, if there is any finite probability at all of picking his favorite, either $p>0$ or $p=0$ , and assuming an uniform probability for all possible choices, which is all positive integers, then we can't get a total of $1$ , which is a contradiction. Getting a total of $1$ is a necessary condition for any probability distribution, discrete or continuous.

Michael Mendrin - 4 years, 8 months ago

Random Integers

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