Is 1 = 0.99999999999...(recurring)

[dingo]

I say not.

Here is my logic.


Let the set of numbers containing all positive integers be called S1

Let the set of numbers containing all non-negative integers be called S2

S2 contains zero, S1 does not.

S1 does not contain the same number of items as S2.

The difference between the two is almost insignificant, but it exists.
If you do number matching between S1 and S2, then you will be left with one un-matched number is S1.

[thehod]

Mathmatically, of course it doesn't.

As far as the rest of the world cares? What's the damn difference?

[dingo]

Mathmatically, of course it doesn't.

As far as the rest of the world cares? What's the damn difference?

See, the thing is, there are some (apparent) mathematical proofs that they are equal.

This started in another thread.

[DarkBlade]

Actually, mathmatically, it does.

x = 0.99999...
10x = 9.9999...
10x − x = 9.9999... − 0.99999...
9x = 9
x = 1

See here: http://polymathematics.typepad.com/polymath/2006/06/no_im_sorry_it_.html

[Night]

I thought you were going to go through the 1/3 (ie .3333333 recurring)times 3 thing which makes more sense than the logic you propose.

[DarkBlade]

I thought you were going to go through the 1/3 (ie .3333333 recurring)times 3 thing which makes more sense than the logic you propose.
I didn't learn that one in HS. Share?

[Dreadstar]

DB got one.

Night's is:

What is .9999999... divided by 3?

[Dreadstar]

Mathmatically, of course it doesn't.

Yeah. It does.

[Agent Helix]

Actually, mathmatically, it does.

x = 0.99999...
10x = 9.9999...
10x − x = 9.9999... − 0.99999...
9x = 9
x = 1

See here: http://polymathematics.typepad.com/polymath/2006/06/no_im_sorry_it_.html

It may just be a lack of sleep talking, but this is actually the most interesting thing I've seen all week.

[dingo]

Actually, mathmatically, it does.

x = 0.99999...
10x = 9.9999...
10x − x = 9.9999... − 0.99999...
9x = 9
x = 1

See here: http://polymathematics.typepad.com/polymath/2006/06/no_im_sorry_it_.html

If I remember correctly the use of the equals sign in those sort of mathematical equation is technically incorrect. It should be an equivelent symbol (three horizontal lines).

For example

x = y
1/x = 1/y

One works for x = 0, the other is undefined (or defined using complex numbers)

[Dreadstar]

I say not.

Here is my logic.


Let the set of numbers containing all positive integers be called S1

Let the set of numbers containing all non-negative integers be called S2

S2 contains zero, S1 does not.

S1 does not contain the same number of items as S2.

The difference between the two is almost insignificant, but it exists.
If you do number matching between S1 and S2, then you will be left with one un-matched number is S1.


I don't get it. Sorry, but you're going to have to do a bit more work to connect that one to the infinite progression .99999... if you don't mind.

[dingo]

DB got one.

Night's is:

What is .9999999... divided by 3?

Lets re-write this in english.

What is (a numer approaching one) divided by three?

A numer approaching one third

[Winslow]

What is .9999999... divided by 3?

Makes sense to me.


DB's link has some other proofs.

[Agent Helix]

Oh, by the way, you're all nerds, and I'm gonna kick your asses in Gym class.

[dingo]

I don't get it. Sorry, but you're going to have to do a bit more work to connect that one to the infinite progression .99999... if you don't mind.

I am furiously trying to find a link to that. ;)

[Winslow]

Lets re-write this in english.

What is (a numer approaching one) divided by three?

A numer approaching one third

Actually no, it is one third.

[DarkBlade]

DB got one.

Night's is:

What is .9999999... divided by 3?

Ahh! I see.

.9999.../3=.333...
.333...=1/3
(1/3)*3=1


Hnh. That one's much more tidy. And probably would have made me scowl at the whiteboard for much less time than the 10x one my teacher shared.

[Dreadstar]

Lets re-write this in english.

What is (a numer approaching one) divided by three?

A numer approaching one third

You know, nevermind.

I quit teaching college level math 30 years ago because of frustrating inability to get through cement semantics.

Now I remember why.

[Dreadstar]

Sorry guys, I'm out, just follow DB's lkink to be educated.

[dingo]

Actually no, it is one third.

Only if you assume that 0.999999.... = 1

Which is what we are arguing.

It using an assumption to prove itself.

In english....

One divided by three is one third.

A number approaching one divided by three is a number approaching one third.

[Winslow]

Only if you assume that 0.999999.... = 1

Which is what we are arguing.

It using an assumption to prove itself.

In english....

One divided by three is one third.

A number approaching one divided by three is a number approaching one third.

Read DB's link dingo.

It's a slam dunk.

[dingo]

Read DB's link dingo.

It's a slam dunk.

I did, both today and a long time ago.

[Mac Danny]

.999999999............ is one for me until I need that extra point .0000000000000000000000000000000000000000000000000 000000000000000................................... 1

[DarkBlade]

It's not an assumption though, not with the algabraic 10x proof up there.

Granted, it's primarily a mental exercise, or something involving energy perhaps that just continues to get exponentially smaller or something?

[Tadhg Adams]

Only if you assume that 0.999999.... = 1

No you're not assuming anything. .999... divided by 3 is .333... That's not an assumption. Start dividing it out, you'll see it's .333... until you get bored of dividing.

[Winslow]

I did, both today and a long time ago.

I don't know what to say . . you can't argue with the mathematical proofs.

I guess you can, but you're pounding your head against the wall of mathematical definitions.

[dingo]

From the link provided by DB


".9 repeating doesn't equal 1, it gets closer and closer to 1."

May I remind you that .9 repeating is a number. That means it has it's place on the number line somewhere. Which means that it's not "getting" anywhere. It doesn't move. It either equals 1 or it doesn't (it does of course), but it doesn't "get" closer to 1.


If I reword that to ".9 repeating doesn't equal 1, it is infintesimally close to 1."
then what is the argument against it now?

[Dreadstar]

No you're not assuming anything. .999... divided by 3 is .333... That's not an assumption. Start dividing it out, you'll see it's .333... until you get bored of dividing.

Unimportant.

He's already incorrectly dismissed .33333... as not equaling 1/3.

[Huh?]

.9999 repeating is 1 mathematically. It is not 1 physically.

It is actually a very good illustration of the fundamental problem with mathematics as a representation of the physical world. You can do things with number that you can't do with objects.

[Ed Cunard]

Unless there's some way this makes my penis bigger mathematically, I don't care.

[The Mirrorball Man]

.9999 repeating is 1 mathematically. It is not 1 physically.

It is actually a very good illustration of the fundamental problem with mathematics as a representation of the physical world. You can do things with number that you can't do with objects.

But in some branches of physics, these "objects" you're talking about are almost pure mathematical constructs anyway.

[Huh?]

