Statistics Question

[StringJunky]

7 hours ago, scherado said:

I didn't accuse you of asking me to pick from the top or bottom--I'm telling you that I can't pick any card which you made clear is what I am supposed to do. This means that I have accuse YOU of "making up stuff. I hope you understand the difference. You appear not to understand the difference.

You are not adding anything constructive to the conversation, you are trashing it. Your objections are inane. Why don't you let him proceed until it's clear where it's going?

[scherado]

2 hours ago, StringJunky said:

You are not adding anything constructive to the conversation, you are trashing it. Your objections are inane. Why don't you let him proceed until it's clear where it's going?

Did you read my objection to the "chose any card" aspect of the description? Do you have a dispute with my objection?

I await your reply. Thanks.

[Strange]

6 minutes ago, scherado said:

Did you read my objection to the "chose any card" aspect of the description? Do you have a dispute with my objection?

Oh, is that what it was supposed to mean. It was not clear. But I don't really understand what your objection is. You have stated a couple of impossible ways of specifying the card you want to choose (the first and the last). Perhaps you are trying to suggest that there is no algorithmic way of specifying the card. This may be true, but I think the axiom of choice still says we can choose a card from an infinite set.

Which makes me think there could be a solution to the problem in terms of set theory (which is often good way of handling infinities) but I have forgotten nearly all I knew of naive set theory! (Which wasn't much to start with...)

[Dubbelosix]

8 hours ago, Lord Antares said:

Yes, it does mathematically. Let's say we have 5 blue balls and 5 red balls. How do you calculate the probability of randomly picking a blue ball? Easy, it's the number of blue balls divided by the number of total balls. 5/10=0.5 or 50% chance. Picking any card from a full deck? 1 divided by 64. It's very basic math.

This is why 1/infinity is mathematically invalid, because you can't divide by infinity, even though I agree with the value of 1/infinity as a practical approach.

Because I feel like you have taken the time to explain why, I believe I understand it now. I read a bit further, found some analogues of the use of infinity, one particle one is that a probability tends zero as something approaches infinity - even though I know people have taken the time to explain it being undefined, if it had been explained clear cut like this, I may have been more willing to soak in what was being said. 

 

''This is an extension of the principle that a finite string of random text has a lower and lower probability of being a particular string the longer it is (though all specific strings are equally unlikely). This probability approaches 0 as the string approaches infinity. Thus, the probability of the monkey typing an endlessly long string, such as of the digits of pi in order, on a 90-key keyboard is (1/90)^∞ which equals (1/∞) which is essentially 0. At the same time, the probability that the sequence contains a particular subsequence (such as the word MONKEY, or the 12th through 999th digits of pi, or a version of the King James Bible) increases as the total string increases. This probability approaches 1 as the total string approaches infinity, and thus the original theorem is correct.''

 

Can someone explain the last part to me, maybe in a clear concise way?

4 hours ago, StringJunky said:

You are not adding anything constructive to the conversation, you are trashing it. Your objections are inane. Why don't you let him proceed until it's clear where it's going?

I don't know why he keeps telling me to pick from the top or bottom of the pack when there is no end points to an infinity? Really weird to me. People wonder why I struggle understanding people sometimes. 

sorry approaches infinity^ fixed

Edited September 30, 2017 by Dubbelosix

[scherado]

12 hours ago, Dubbelosix said:

I don't know why he keeps telling me to pick from the top or bottom of the pack when there is (sic) no end points to an infinity? Really weird to me. People wonder why I struggle understanding people sometimes. 

sorry approaches infinity^ fixed

I think you understand as indicated in the red text.

Now do you understand my objection? There are no end-points to numerical infinity; therefore, I can't "chose any card." There are two cards that I CAN'T chose.

This is progress! Unless I'm the only one who understands numerical infinity.

Edited October 1, 2017 by scherado

[Strange]

14 minutes ago, scherado said:

Now do you understand my objection? There are no end-points to numerical infinity; therefore, I can't "chose any card." There are two cards that I CAN'T chose.

So you can't choose two cards that don't exist? Now that's a rational objection.

