Anonymous at Sun, 10 Mar 2024 02:27:31 UTC No. 16065897

>>16065884
>infinite
stopped reading there

Anonymous at Sun, 10 Mar 2024 02:32:23 UTC No. 16065905

>>16065884
Does the machine operate on a function of addition subtraction multiplication or division?

Anonymous at Sun, 10 Mar 2024 02:35:25 UTC No. 16065913

>>16065884
Easy.
5,000,000/∞ = 0
So 0% chance

Anonymous at Sun, 10 Mar 2024 02:49:18 UTC No. 16065928

>>16065905
I dunno, your choice.

>>16065913
How far does this 0% chance extend then?

Anonymous at Sun, 10 Mar 2024 02:51:16 UTC No. 16065930

>>16065884
You're going to have to define "at random" because I don't know what you mean by it.

Anonymous at Sun, 10 Mar 2024 02:52:55 UTC No. 16065933

>>16065884
>all numbers have an equal chance of being chosen.
>>16065913
>So 0% chance

Anonymous at Sun, 10 Mar 2024 06:22:21 UTC No. 16066051

>>16065930
An arbitrary integer between 0 and infinity

Anonymous at Sun, 10 Mar 2024 06:49:24 UTC No. 16066079

>>16065928
>How far does this 0% chance extend then?
everywhere, since every number is finite

Anonymous at Sun, 10 Mar 2024 07:28:42 UTC No. 16066099

As the number of integers increases, the likelihood of selecting any given number approaches 0. Of course your hypothetical number machine would have a chance of getting <5 000 000, but with an infinite number of leading digits behind it, each with a 90% chance of being nonzero (and therefore making the number >5 000 000), there is an infinity*0.9 chance that the number will be >5 000 000. Infinity times 0.9 is still infinity, so there is an infinite chance that the number is >5 000 000

Anonymous at Sun, 10 Mar 2024 08:02:48 UTC No. 16066118

>>16065884
between 0% and -0%

Anonymous at Sun, 10 Mar 2024 09:41:18 UTC No. 16066167

{1,...,n} is a finite set and the natural numbers are infinite. So, if a natural number is chosen uniformly at random, the probability is 0 that it will be an element of {1,...,n}.

Anonymous at Sun, 10 Mar 2024 09:52:53 UTC No. 16066174

>random
You keep using that word, but I don't think you know what it means

Anonymous at Sun, 10 Mar 2024 09:55:02 UTC No. 16066177

>>16066051
That doesn't mean anything. In particular, that doesn't mean a probability measure because there isn't one with equal chances on the integers. At [math]Pr[n] = 0[/math] for each [math]n[/math] it would still violate violate the countable additivity requirement. So, go on, define "arbitrary" precisely enough for the question to be answered.

Anonymous at Sun, 10 Mar 2024 16:46:01 UTC No. 16066603

>>16065913
>5,000,000/∞ = 0
Multiplying both sides by ∞ we get:
5,000,000 = 0
which is incorrect.

Anonymous at Sun, 10 Mar 2024 17:43:08 UTC No. 16066691

>>16065884
>all numbers have an equal chance of being chosen.
false assumption

Anonymous at Sun, 10 Mar 2024 17:44:32 UTC No. 16066694

the machine needs infinite resources to make the choice. it will always blow the fuck up before making a choice thus making all numbers have a 0% chance of being chosen.

Anonymous at Mon, 11 Mar 2024 04:06:19 UTC No. 16067564

>>16065897
Tsmt

Anonymous at Mon, 11 Mar 2024 04:09:36 UTC No. 16067572

>>16065884
100% chance
>you have a machine that selects one at random
So my machine already selected 1.

Anonymous at Mon, 11 Mar 2024 12:01:06 UTC No. 16067978

>>16066051
If the selection is arbitrary, then the chance is also arbitrary.

🗑️ Anonymous at Mon, 11 Mar 2024 12:29:03 UTC No. 16068004

[math]P(A=n\le5000000|B=n\sim U(- \infty, \infty)), n\in\mathbb{Z}[/math]

Anonymous at Mon, 11 Mar 2024 12:49:14 UTC No. 16068023

[math]P(A=n\le5,000,000|B=n\sim U(0,\infty)), n\in\mathbb Z[/math]

Anonymous at Mon, 11 Mar 2024 13:01:27 UTC No. 16068029

>>16066694
This is the real answer. Everyone else is having a psychosis