Unless there's some way this makes my penis bigger mathematically, I don't care.I am afraid, for you, that 2" will always be 2" ... sorry.

[Huh?]

But in some branches of physics, these "objects" you're talking about are almost pure mathematical constructs anyway.Very true, but it is universally recognized that a .9999 repeating piece of an apple will never be the same thing as 1 apple.

[Tadhg Adams]

If I reword that to ".9 repeating doesn't equal 1, it is infintesimally close to 1."
then what is the argument against it now?

The same mathematical proofs you're already ignoring.

[dingo]

Here (http://en.wikipedia.org/wiki/Infinity#Infinity_in_set_theory) is a link to the theory I have seen used to argue this.

I know it is Wiki, but it was the first I could find.

Basically it means that there are differences in infinities that cannot be paired up on a one-to-one basis.

From Wiki


Dedekind's approach was essentially to adopt the idea of one-to-one correspondence as a standard for comparing the size of sets, and to reject the view of Galileo (which derived from Euclid) that the whole cannot be the same size as the part. An infinite set can simply be defined as one having the same size as at least one of its " proper" parts; this notion of infinity is called Dedekind infinite.


So in my first example....

S1= all positive integers.
S2 = all non negative integers (i.e. S1 and the number 0)

S1 is conceptually equivelent to the number 0.9999999...
S2 is conceptually equivelent to the number 1

Does this make sense?

[Paragon Kobold]

The key lies in finding the solution to 1 - 0.99999...

:)

[Dreadstar]

Very true, but it is universally recognized that a .9999 repeating piece of an apple will never be the same thing as 1 apple.

That's because infinity is a *concept* and the *concept* does not exist in reality.

The biggest problem with understanding this entire execise is exactly that: Inability or unwillingness to accept infinity as a concept.

[Winslow]

If I reword that to ".9 repeating doesn't equal 1, it is infintesimally close to 1."
then what is the argument against it now?

The problem is, "infintesimally close", has no defined meaning.

The proofs are conclusive.

Edit: Dread said it better.

[dingo]

Unimportant.

He's already incorrectly dismissed .33333... as not equaling 1/3.

No I haven't.

Please read what I have said again...

If 1 divided by three equals one third, then a number infintesimally smaller than one divided by three equals a number infintesimally smaller than one third.

[Dreadstar]

Here (http://en.wikipedia.org/wiki/Infinity#Infinity_in_set_theory) is a link to the theory I have seen used to argue this.

I know it is Wiki, but it was the first I could find.

Basically it means that there are differences in infinities that cannot be paired up on a one-to-one basis.

From Wiki



So in my first example....

S1= all positive integers.
S2 = all non negative integers (i.e. S1 and the number 0)

S1 is conceptually equivelent to the number 0.9999999...
S2 is conceptually equivelent to the number 1

Does this make sense?

I guess I'm back in.



If .99999... is a "PART" of 1, give me a number between .99999... and 1. It should be easy if it's only a part.

[The Mirrorball Man]

That's because infinity is a *concept* and the *concept* does not exist in reality.
What do you mean? Do concepts exist somewhere _outside_ of reality?

[Dreadstar]

No I haven't.

Please read what I have said again...

If 1 divided by three equals one third, then a number infintesimally smaller than one divided by three equals a number infintesimally smaller than one third.

What is 1 divided by 3?

[Dreadstar]

What do you mean? Do concepts exist somewhere _outside_ of reality?

Now I have to define "concept?"

[dingo]

I guess I'm back in.



If .99999... is a "PART" of 1, give me a number between .99999... and 1. It should be easy if it's only a part.

I'm sorry I don't understand what you mean by PART.

I am not trying to be obtuse, I just don't want to go off on a tangent.

[dingo]

What is 1 divided by 3?

One third.

[Ed Cunard]

I am afraid, for you, that 2" will always be 2" ... sorry.

Sorry? You type something like that and say "sorry?"

You should be saying "congratulations!" Two inches? IT IS WORKING ALREADY!

[Dreadstar]

The key lies in finding the solution to 1 - 0.99999...

Another one, thank you.

[The Mirrorball Man]

Now I have to define "concept?"

Oh yeah, you bet.

[Dreadstar]

One third.

Don't be obtuse. Express it as a decimal or go away.

[dingo]

Don't be obtuse. Express it as a decimal or go away.

Look, I am really trying hard not to take this personally.

No I will not express it as a decimal. Here is why.

Expressing one third as a decimal involves the human mind interpreting an aspect of infinity.

The entire basis of what I am saying is that when you stop assuming infinty just means "the biggest number ever" and start thinking about how infinity applies to the concept of one-to-one pairing, then 1 may not equal 0.99999...

[Dreadstar]

I'm sorry I don't understand what you mean by PART.

Unbelievable:

"Dedekind's approach was essentially to adopt the idea of one-to-one correspondence as a standard for comparing the size of sets, and to reject the view of Galileo (which derived from Euclid) that the whole cannot be the same size as the part. An infinite set can simply be defined as one having the same size as at least one of its " proper" parts; this notion of infinity is called Dedekind infinite."?

I am not trying to be obtuse, I just don't want to go off on a tangent.

I'm sorry, but I really think you are being deliberately obtuse. It's the only explanation I can see.

[Huh?]

I'm sorry I don't understand what you mean by PART.

I am not trying to be obtuse, I just don't want to go off on a tangent.Since .999... is infinite, you can never name something between it and 1. Therefore, in math they are equal.

[Dreadstar]

Expressing one third as a decimal involves the human mind interpreting an aspect of infinity.


If you refuse to conceptualize infinity then this whole thread is useless.
If you cannot conceptualize infinity then this whole thread is useless.
If infinity does not exist to you then this whole thread is useless.

[Dreadstar]

Since .999... is infinite, you can never name something between it and 1. Therefore, in math they are equal.

Number 4, in the series.

[dingo]

Since .999... is infinite, you can never name something between it and 1. Therefore, in math they are equal.

Thankyou. And no, in spite of what it looks like, I was not being deliberately obtuse. Dreadstar, you have a point here that I cannot see an answer to. I will think about it.

If you refuse to conceptualize infinity then this whole thread is useless.
If you cannot conceptualize infinity then this whole thread is useless.
If infinity does not exist to you then this whole thread is useless.

I do not refuse to conceptualise infinity, I am saying that we should conceptualise infinity in terms of having diferent levels of infinity, which is very hard to do.

[The Mirrorball Man]

If you refuse to conceptualize infinity then this whole thread is useless.
If you cannot conceptualize infinity then this whole thread is useless.
If infinity does not exist to you then this whole thread is useless.

Well, almost all threads are useless, but that would be an interesting way to prove it once and for all.

[The Mirrorball Man]

Thankyou. And no, in spite of what it looks like, I was not being deliberately obtuse. Dreadstar, you have a point here that I cannot see an answer to. I will think about it.



