My Math Forum Probability paradoxes and other difficulties

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 April 14th, 2017, 11:21 AM #11 Banned Camp   Joined: Mar 2017 From: . Posts: 338 Thanks: 8 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.
April 15th, 2017, 12:41 AM   #12
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 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?

 April 15th, 2017, 04:52 AM #13 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,305 Thanks: 2443 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.
April 15th, 2017, 05:38 AM   #14
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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.

April 15th, 2017, 12:02 PM   #15
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 Originally Posted by Mariga 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.

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?

April 15th, 2017, 01:05 PM   #16
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 Originally Posted by v8archie 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.

 April 15th, 2017, 01:24 PM #17 Senior Member     Joined: Sep 2015 From: USA Posts: 1,931 Thanks: 999 [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.
April 15th, 2017, 01:40 PM   #18
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 Originally Posted by AplanisTophet $\frac{1}{\infty} = 0$ 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

April 15th, 2017, 01:52 PM   #19
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 Originally Posted by AplanisTophet $\frac{1}{\infty} = 0$ 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.

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 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.

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 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.

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 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.

April 15th, 2017, 02:15 PM   #20
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 Originally Posted by AplanisTophet 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.

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