My Math Forum  

Go Back   My Math Forum > High School Math Forum > Probability and Statistics

Probability and Statistics Basic Probability and Statistics Math Forum


Thanks Tree15Thanks
Reply
 
LinkBack Thread Tools Display Modes
April 14th, 2017, 11:21 AM   #11
Member
 
Joined: Mar 2017
From: Nairobi, Kenya.

Posts: 98
Thanks: 2

Math Focus: Number theory
Good example.. the card problem. Lets say there are 52 cards. You are told to choose one at random. The probability of choosing a card is 1. The probability of choosing an ace is 1/52.

What the solution to the question was to clarify is that it is not a paradox.
Perhaps it was wrongly phrased by using the word random by which you have deduced it is impossible to pick any number randomly in that case. Either way we arrive to the same conclusion. The problem is not a paradox as the writer claimed it is.
Mariga is offline  
 
April 15th, 2017, 12:41 AM   #12
Senior Member
 
Joined: Jun 2015
From: England

Posts: 558
Thanks: 145

Quote:
Most particularly how can something be random and have a probability of 1?

You have completely failed to answer, or even offer any answer to that question I asked.

Why not?
I am still waiting for your answer.
studiot is offline  
April 15th, 2017, 04:52 AM   #13
Math Team
 
Joined: Dec 2013
From: Colombia

Posts: 6,557
Thanks: 2148

Math Focus: Mainly analysis and algebra
It is perfectly possible to randomly pick a natural number. Two such probability distributions for this are given in the other thread.

What is not possible is a uniform probability distribution over an infinite domain. Over an infinite domain, the probability of picking a number $n$ must tend to zero as $|n| \to \infty$. This is required for the cumulative probability function to converge to unity without which we don't have a probability distribution at all.
Thanks from Joppy

Last edited by v8archie; April 15th, 2017 at 04:55 AM.
v8archie is offline  
April 15th, 2017, 05:38 AM   #14
Member
 
Joined: Mar 2017
From: Nairobi, Kenya.

Posts: 98
Thanks: 2

Math Focus: Number theory
Quote:
Originally Posted by studiot View Post
I am still waiting for your answer.
I don't really know whether your question is really relevant here but I'll still answer it. You can choose something randomly and it's probability would be 1 if we only have one choice.
Read #11. There are two different probabilities we are considering.
Mariga is offline  
April 15th, 2017, 12:02 PM   #15
Senior Member
 
Joined: Jun 2015
From: England

Posts: 558
Thanks: 145

Quote:
Originally Posted by Mariga View Post
I don't really know whether your question is really relevant here but I'll still answer it. You can choose something randomly and it's probability would be 1 if we only have one choice.
Read #11. There are two different probabilities we are considering.
No this is not an answer or an explanation or a proof or a derivation.

It is an assertion, one you have already made.

I asked for an explanation since I assert the opposite.

Yes I agree it is possible to construct a process wherein the outcome is randomly dependent on a list of possiblities with preassigned probabilites, one of which is 1.

To me this would also indicate that the process was rigged.

But I do not agree when there is only one choice as in the last sweet in my finite bag of sweets.
In that case the process is not random, it is deterministic.

Do you think a process can simultaneously be both random and deterministic?



In response to your second point,

I do not understand the point of saying you are told to chose an outcome and then saying therefore the probability of there being an outcome is 1.

Yes this is true at the end of the process, but so what.

What if you never reach the end of the process?
studiot is offline  
April 15th, 2017, 01:05 PM   #16
Senior Member
 
Joined: Jun 2014
From: USA

Posts: 210
Thanks: 6

Quote:
Originally Posted by v8archie View Post
It is perfectly possible to randomly pick a natural number. Two such probability distributions for this are given in the other thread.

What is not possible is a uniform probability distribution over an infinite domain. Over an infinite domain, the probability of picking a number $n$ must tend to zero as $|n| \to \infty$. This is required for the cumulative probability function to converge to unity without which we don't have a probability distribution at all.
$\frac{1}{\infty} = 0$ is not technically true. It is undefined.

For example, consider:

$0 = 0 + 0 + 0 + ... = \frac{1}{\infty} + \frac{1}{\infty} + \frac{1}{\infty} + ... = \frac{1}{\infty} ( \, 1 + 1 + 1 + ... ) \, = \frac{\infty}{\infty} = 1$

Also, given:

1) the axiom of choice,
2) a randomly selected infinite binary sequence (created via the theoretical flipping of a coin infinitely many times), and
3) a definition of "select an element of an infinite set uniformly at random" that simply means, in layman's terms, that one and only one element of an infinite set will be selected using a process where all elements of the set have an equal chance of being selected, then

it is possible to select a natural number uniformly at random as well as a real number (and not just a real number on a closed interval, I mean from all of $\mathbb{R}$).