I do not refuse to conceptualise infinity, I am saying that we should conceptualise infinity in terms of having diferent levels of infinity, which is very hard to do.

I still don't fully understand what you mean. I'm not a mathematician, but from what I remember from school, there are "countable" infinites and "uncountable" infinites. 0.99999... is uncountable. Even if it weren't, I'm not sure why it would lead one to think that it isn't equal to 1. Or were you thinking of something else?

[Dreadstar]

Thankyou. And no, in spite of what it looks like, I was not being deliberately obtuse. Dreadstar, you have a point here that I cannot see an answer to. I will think about it.

I *do* hope you think about it, but keep the idea behind the scientific method in mind. Don't try to find answers to fit what you THINK is right. Just find the answers.


I do not refuse to conceptualise infinity, I am saying that we should conceptualise infinity in terms of having diferent levels of infinity, which is very hard to do.

No, I understand the idea of "different levels of infinity" from a theoretical math standpoint. The problem is, from a purely straight up and lay it on the line mathematical standpoint there is no difference between "infinity" and "1,356 times infinity." This is one of the main failures in grasping the concept of infinity. It always HAS been. See, infinity, like ZERO, is a conceptual "number." Different rules apply to concepts. The fact that zero isn't a number, but is rather the numerical depiction of a null, is the reason you cannot divide by it.

[Matt Algren]

This is a whole lot more fun that following a politics fight thread. Seriously. I love math proofs. I am a geek.

If 1/7 + 6/7 = 7/7 = 1, and
0.14285714285714285714285714285714... + 0.85714285714285714285714285714286... = 0.999999999999999999999...
Then 1 necessarily = 0.999999999999999999999... .

Dingo, follow DarkBlade's link. It's all there, including some back and forth in the comments section that mirrors this discussion.

[Paragon Kobold]

I don't see how differences between infinite sets have any bearing on the provable fact that 1 = 0.9999999999...

[Dreadstar]

I don't see how differences between infinite sets have any bearing on the provable fact that 1 = 0.9999999999...

He was trying to intimate that by Dedekind's reasoning, .999999... is a part of 1, and not 1.

I think.

If not, then I still don't understand it myself.

[Dreadstar]

Now that I've had my coffee and a good constitutional (*ahem*), I'm getting the wild notion of getting into the REAL flamewar territory and going for the Monte Hall problem!

[dingo]

I don't see how differences between infinite sets have any bearing on the provable fact that 1 = 0.9999999999...

What I am trying (unsucsessfully) to get across is this.

Imagine a number that is infinetly small, but greater than zero...
which I know acording to this debate would be equal to zero, I am not trying to prove anything, just explain my logic.

Now, the number of these that you would have to add together to get 1 (infinity) would be a different order of infinity than you would need to add together to get 0.9999999....

[dingo]

He was trying to intimate that by Dedekind's reasoning, .999999... is a part of 1, and not 1.

I think.

If not, then I still don't understand it myself.

That is a lot more succinct than my mental concept, which I will be the first to admit swings around on it's definition of infinity.

[Sir Tim Drake]

Where is coke & comics when you need him?

[dingo]

Where is coke & comics when you need him?

Sleeping. The lazy bugger.

[Agent Helix]

http://www.luminomagazine.com/mw/storyimages/347_wide.jpg
NEEEEEEEEEEEERDS!

[Dreadstar]

Now, the number of these that you would have to add together to get 1 (infinity) would be a different order of infinity than you would need to add together to get 0.9999999....

OK, I think I can sort this.

You want to see the "infinity" relating to the VALUE of the number.

The infinity of .999999... and .000000000...1 does NOT refer to the value of the number, but rather the PROGRESSION of digits REPRESENTING that value.

In this case, the infinite progression .9999999... is equivalent to the infinte progression .000000...1 --- in digital representation. i.e. they have the same number of places right of the decimal point: infinity.

[dingo]

Now that I've had my coffee and a good constitutional (*ahem*), I'm getting the wild notion of getting into the REAL flamewar territory and going for the Monte Hall problem!

Is that the one with probabilities that change depending on whether or not you define the events as discrete or not?

[Winslow]

All I can say, is that as a future high school math teacher, this thread is making me very sadistic.

[dingo]

OK, I think I can sort this.

You want to see the "infinity" relating to the VALUE of the number.

The infinity of .999999... and .000000000...1 does NOT refer to the value of the number, but rather the PROGRESSION of digits REPRESENTING that value.

In this case, the infinite progression .9999999... is equivalent to the infinte progression .000000...1 --- in digital representation. i.e. they have the same number of places right of the decimal point: infinity.

Yes. Pretty much.

See, where I get really confused, is that logically, there would be no difference between...

0.00000000...1
and
0.00000000...2

Edit: even though neither is really a logical representation of the concept I have in mind.

[Dreadstar]

Is that the one with probabilities that change depending on whether or not you define the events as discrete or not?

Yes, basically.

[Tadhg Adams]

All I can say, is that as a future high school math teacher, this thread is making me very sadistic.

As a current high school employee. I say good luck at teaching kids algebraic expressions let alone anything more complex.

[Dreadstar]

Yes. Pretty much.

See, where I get really confused, is that logically, there would be no difference between...

0.00000000...1
and
0.00000000...2

There isn't, I don't think. Though I could be wrong. This gets into theoretical I think. I'm too old to remember too much of the theoretical.

[Winslow]

As a current high school employee. I say good luck at teaching kids algebraic expressions let alone anything more complex.

Oh, I was thinking more along the lines of messin' with their heads.

When Dread first posed this (i.e. 1=.9999~) in the "Nerd Commandments" thread, it confused me, but once I read DB's link . . it was pretty clear.

[dingo]

Can you do maths with an infinity symbol?
I may be making a fool of myself if you can't but....

I am going to use ampersand (&) for 'infinity' symbol
First an assumption.
1- (1/&) is equivelent to 0.9999999999...

so.

1 = 1 - (1/&)
take 1 from each side gives you.
0 = -(1/&)
multiply both by -1
0 = 1/&
multiply both by &
0 = 1


Edit: The above comes under the heading of "I know this can't be right, please tell me why".
Edit 2: Stupid typo.

[Dreadstar]

The failure is that symbolics *must* take into account situations where conceptual numbers occur.

if 0 = 1/&

then the number 1/& becomes the conceptual i.e. ZERO, a null set. & times 0 equals 0.

Secondly, in the final line, if you did not know that 1/& was equal to zero, 1/& multiplied by & doesn't equal &. It equals 1. &/&.

So, you either have an error or you've just proven that 0 = 1.

[dingo]

The failure is that symbolics *must* take into account situations where conceptual numbers occur.

if 0 = 1/&

then the number 1/& becomes the conceptual i.e. ZERO, a null set. & times 0 equals 0.