14 minutes ago, scherado said:

Unless I'm the only one who understands numerical infinity.

There is no such thing as "numerical infinity". You may be the only person who doesn't understand.

[Lord Antares]

44 minutes ago, scherado said:

There are two cards that I CAN'T chose.

That's not quite right. Following your logic, there are more than two cards you cannot choose. In fact, there's an infinite amount of them. What about the second to last card? Third to last? Fourth? Etc.

47 minutes ago, scherado said:

There are no end-points to numerical infinity; therefore, I can't "chose any card."

That's not quite right either. For example, what are the end points of the set of positive integers? 1 and infinity, right? You cannot identify the last one but you can clearly identify the first.
If you want to extend it further and compare it with our case of the cards, imagine if, instead of standing in an undefined point along the line of infinite cards, the cards are placed on the table, stacked on one another, extending into infinity. The first one touching the table can be called the first card, while you cannot find the last one. Obviously, ignore the physical impossibilities of this example.

[studiot]

 

1 hour ago, Lord Antares said:

That's not quite right. Following your logic, there are more than two cards you cannot choose. In fact, there's an infinite amount of them. What about the second to last card? Third to last? Fourth? Etc.

That's not quite right either. For example, what are the end points of the set of positive integers? 1 and infinity, right? You cannot identify the last one but you can clearly identify the first.
If you want to extend it further and compare it with our case of the cards, imagine if, instead of standing in an undefined point along the line of infinite cards, the cards are placed on the table, stacked on one another, extending into infinity. The first one touching the table can be called the first card, while you cannot find the last one. Obviously, ignore the physical impossibilities of this example.

Good thinking +1

Edited October 1, 2017 by studiot

[scherado]

5 hours ago, scherado said:

There are two cards that I CAN'T chose.

4 hours ago, Lord Antares said:

 

That's not quite right. Following your logic, there are more than two cards you cannot choose. In fact, there's an infinite amount of them. What about the second to last card? Third to last? Fourth? Etc.

Doesn't the "logic" you assert to have followed from mine prevent the availability of the first card? I have the answer to this question, but will wait for your answer, if that is possible.

Further, your words, "there are more than two cards," most certainly implies that you you agree that I can not chose the two I specify.

I think I'll pause here before I address what remains of your reply, for obvious reasons. I've copied that below for convenience.

4 hours ago, Lord Antares said:

That's not quite right either. For example, what are the end points of the set of positive integers? 1 and infinity, right? You cannot identify the last one but you can clearly identify the first.
If you want to extend it further and compare it with our case of the cards, imagine if, instead of standing in an undefined point along the line of infinite cards, the cards are placed on the table, stacked on one another, extending into infinity. The first one touching the table can be called the first card, while you cannot find the last one. Obviously, ignore the physical impossibilities of this example.

Incidentally, do you think my explanation of the mis-use of "statistics" in the thread's title is "not quite right?"

[Lord Antares]

I have no idea what you are on about. This is a hypothetical scenario, where obviously you cannot specify a card out of an infinite deck because of the impossibility of having an infinite deck. It is implied that the card is chosen randomly. But all of that is besides the point.

1 hour ago, scherado said:

Doesn't the "logic" you assert to have followed from mine prevent the availability of the first card? I have the answer to this question, but will wait for your answer, if that is possible.

Further, your words, "there are more than two cards," most certainly implies that you you agree that I can not chose the two I specify.

 

No, why? In my example of the card deck, you can actually specify an infinite number of cards and you can't specify an infinite number of cards. You see, you can start from the first (the one touching the table) and choose any card in any place relative to the first card. But you can't specify any card in any place relative the the last one. Funny case.

[Strange]

There are an infinite number of integers. There is therefore no first or last integer. You can still "pick a number, any number".

[Lord Antares]

1 hour ago, Strange said:

There are an infinite number of integers. There is therefore no first or last integer. You can still "pick a number, any number".

There is an infinite amount of positive integers, yet there is a first integer. But that's all besides the point. He's complaining about a physical impossibility in a mathematical and hypothetical scenario.