I am happy to post such a proof sometime in the next two weeks. It involves partitioning [0, 1] into an uncountable number of countable subsets, using choice to select a single element from each subset, and then defining a bijection between $\mathbb{N}$ and each subset based on the element selected from it using choice. I struggled with this a while back with Maschke, but have a proof now that works with the above criteria.
AplanisTophet is offline  
April 15th, 2017, 01:24 PM   #17
Senior Member
 
romsek's Avatar
 
Joined: Sep 2015
From: CA

Posts: 1,111
Thanks: 580

[QUOTE=AplanisTophet;567341]$\frac{1}{\infty} = 0$ is not technically true. It is undefined.

For example, consider:

$\frac{\infty}{\infty} = 1$

no, you cannot just aribtarily assert this.
romsek is online now  
April 15th, 2017, 01:40 PM   #18
Senior Member
 
Joined: Jun 2014
From: USA

Posts: 210
Thanks: 6

[QUOTE=romsek;567343]
Quote:
Originally Posted by AplanisTophet View Post
$\frac{1}{\infty} = 0$ is not technically true. It is undefined.

For example, consider:

$\frac{\infty}{\infty} = 1$

no, you cannot just aribtarily assert this.
I didn't. $\frac{\infty}{\infty} = 1$ is also undefined.

You don't have to take my word for it though:

https://www.mathsisfun.com/calculus/...-infinity.html
AplanisTophet is offline  
April 15th, 2017, 01:52 PM   #19
Senior Member
 
romsek's Avatar
 
Joined: Sep 2015
From: CA

Posts: 1,111
Thanks: 580

Quote:
Originally Posted by AplanisTophet View Post
$\frac{1}{\infty} = 0$ is not technically true. It is undefined.

For example, consider:

$0 = 0 + 0 + 0 + ... = \frac{1}{\infty} + \frac{1}{\infty} + \frac{1}{\infty} + ... = \frac{1}{\infty} ( \, 1 + 1 + 1 + ... ) \, = \frac{\infty}{\infty} = 1$
despite being couched with the term "consider" this sure looks like an assertion to me.

Quote:
I am happy to post such a proof sometime in the next two weeks.
of course you are. I will gladly pay you on Tuesday.

Quote:
It involves partitioning [0, 1] into an uncountable number of countable subsets, using choice to select a single element from each subset, and then defining a bijection between $\mathbb{N}$ and each subset based on the element selected from it using choice.
of course it does. It's clearly much more valid now than it was a few weeks back.

Quote:
I struggled with this a while back with Maschke, but have a proof now that works with the above criteria.
of course you do. Maybe Kellyanne and Sean can announce it for you.
romsek is online now  
April 15th, 2017, 02:15 PM   #20
Senior Member
 
Joined: Aug 2012

Posts: 1,165
Thanks: 257

Quote:
Originally Posted by AplanisTophet View Post
it is possible to select a natural number uniformly at random...
If your probability distribution is countably additive then you can't possibly do that. If the probability of a singleton is zero, then the total probability of all the natural numbers must be zero; and if it's positive, the total probability of all the natural numbers must be infinite. So either you are mistaken or you've broken math. I'll let you form your own conclusion.

However the Vitali set is always a fascinating topic of discussion and actually bears on the Banach-Tarski paradox, which has been a topic of conversation lately. So feel free to have another go at it and I'm sure I'll be unable to resist chiming in.
Maschke is online now  
Reply

  My Math Forum > High School Math Forum > Probability and Statistics

Tags
difficulties, paradoxes, probability


« Probability - Game | - »

Thread Tools
Display Modes


Similar Threads
Thread Thread Starter Forum Replies Last Post
Infinity just has too many paradoxes to be a real thing uperkurk Applied Math 11 September 30th, 2013 03:10 PM
Euler's Paradoxes unm Number Theory 0 December 3rd, 2012 07:05 PM
Greatest of all paradoxes krausebj0 New Users 0 November 25th, 2011 04:50 PM
Resolution of Russell's and Cantor's paradoxes DaniilTeplitskiy Applied Math 35 August 30th, 2011 11:58 PM
Integration difficulties JohnTan Calculus 4 February 15th, 2010 04:03 PM





Copyright © 2017 My Math Forum. All rights reserved.