But since I am attempting to prove or disprove whether or not 1/& is equal to 0 then having 0 = 1/& as an axiom* kind of makes the whole exercise pointless doesn't it? (regardless of any bias on my part)

How can I get around that?

* Am I using that word right? I mean fundamental assumption.


Secondly, in the final line, if you did not know that 1/& was equal to zero, 1/& multiplied by & doesn't equal &. It equals 1. &/&.

So, you either have an error or you've just proven that 0 = 1.

Sorry, that was a typo.

[Dreadstar]

But since I am attempting to prove or disprove whether or not 1/& is equal to 0 then having 0 = 1/& as an axiom* kind of makes the whole exercise pointless doesn't it? (regardless of any bias on my part)

How can I get around that?

* Am I using that word right? I mean fundamental assumption.

I'm sorry, I can't answer that.

See, in shorthand (in my memory, at least) anything divided by zero is infinity. anything divided by infinity is zero. The practical side of math says you CANNOT do that. The conceptual side says that conceptual numbers such as 0 and & must be treated as special cases. in other words, you can't do practical math with conceptual numbers.

BTW, what I wrote in the last post was in regarding the (1/& * &) as a varible, as in (1/X * X). As I see it, dividing ANYTHING by infinity equals zero, incuding infinity. So &/& = 0. You really have to be careful of special cases. Special cases is why that old "I can prove that 1 = 2" proof fails. In fact it fails for precisely the same reason: Failure to take into account that even though you're using variables, as soon as you run into a special case (in this case, dividing by zero) you must take that into account. It's the reason why when you graph a function like 1/X, the graph may have a gap at zero. Does this make the function incorrect? or false or whatever the word is? No. What it says is: "Except at zero, which is a special case."

[dingo]

So basically I can't get around that axiom any more readily than I could get around divide by zero.
I don't understand complex numbers at all.

[HomerJay]

BTW, what I wrote in the last post was in regarding the (1/& * &) as a varible, as in (1/X * X). As I see it, dividing ANYTHING by infinity equals zero, incuding infinity. So &/& = 0. You really have to be careful of special cases. Special cases is why that old "I can prove that 1 = 2" proof fails. In fact it fails for precisely the same reason: Failure to take into account that even though you're using variables, as soon as you run into a special case (in this case, dividing by zero) you must take that into account. It's the reason why when you graph a function like 1/X, the graph may have a gap at zero. Does this make the function incorrect? or false or whatever the word is? No. What it says is: "Except at zero, which is a special case."
http://img.photobucket.com/albums/v655/HomerJay64/scanners-exploding-head-3.jpg

[Dreadstar]

http://img.photobucket.com/albums/v655/HomerJay64/scanners-exploding-head-3.jpg


http://i34.photobucket.com/albums/d108/Aliaswit/111111a.jpg + http://i34.photobucket.com/albums/d108/Aliaswit/1111b.jpg = http://i34.photobucket.com/albums/d108/Aliaswit/22222222.jpg

Better?

[Guapo Méndez]

http://i34.photobucket.com/albums/d108/Aliaswit/111111a.jpg + http://i34.photobucket.com/albums/d108/Aliaswit/1111b.jpg = http://i34.photobucket.com/albums/d108/Aliaswit/22222222.jpg

Better?


Now that I understand.
That I can get under.
That's some tangible proof.
That's something that brings a smile to my face.

etc, etc.

[Joe Acro]

Actually, mathmatically, it does.

x = 0.99999...
10x = 9.9999...
10x − x = 9.9999... − 0.99999...
9x = 9
x = 1

See here: http://polymathematics.typepad.com/polymath/2006/06/no_im_sorry_it_.html
I keep looking at this and thinking it proves nothing.

The system of solving is incorrect.

[Michael P]

Now that I understand.
That I can get under.
That's some tangible proof.
That's something that brings a smile to my face.

etc, etc.

There's a concept I can grasp.

[Guapo Méndez]

Not too mention that in spanish, "sine" is "seno", which means breast. "Theta" in spanish is writen "teta" which also means breast.

You can imagine what a hoot math classes were when the teacher asked us to find the "sine" of theta. Especially when we asked the girls in our class if they could show us their "sine" or at least, their "theta."

[HomerJay]

http://i34.photobucket.com/albums/d108/Aliaswit/111111a.jpg + http://i34.photobucket.com/albums/d108/Aliaswit/1111b.jpg = http://i34.photobucket.com/albums/d108/Aliaswit/22222222.jpg
Better?
Much.
Now that I understand.
That I can get under.
That's some tangible proof.
That's something that brings a smile to my face.

etc, etc.

There's a concept I can grasp.
*insert joke re: the number 69 here*

[Sir Tim Drake]

*insert joke re: the number 69 here*

What's the square root of 69?

8 something.

[jessecuster3]

What's the square root of 69?

8 something.

You what a cinderella 69 is ?


a 69 and then at midnight turns into a 6 pack

[Chris Nowlin]

I've been waiting months for a thread about math!

[jessecuster3]

I've been waiting months for a thread about math!

Uh oh, I think we all about to get edjumicated.

[Chris Nowlin]

Actually, mathmatically, it does.

x = 0.99999...
10x = 9.9999...
10x − x = 9.9999... − 0.99999...
9x = 9
x = 1

See here: http://polymathematics.typepad.com/polymath/2006/06/no_im_sorry_it_.html

You're absolutely correct. And this is the cleanest way to see it. The fact that 1/3=.333333... more properly follows from this observation, rather than the reverse.

The other way to see this is, if you've taken any calculus, is to think of the number .999999 as the limit of the sequence:

9/10,99/100,999/1000,...

As you go to infinity, the limit is .99999..., by definition. That's the only thing the crazy ellipse can mean.

But, one can prove mathematically that the limit is 1. The point being numbers like 99999999/10000000 (pretend the number of zeroes and nines match) are damn close to one and getting arbitrarily closer.

One sequence can only have one limit point. So 1=.999999.... Equality is correct.

[HomerJay]

I've been waiting months for a thread about math!
That's a first-ballot "nerdiest posts of all-time" hall-of-famer.

[Dreadstar]

The other way to see this is, if you've taken any calculus, is to think of the number .999999 as the limit of the sequence:

Path number 5 in the thread, Calculus.

[Ray R.]

You're absolutely correct. And this is the cleanest way to see it. The fact that 1/3=.333333... more properly follows from this observation, rather than the reverse.

The other way to see this is, if you've taken any calculus, is to think of the number .999999 as the limit of the sequence:

9/10,99/100,999/1000,...

As you go to infinity, the limit is .99999..., by definition. That's the only thing the crazy ellipse can mean.

But, one can prove mathematically that the limit is 1. The point being numbers like 99999999/10000000 (pretend the number of zeroes and nines match) are damn close to one and getting arbitrarily closer.

One sequence can only have one limit point. So 1=.999999.... Equality is correct.