[studiot]

1 hour ago, Strange said:

There are an infinite number of integers. There is therefore no first or last integer. You can still "pick a number, any number".

Antares was correct in that he specified a half open set - the positive integers.

This does indeed have a least or first member - unity.

 

Edit I see I cross posted.

 

Edited October 1, 2017 by studiot

[Strange]

Just now, studiot said:

Antares was correct in that he specified a half open set - the positive integers.

I was commenting on the ridiculousness and irrelevance of Scherzo's objection. Obviously, Lord Antares is also correct that an infinite set can have a start.

[scherado]

4 hours ago, Lord Antares said:

But you can't specify any card in any place relative the the last one

When I wrote yesterday, "My...choice is the one a[t] the top of the stack" I meant what you wrote today, "the last one" (your words) in your post today (about 4 hours ago) This means you are late to the game. Not so funny case.

I will repeat: "choose any card" can't be part of the OP if I can't choose the one that represents the card we know doesn't exist out in infinity-land.

Edited October 1, 2017 by scherado

[Lord Antares]

I seriously don't know what you're talking about.

 

[Strange]

9 minutes ago, scherado said:

I will repeat: "choose any card" can't be part of the OP if I can't choose the one that represents the card we know doesn't exist out in infinity-land.

How about if we change it to: "choose any card that exists"? Or are you claiming that an infinite deck of cards has no cards that exist?

[DrKrettin]

14 minutes ago, Lord Antares said:

I seriously don't know what you're talking about.

 

It's time you joined the IgnoredByScherado club 

[scherado]

2 minutes ago, DrKrettin said:

It's time you joined the IgnoredByScherado club 

He may have submitted his application today,  we shall see: I made the point yesterday that he is making today, if you get my drift.

[DrKrettin]

7 minutes ago, scherado said:

He may have submitted his application today,  we shall see: I made the point yesterday that he is making today, if you get my drift.

This is actually fascinating, because I have never encountered anybody online who is so absurdly arrogant as to attach so much importance to their "ignore list" on the assumption that it might upset anybody included on it. That, plus the obsession with irrelevant detail about some BA qualification (I dare not be more specific) points to some psychological state which needs some clinical attention. I find myself thinking about Trump for some reason.

[swansont]

46 minutes ago, DrKrettin said:

It's time you joined the IgnoredByScherado club 

 

42 minutes ago, scherado said:

He may have submitted his application today,  we shall see: I made the point yesterday that he is making today, if you get my drift.

 

!

Moderator Note

Stick to the subject, please

 

[MigL]

Aside from Scherado confusing the problem, it actually seems simple enough...

Lets simplify by assigning odd numbers to all the blank cards, and even numbers to all the face cards ( Strange has already alluded to this )
All odd numbers, and also even numbers, are countably infinite, which means they can be placed on a one-to-one correspondence with the set of all natural numbers. Which means they are of the same class infinity ( see Cantor ). IE there is no difference in the number of even and odd choices.

The problem then reduces to , 'what is the probability of any number being even ?'. And the probability calculation is not one of dividing by infinity but of having one of two choices odd or even.

Physicists often simplify problems to rework intractable situations, and last time I checked, so did mathematicians...

Edited October 1, 2017 by MigL

[Strange]

2 minutes ago, MigL said:

... but of having one of two choices odd or even.

Or zero.  

Which is why the house always wins.

[MigL]

Not quite...

At  the casino, or the lottery, there are more ways to lose than win.
In this case the two infinities are equal, and, as there are no other choices, the probability is 1 in 2.
( Infinities cannot be a divisor, but the can be cancelled, and it's done all the time when calculating limiting values )
 

[Dubbelosix]

24 minutes ago, MigL said:

Aside from Scherado confusing the problem, it actually seems simple enough...

Lets simplify by assigning odd numbers to all the blank cards, and even numbers to all the face cards ( Strange has already alluded to this )
All odd numbers, and also even numbers, are countably infinite,....

Now this is more interesting - it looks like people are working with their good ol' brain boxes. 

I actually like your solution as well to my problem. If the solution holds, alas, you do not win a prize.