Staredit Network > Forums > Lite Discussion > Topic: A question (or 2!)
A question (or 2!)
Nov 5 2011, 7:06 am
By: Sacrieur
Pages: 1 2 35 >
 

Nov 5 2011, 7:06 am Sacrieur Post #1

Still Napping

What is the probability of choosing the number 42 from the set of all integers?

Furthermore, what is the probability of choosing the number 42 from the set of all real numbers?



None.

Nov 5 2011, 8:23 am Madroc Post #2



Okay so the chance of choosing choosing 42 out of all integers or real numbers is 1/x, where x is the number of real numbers or integers (both infinity). Take the limit as x goes to infinity, and you get lim x --> inf, 1/x = 0.

The chance of choosing 42 in either case is 0.




None.

Nov 5 2011, 11:00 am The Starport Post #3



Maths: srs bsns!



None.

Nov 5 2011, 2:08 pm Aristocrat Post #4



Quote from Madroc
Okay so the chance of choosing choosing 42 out of all integers or real numbers is 1/x, where x is the number of real numbers or integers (both infinity). Take the limit as x goes to infinity, and you get lim x --> inf, 1/x = 0.

The chance of choosing 42 in either case is 0.
But it can't be zero, because it's possible to choose 42. Therefore, it must be a number above 0.



None.

Nov 5 2011, 2:21 pm JaFF Post #5



Assuming all elements in your sets are equally likely to be chosen, the answer is 0 for both cases.



None.

Nov 5 2011, 2:27 pm BeDazed Post #6



Quote
But it can't be zero, because it's possible to choose 42. Therefore, it must be a number above 0.
The probabability of choosing 42 out of real numbers in the given interval [42-0, 42+0] is 0.



None.

Nov 5 2011, 2:30 pm Aristocrat Post #7



Quote from BeDazed
Quote
But it can't be zero, because it's possible to choose 42. Therefore, it must be a number above 0.
The probabability of choosing 42 out of real numbers in the given interval [42-0, 42+0] is 0.
[citation needed]



None.

Nov 5 2011, 2:55 pm ClansAreForGays Post #8



So there's no chance of picking any number at random out of infinity?




Nov 5 2011, 3:39 pm The Starport Post #9



Limit near 0 is not the same as 0.

Close enough for me not to give a shit, though.

Post has been edited 3 time(s), last time on Nov 5 2011, 3:55 pm by Tuxedo-Templar.



None.

Nov 5 2011, 3:49 pm Sacrieur Post #10

Still Napping

How could it be zero? It is completely possible that I could draw a random number out of the hat and it be 42.



None.

Nov 5 2011, 4:01 pm poison_us Post #11

Back* from the grave

Quote from name:Tuxedo-Templar
Limit near 0 is not the same as 0.

Close enough for me not to give a shit, though.
This. The chance is always greater than 0, but as you add more numbers it gets infinitely close to 0. This is like the difference between .9 repeating and 1.

Yay, statistics!





Nov 5 2011, 4:35 pm BeDazed Post #12



Actually, if you had infinite life, infinite energy, and infinite will to pull out integers from a magic hat containing all integers, until the integer 42 has been pulled out, then I think the probability wouldn't be zero. Though this statement is the one thing that really needs citation.

Real number has an infinite amount of numbers in any given interval. And it's pretty obvious to anyone that +0, -0 on the notation is not an actual zero but limit near zero, otherwise I wouldn't have had the need to put such efforts in putting the zeros.

Post has been edited 1 time(s), last time on Nov 6 2011, 10:39 am by BeDazed.



None.

Nov 6 2011, 10:32 am ShadowFlare Post #13



Impossible to do without a fixed range, so the probability that you could do this is 0. :P



None.

Nov 6 2011, 4:59 pm Moose Post #14

We live in a society.

Quote from Sacrieur
How could it be zero? It is completely possible that I could draw a random number out of the hat and it be 42.
Your hat of infinite size containing an infinite number of markers with numbers on them?

Let me know where you managed to get one of those.




Nov 6 2011, 5:00 pm Roy Post #15

An artist's depiction of an Extended Unit Death

You're asking for a number for 1/infinity, which is like asking for a number for 1/0; you're not going to get a number.

The answer is infinitesimal. Those saying limit zero are also right.