That deserves a polite golf clap. Nice show.

I hated math until I got to college and started doing derivatives and advanced calculus. It's actually like jogging or weight-lifting for your brain. (And I'm out of shape....)

[Chris Nowlin]

That's because infinity is a *concept* and the *concept* does not exist in reality.

The biggest problem with understanding this entire execise is exactly that: Inability or unwillingness to accept infinity as a concept.

Reality sucks. Math is cool. Reality must be left at home, where it belongs.

[Chris Nowlin]

So in my first example....

S1= all positive integers.
S2 = all non negative integers (i.e. S1 and the number 0)

S1 is conceptually equivelent to the number 0.9999999...
S2 is conceptually equivelent to the number 1

Does this make sense?

I actually do see what you're saying. As an undergraduate, I spent a lot of brainpower thinking about this. Hypothesizing that the difference between 1 and .9999... would be the smallest positive number.

Unfortunately, the whole of calculus is based on the idea that there is no smallest positive number. Eventually, I grew old and went to graduate school, and lost all sense of wonder, and I accepted that .9999...=1 and I would let them teach me the math they way "the man" thought it was. But that doesn't mean you should give up. Math and science aren't about accepting what is known, but challenging. That's where innovation comes from.


I do not refuse to conceptualise infinity, I am saying that we should conceptualise infinity in terms of having diferent levels of infinity, which is very hard to do.

Actually, there are, which is really, really cool, but the number you're talking about can be realised with a countably infinite sequence, and that's the smalllest kind of infinity.

[Gilda Dent]

That deserves a polite golf clap. Nice show.

I hated math until I got to college and started doing derivatives and advanced calculus. It's actually like jogging or weight-lifting for your brain. (And I'm out of shape....)

I loved math until I got to college and ran into calculus and discovered I'd forgotten everything I learned in trig, despite getting an A. It's why I switched from math to elementary eduction and reading.

[Shellhead]

This thread reminds me of a party back when I was in college. A friend of mine was visiting from Purdue, where he majored in both electrical engineering and physics. After he had a few drinks, he was talking to somebody who was struggling with a class. Within minutes, he was scribbling formulas on napkins and raving about "Marvelous Diffy Q!" aka differential equations. I guess he had a great time, at least until the puking started.

[Guapo Méndez]

This thread reminds me of a party back when I was in college. A friend of mine was visiting from Purdue, where he majored in both electrical engineering and physics. After he had a few drinks, he was talking to somebody who was struggling with a class. Within minutes, he was scribbling formulas on napkins and raving about "Marvelous Diffy Q!" aka differential equations. I guess he had a great time, at least until the puking started.

What? He couldn't figure the numbers on the projectile vomit trajectory?

[Shellhead]

What? He couldn't figure the numbers on the projectile vomit trajectory?

No, he learned a hard lesson about drinking heavily on an empty stomach. They really should cover that during freshmen orientation.

[Loren]

I say not.

Here is my logic.


Let the set of numbers containing all positive integers be called S1

Let the set of numbers containing all non-negative integers be called S2

S2 contains zero, S1 does not.

S1 does not contain the same number of items as S2.

The difference between the two is almost insignificant, but it exists.
If you do number matching between S1 and S2, then you will be left with one un-matched number is S1.

Actually, that's not correct. S1 and S2 are both countably infinite. They can be matched one-to-one, and are the "same" infinity, which is called Aleph-null (http://en.wikipedia.org/wiki/Aleph_number#Aleph-null). And despite your gut feeling above, the set of all integers (all positive AND negative integers combined) is the same infinity as well.

On the other hand, the set of all *real* numbers is a "larger" infinity. Real numbers can't be matched one-to-one with integers.

[Guapo Méndez]

And 1/3rd of Infinity is still Infinity.

[Chris Nowlin]

The other thing missing in this thread was a precise definition of .99999...

You're trying to work with an intuitive idea of what that notation should mean, when a precise mathematical one is needed. Otherwise, you're relying on intuition, and when it comes to math, especially dealing with the infinite, intuition is rubbish and needs to be left at home. Precision and careful logic are needed.

Any actual definition for that symbol I can write down for that symbol, .999..., would immediately imply that it is 1 (such as the limit of the sequence I mentioned above). You may come up with a new definition for that symbol that matches your intuition and makes it strictly less than one. But without precision, we only have mental masturbation.

[Chris Nowlin]

And 1/3rd of Infinity is still Infinity.

For infinity, think about being arbitrarily large. The best way to think about things that can be arbitrarily large, as opposed to things that are comparatively small, is to imagine my penis, then contrast it with Ed's.

[Loren]

And 1/3rd of Infinity is still Infinity.

Exactly. In fact, that's another way to phrase the relationship between the set of all integers, and the set of all positive integers. Since the first set has all the members of the second set, PLUS negative counterparts to every number in the second set, one would think the first set would be 2X the size of the second set.

The only problem is, 2x infinity is still infinity.

And for more fun, the set of all rational numbers (integers plus fractions) is the same infinity as the set of all integers. Even though there are an infinite number of fractions between 1 and 2.

[Loren]

Unless there's some way this makes my penis bigger mathematically, I don't care.

Measure it in Base 5. In centimeters.

[Chris Nowlin]

And for more fun, the set of all rational numbers (integers plus fractions) is the same infinity as the set of all integers. Even though there are an infinite number of fractions between 1 and 2.

Another way to say that is to note that the rational numbers take up exactly 0% of the real number line.

On top of that, every number man could presently hope to write down, (8, 2/3, pi, pi+2, ln(7), e^pi, etc.), that whole set takes up 0% of the number line.

We've gotten a handle on 0% of the real numbers out there!

[Ray R.]

I loved math until I got to college and ran into calculus and discovered I'd forgotten everything I learned in trig, despite getting an A. It's why I switched from math to elementary eduction and reading.

I wish I had made that switch from law to elementary education and reading. Not too late, I suppose.

And I never could have majored in math. I just took an advanced calculus for non-math majors my senior year to fill out the course load and because it only met once a week (plus I think it was pass/fail). So the pressure was definitely off.

I enjoyed it. How's it go....nice place to visit, but I wouldn't want to live there. Admire the thinking processes, though.

[Dreadstar]

So, you go on Let's Make a Deal! and Monte offers you this deal:

You may choose one of three doors. Behind one door is the grand prize. Behind 2 doors are goats. After you choose, I will open one door to reveal a goat. I will then have the option to change your pick. Assuming you're not Tin Man and immediately go for the revealed goat, what do you do?

A- Keep your door.

B- Switch doors.

C - It doesn't matter.

[Chris Nowlin]

Which door was opened? Which one had I chosen?

[Huh?]

So, you go on Let's Make a Deal! and Monte offers you this deal:

You may choose one of three doors. Behind one door is the grand prize. Behind 2 doors are goats. After you choose, I will open one door to reveal a goat. I will then have the option to change your pick. Assuming you're not Tin Man and immediately go for the revealed goat, what do you do?

A- Keep your door.

B- Switch doors.

C - It doesn't matter.C. It doesn't matter

[Chris Nowlin]

So, the goat is behind door A, or it isn't. But the odds of it being back there just jumped from 33% to 50%. But the goat hasn't moved. The answer is already determined. So really the odds are either 100% or 0%.

[Dreadstar]

Answer:

B. ALWAYS switch doors.

It's not intuitive, but yeah, you have a 66% chance by picking the remaining door instead of the 33% you started with.

[Huh?]

So, the goat is behind door A, or it isn't. But the odds of it being back there just jumped from 33% to 50%. But the goat hasn't moved. The answer is already determined. So really the odds are either 100% or 0%.No, the odd's that the goat is behind a went from 66% to 50%.

[Huh?]

Answer:

B. ALWAYS switch doors.

It's not intuitive, but yeah, you have a 66% chance by picking the remaining door instead of the 33% you started with.Does not compute for me. What difference does it make?

[Chris Nowlin]

No, the odd's that the goat is behind a went from 66% to 50%.

Oh, yeah. I forgot there were 2 goats. I suck.

I was thinking about the prize. I forgot the goat wasn't the prize

[Chris Nowlin]

Answer:

B. ALWAYS switch doors.

It's not intuitive, but yeah, you have a 66% chance by picking the remaining door instead of the 33% you started with.

You're correct, mostly. You're not factoring in the divine inspiration that led me to be 100% about the door I chose the first time.

From your first pick, odds that you were right were 33%.

Suppose you were wrong. Odds of that are 66%. Now Dread is showing you where the goat is. If you were wrong to begin with, the other door is correct.

However, it's not taking into account that sometimes you want to put it all on 34 black.

[Dreadstar]

3 doors. If you choose one, what are your chances of having the prize? 1 in 3. What are the chances that Monte will have the prize? 2 in 3. Monte will ALWAYS have a goat to show you. ALWAYS. This does not change the outcome of the game, and thus he can show you the goat with impugnity. Since showing you the goat has NO bearing on the outcome (it's 100% positive that Monte will have a goat to show you, every time), Monte STILL has the 2 in 3 odds of having the prize.

Thus, if you switch you double your odds.

[jessecuster3]

OOO I like this game.


So I have 3 doors, but only one door has the prize. So I have a 33% chance of getting it right. You show me one wrong door, so my chances are now 50%. Now I have 2 doors to choose from, the one I picked originally or the one un-picked one. How much is the banker offering? Oh wait wrong show. So why would you switch doors, it shouldn't matter at this point ?

[Chris Nowlin]

Can you do maths with an infinity symbol?
I may be making a fool of myself if you can't but....

I am going to use ampersand (&) for 'infinity' symbol
First an assumption.
1- (1/&) is equivelent to 0.9999999999...

so.

1 = 1 - (1/&)
take 1 from each side gives you.
0 = -(1/&)
multiply both by -1
0 = 1/&
multiply both by &
0 = 1


Edit: The above comes under the heading of "I know this can't be right, please tell me why".
Edit 2: Stupid typo.



In general, you can't treat 1/infinity or infinity as a number. You loosely can, and much does work. But your error there is in assuming 0*&=0, because you think 0*anything=0.

As has been noted, infinity*anything=infinity. So &*7=&, etc.

So when you multiply 0*&, you get immovable object times irresistible force. It's not clear whether is should be 0, or infinity, or somewhere in between.

It can be anything.

Think of 1/x and 7x, and let x approach infity. Either way 1/x*7x=7. Yet, in limits, 1/x->0, and 7x->infinity, so you should get 0*infinity. I showed that 0*infinity=7.

That's kind of where your fallacy occurred.

Oh, and in assuming that 0 does not equal 1. Never assume.

[Dreadstar]

... So why would you switch doors, it shouldn't matter at this point ?

I would switch doors because I get Monty's 67% chance instead of my 33%.

[Chris Nowlin]

I would switch doors because I get Monty's 67% chance instead of my 33%.

Ignoring the fact that I'm always right to begin with; I always beat the odds!

[Dreadstar]

As has been noted, infinity*anything=infinity. So &*7=&, etc.

So when you multiply 0*&, you get immovable object times irresistible force. It's not clear whether is should be 0, or infinity, or somewhere in between.

Really? That's changed since the 70s then. Back then it was 0 * & = 0 because you're saying "I have no infinities" or something of that sort.

[Chris Nowlin]

Really? That's changed since the 70s then. Back then it was 0 * & = 0 because you're saying "I have no infinities" or something of that sort.

Well, it's just usually said to not mean anything. But you write down things like that loosely doing calculus all the time.

The difference is, in calculus, you're talking about "essentially zero" times "nearly infinite" when you write such a thing, and thus they balance each other out.

Beyond that, you can make up your own convention depending on the context. There's no absolutes beyond the notation and what it's mean to represent. Multiplication is defined on all real numbers, of which infinity is not one. If you want to throw a new number into the set, and define multiplication with it, people are free to do so however they wish.

I see the sense in what you're saying. An "infinite amount of nothing" just sounds like a lot of nothing.

[jessecuster3]

I would switch doors because I get Monty's 67% chance instead of my 33%.

Ok how about with 3 card monty, let's make it an Ace and two Queens. After I shuffle them up their position is now static. My dealer flips up one Queen, now you are saying the odds are different for which card I choose ?

[Huh?]

I would switch doors because I get Monty's 67% chance instead of my 33%.Except, your odds of being wrong are 50% no matter which door you pick. The only chance that has changed is the chance that you are wrong, and it went down. Right?

[Chris Nowlin]

Ok how about with 3 card monty, let's make it an Ace and two Queens. After I shuffle them up their position is now static. My dealer flips up one Queen, now you are saying the odds are different for which card I choose ?

If you had selected a card for him to not flip first.

Careful there, though. The odds depends on perspective. From your point-of-view, the odds are different.

Suppose I could peek under all the cards, then make my choice. I'd know 100% were the ace was. You'd still have a 66% chance. Odds aren't a function of the cards, but of you and your knowledge

[jessecuster3]

If you had selected a card for him to not flip first.

Careful there, though. The odds depends on perspective. From your point-of-view, the odds are different.

Suppose I could peek under all the cards, then make my choice. I'd know 100% were the ace was. You'd still have a 66% chance. Odds aren't a function of the cards, but of you and your knowledge

So then you are saying Monte Hall has advance knowledge of what's behind the doors, thats a whole different wrinkle I didn't even consider, because a ton of other factors come in then.

[Dreadstar]

Ok how about with 3 card monty, let's make it an Ace and two Queens. After I shuffle them up their position is now static. My dealer flips up one Queen, now you are saying the odds are different for which card I choose ?