Nov 6 2011, 5:23 pm Sacrieur Post #16

Still Napping

It appears our attempts to dissuade people from the answer was unsuccessful. This is a very counter-intuitive problem to most people, and the fact nearly all of the responses claimed zero is somewhat reassuring in humanity. Although, allow me to clear up some points.

1/n, n → ∞ is 0. This zero is not special in any way, and is the same as all other zeroes. That is to say the probability of choosing the number 42 is zero, even if you choose the number. It can happen, but it is statistically impossible.

The set of all integers is an aleph-null set (a designation of infinite sets). This is because the set is composed of countable entities. If we expand the set to all real numbers, we are working in aleph-one: which is a degree higher. Regardless, the probability is the same for both the aleph-null and aleph-one set.


Quote from BeDazed
Quote
But it can't be zero, because it's possible to choose 42. Therefore, it must be a number above 0.
The probabability of choosing 42 out of real numbers in the given interval [42-0, 42+0] is 0.

I understand what you're saying, but it is the same as [42, 42], which only contains one number (1/n → ∞ doesn't change how the zero behaves). If you wished to designate a neighborhood around 42, the proper calculus designation would be (41, 43).


Quote from BeDazed
Actually, if you had infinite life, infinite energy, and infinite will to pull out integers from a magic hat containing all integers, until the integer 42 has been pulled out, then I think the probability wouldn't be zero. Though this statement is the one thing that really needs citation.

This actually can be solved by playing around with limits and convergence tests. Infinite trials would result in a 100% chance of success in an aleph-null set. In an aleph-one set, the chance of success would still be zero.



Quote from ShadowFlare
Impossible to do without a fixed range, so the probability that you could do this is 0. :P

I think you mean set.



Quote from Roy
You're asking for a number for 1/infinity, which is like asking for a number for 1/0; you're not going to get a number.

The answer is infinitesimal. Those saying limit zero are also right.

Actually limits demonstrate that 1/∞ is equal to zero. The answer is zero, through the same logic that 0.999 ... = 1



None.

Nov 6 2011, 8:45 pm Azrael Post #17



That's a fantastic game of Theorycraft you're playing there, but let's put this into perspective by going back to the originally proposed situation.

The scenario was, you build a computer to simulate the current state of the universe. Within that simulation, a simulated you has done the same thing at the same time. The computer is powerful to the point of being able to create an infinite number of universes-within-universes in this manner. Each version of you believes they are the real one.

You said that there was a literal 0% chance of being the "real" one, that it was equivalent to choosing a number from an infinite set, so there was no "real" one.

However, one second before flipping the switch to turn the simulator on, there was only one universe, and that remained perfectly intact and unchanged upon turning the machine on.

I propose that the mathematics presented in this thread are completely arbitrary and irrelevant to the situation. If you wanted to apply them, then you would have to look at it as a single real universe which is separate from an infinite set of simulated ones. Lumping the real universe in with all the fakes and then saying it's impossible to be in the real one because lolmath is completely nonsensical on every level.




Nov 6 2011, 9:40 pm DT_Battlekruser Post #18



An event with probability zero occurs almost never. This is NOT the same thing as an impossible event, which is forbidden from occurring (such as picking 3.2 from the set of integers). As was said, if you pick a number at random from the reals, whatever number you will pick will have had a probability of zero of being chosen.

Here's a more interesting thought. Between any two irrational numbers exists a rational number, yet if you choose a real number at random, it will almost surely (meaning "with probability 1") be irrational.




None.

Nov 7 2011, 3:08 am ShadowFlare Post #19



Quote from Sacrieur
Quote from ShadowFlare
Impossible to do without a fixed range, so the probability that you could do this is 0. :P

I think you mean set.
lol, don't need to be technical about the terms. I was just making a joke about that you couldn't set up a test for it.



None.

Nov 8 2011, 8:13 am Vrael Post #20



Quote from JaFF
Assuming all elements in your sets are equally likely to be chosen, the answer is 0 for both cases.
I don't agree with this assumption.

For example, the popularity of the book "The Hitchhikers Guide to the Galaxy" will skew the probability in favor of the number 42. I would also impose a skew towards numbers within the order 10^2, I would disregard all irrational numbers except the well known ones like pi and e, I would favor any multiple of 10 or 5, I would disregard any rational number requiring a decimal expansion of more than 10 or so digits, ect ect ect.

I also don't agree that the question is fully defined.

Post has been edited 1 time(s), last time on Nov 8 2011, 8:21 am by Vrael.



None.

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