No. I'm saying that *IF* the dealer is straight and there are inded an ace and 2 queens on the table, EVERY TIME, and you choose one of 3, and AFTER you choose, he ALWAYS shows you a queen and then gives you the opportunity to change your mind, change it, you double your odds.

Try it on a friend. Give it a fair reading. Do exactly as I have instructed. see who gets the Ace more often.

[Huh?]

So then you are saying Monte Hall has advance knowledge of what's behind the doors, thats a whole different wrinkle I didn't even consider, because a ton of other factors come in then.But that doesn't affect any of your probabilities, because you don't know where anything is until he shows you.

[Huh?]

No. I'm saying that *IF* the dealer is straight and there are inded an ace and 2 queens on the table, EVERY TIME, and you choose one of 3, and AFTER you choose, he ALWAYS shows you a queen and then gives you the opportunity to change your mind, change it, you double your odds.

Try it on a friend. Give it a fair reading. Do exactly as I have instructed. see who gets the Ace more often.I'm sorry Dread, but the more times you do it, the closer you will approach a 50/50 split.

[Dreadstar]

Except, your odds of being wrong are 50% no matter which door you pick. The only chance that has changed is the chance that you are wrong, and it went down. Right?

Nope. The goat-door that's revealed does NOT affect the original 2 in 3 chance that Monty had. He STILL has a 2 in 3 chance, because he ALWAYS has a goat. ALWAYS.

You aren't getting one door for one door. You're getting 2 doors for one door.

Seriously. Try it on someone. While you're doing it, the 2 in 3 chnce never changing will eventually become clear.

[Dreadstar]

I'm sorry Dread, but the more times you do it, the closer you will approach a 50/50 split.

Nope. Seriously. I'd expect that if you do it 100 times, you'd be in the neighborhood of 45/55 to 70/30.

After 1000 times you'd be a lot closer to the 66/34 neighborhood.

Really, try it.

[jessecuster3]

No. I'm saying that *IF* the dealer is straight and there are inded an ace and 2 queens on the table, EVERY TIME, and you choose one of 3, and AFTER you choose, he ALWAYS shows you a queen and then gives you the opportunity to change your mind, change it, you double your odds.

Try it on a friend. Give it a fair reading. Do exactly as I have instructed. see who gets the Ace more often.

That makes no sense, maybe I watch too much Dead or No Deal, but the chances have to go up with every piece you eliminate. Whether I can switch cases will not affect that number because it will still be an unknown. I mean, in this situation if 1 queen is eliminated, its a coin flip, no matter which card you choose, because you cannot get more information than is altready present.

[Chris Nowlin]

So, it's bizarre, and at some point, leaves math and becomes perspective. See the odds from the beginning of the game, not the middle.

Suppose you know in advance you'll use Dread's plan.

Then you're betting on getting goat, rather than the prize. Because, if you plan to do it Dread's way from the start, if you pick a goat to begin with, you win.

If you pick the prize to begin with, you lose.

You have a better chance of choosing the goat. You have three choices. 2 win.

On the other hand...

Say you don't know if you'll do it Dread's way or not. You make a choice. Now you have 2 doors in front of you. One is prize. One is a goat. There are now 2 choices. 1 wins. Sounds like 50%. It's kind of a point-of-view.

[Dreadstar]

That makes no sense, maybe I watch too much Dead or No Deal, but the chances have to go up with every piece you eliminate. Whether I can switch cases will not affect that number because it will still be an unknown. I mean, in this situation if 1 queen is eliminated, its a coin flip, no matter which card you choose, because you cannot get more information than is altready present.

The thing is, the information IS already present. Monty has a 2 in 3 chance. Showing you a guaranteed queen does not change that chance in this case. It's the "if" that's holding you up.

Try it. Be honest. Be fair. You might eventually see what I'm talking about.

[Huh?]

Nope. The goat-door that's revealed does NOT affect the original 2 in 3 chance that Monty had. He STILL has a 2 in 3 chance, because he ALWAYS has a goat. ALWAYS.It doesn't affect Monty's chance, but it changes my starting set. I go from a set of 3 to a set of 2, and the chance I am wrong when choosing between two doors is 50%. The chance Monty is right is still 66%.

No matter what I do I have a 50% chance of being wrong, and a 50% chance of being right.

Monty has a 66% chance of being right and a 33% chance of being wrong.

If I switch doors, my odds do not change, and Monty's don't either.

If we repeat this ad infinitum, I will win 33% of the time whether I switch doors or not. So how does switching doors make a difference?

[Matt Algren]

Nope. Seriously. I'd expect that if you do it 100 times, you'd be in the neighborhood of 45/55 to 70/30.

After 1000 times you'd be a lot closer to the 66/34 neighborhood.

Really, try it.Dread's right. (I feel dirty just typing it,) We did this in a statistics class I took a couple years ago. The more games, the closer you get to a 2/3 chance when you switch cards after the first card is shown.

[Chris Nowlin]

It doesn't affect Monty's chance, but it changes my starting set. I go from a set of 3 to a set of 2, and the chance I am wrong when choosing between two doors is 50%. The chance Monty is right is still 66%.

No matter what I do I have a 50% chance of being wrong, and a 50% chance of being right.

Monty has a 66% chance of being right and a 33% chance of being wrong.

If I switch doors, my odds do not change, and Monty's don't either.

If we repeat this ad infinitum, I will win 33% of the time whether I switch doors or not. So how does switching doors make a difference?

Really, though, you are changing the game with Dread's plan. You are essentiallybetting on Monty instead of yourself by following it.

[Dreadstar]

So, it's bizarre, and at some point, leaves math and becomes perspective.

No. It doesn't. It's pure math.

Say you don't know if you'll do it Dread's way or not. You make a choice. Now you have 2 doors in front of you. One is prize. One is a goat. There are now 2 choices. 1 wins. Sounds like 50%. It's kind of a point-of-view.

In order for it to work you MUST go by the rules set froth at the beginning. Let's forget the second choice for a second. Toss it. You choose and you're STUCK with that choice.

3 doors. Choose 1. Who has the winner more often?

Monty, 2 out of 3.

The reveal means NOTHING. It's a misdirection, a red herring.

[Chris Nowlin]

Dread's right.

Now that's statistically improbable, if not impossible.

[Dreadstar]

Really, though, you are changing the game with Dread's plan. You are essentiallybetting on Monty instead of yourself by following it.

DINGDINGDINGDING!

[Dreadstar]

Now that's statistically improbable, if not impossible.

Even a stopped clock is right twice a day.

[Chris Nowlin]

No. It doesn't. It's pure math.

More likely pure statistics.



In order for it to work you MUST go by the rules set froth at the beginning. Let's forget the second choice for a second. Toss it. You choose and you're STUCK with that choice.

3 doors. Choose 1. Who has the winner more often?

Monty, 2 out of 3.

The reveal means NOTHING. It's a misdirection, a red herring.

True. From the perspective of forgetting the second choice, and seeing it as a premade decision, you are absolutely correct.

Except for failing to take into account the feeling I have; that I just know where the prize is.

[jessecuster3]

Even a stopped clock is right twice a day.

Not if its a military clock....

[Matt Algren]

That makes no sense, maybe I watch too much Dead or No Deal, but the chances have to go up with every piece you eliminate. Whether I can switch cases will not affect that number because it will still be an unknown. I mean, in this situation if 1 queen is eliminated, its a coin flip, no matter which card you choose, because you cannot get more information than is already present.I think I see the problem with your reasoning. Intuitively, you would think that after removing the first queen, you have one of two chances to win. But you don't. The same three options exist: Ace, Queen 1, Queen 2. Conceptually, if you really wanted to, you could choose the Queen that's already been revealed.

So you aren't choosing between Ace or Queen 2, you're switching your first one-of-three pick for Monty's first two-of-three pick. It's hard to visualize.

[jessecuster3]

No. It doesn't. It's pure math.



In order for it to work you MUST go by the rules set froth at the beginning. Let's forget the second choice for a second. Toss it. You choose and you're STUCK with that choice.

3 doors. Choose 1. Who has the winner more often?

Monty, 2 out of 3.

The reveal means NOTHING. It's a misdirection, a red herring.


See this is the real place I am hung up. You are really saying I only have the choice of two things not three. This is effectively like flipping a coin, and we all know that probability.

[Matt Algren]

You are essentially betting on Monty instead of yourself by following it.That's exactly the conceptual hurdle you have to jump. Exactly right. Jesse, this is it.
Except for failing to take into account the feeling I have; that I just know where the prize is.And that's what games like Deal or No Deal count on. A-type personalities who will go with their gut feeling or emotion instead of playing the numbers.

[Huh?]

I think I see the problem with your reasoning. Intuitively, you would think that after removing the first queen, you have one of two chances to win. But you don't. The same three options exist: Ace, Queen 1, Queen 2. Conceptually, if you really wanted to, you could choose the Queen that's already been revealed.

So you aren't choosing between Ace or Queen 2, you're switching your first one-of-three pick for Monty's first two-of-three pick. It's hard to visualize.But Monty isn't picking anything. I am sorry, but your chance of winning does not change whether you switch or not. You are only going to win this game 33% of the time whether you switch or not.

[Chris Nowlin]

DINGDINGDINGDING!


Really, you should take the second choice into account though.

You have 6 possibilites. Door A then switch, Door A then stay, Door B then switch, etc.

Without loss of generality, assume A has the prize.

You win if you:

pick A and stay
pick B and switch
pick C and switch

You lose if you

pick A and switch
pick B and stay
pick C and stay.

From that POV, it looks like staying causes you to lose 2 out of 3 times. And switching causes you to win 2/3 times.

[Dreadstar]

See this is the real place I am hung up.

Yes.

You are really saying I only have the choice of two things not three.

No, I'm not. See, you JUMP to that conclusion BECAUSE of the reveal. The reveal means NOTHING. It's a fake-out. Again, let's look at it *without* the option to change your mind.

3 doors, choose a door. Your stuck with it.

What are your odds? What are Monty's odds? Which would you rather have? Now, here's what throws you off: HE ALWAYS has a loser. ALWAYS. If he shows you a loser, it does not change the original odds.

This is effectively like flipping a coin, and we all know that probability.

Try it.

[Chris Nowlin]

And that's what games like Deal or No Deal count on. A-type personalities who will go with their gut feeling or emotion instead of playing the numbers.

Cool. Can I be an A-type personality and a mathematician?

[Paragon Kobold]

See this is the real place I am hung up. You are really saying I only have the choice of two things not three. This is effectively like flipping a coin, and we all know that probability.

Not if you have an unbalanced coin. Two possibilites need not always be a fifty-fifty chance.

[Matt Algren]

Cool. Can I be an A-type personality and a mathematician?
The universe would implode.

[Dreadstar]

But Monty isn't picking anything. I am sorry, but your chance of winning does not change whether you switch or not. You are only going to win this game 33% of the time whether you switch or not.

If you were a betting man, I'd take your money 2 out of 3 times, over a long period of betting.

[Shellhead]

I haven't watched Monte in decades, so I have a question to ask. Does he always show a "queen" after the initial pick and before the final decision to stay or switch? (using the 3-card monte ace-queen-queen setup).

If so, then the odds are always 50-50. The spare queen is irrelevant, because it will be revealed and eliminated before the final decision is made, meaning that the final (and only relevant) choice will always be between an ace and a queen. Which would be 50-50 odds.

But if there is a scenario where sometimes Monte doesn't show one of the cards, then I could see where the odds really start at 1 in 3, because the chooser doesn't know at first if there will be a chance to switch.

[jessecuster3]

And that's what games like Deal or No Deal count on. A-type personalities who will go with their gut feeling or emotion instead of playing the numbers.

No it is so much more than that, the banker makes the show, because now lets say the same scenario as the monty hall thing, only the prize has $1 million, but still 2 goats. Now he shows me a goat, so I either have a goat or $ 1 million left, with the option to switch what I chose. But, my choices in Deal or No Deal land, I have to choose between 2 cases left, switch my choice or take $625K and walk away without revealing either case.

[Matt Algren]

But Monty isn't picking anything. I am sorry, but your chance of winning does not change whether you switch or not. You are only going to win this game 33% of the time whether you switch or not.Like C&C said, you're betting on the odds of Monty having the right card (before the reveal) versus you having the right card. The reveal is just there to push emotions around.

[Dreadstar]

it looks like staying causes you to lose 2 out of 3 times. And switching causes you to win 2/3 times.

This is absolutely correct, no matter how you want to complicate it.

[Dreadstar]

I haven't watched Monte in decades, so I have a question to ask. Does he always show a "queen" after the initial pick and before the final decision to stay or switch? (using the 3-card monte ace-queen-queen setup).

If so, then the odds are always 50-50. The spare queen is irrelevant, because it will be revealed and eliminated before the final decision is made, meaning that the final (and only relevant) choice will always be between an ace and a queen. Which would be 50-50 odds.

But if there is a scenario where sometimes Monte doesn't show one of the cards, then I could see where the odds really start at 1 in 3, because the chooser doesn't know at first if there will be a chance to switch.

Try it.

Seriously.

[Chris Nowlin]

No it is so much more than that, the banker makes the show, because now lets say the same scenario as the monty hall thing, only the prize has $1 million, but still 2 goats. Now he shows me a goat, so i either have a goat or $ 1 million left, with the option to switch what I chose. But, my choices in Deal or No Deal land, I have to choose between 2 cases left, switch my choice or take $625K and walk away without revealing either case.

So, the thing is, odds only mean anything if you do it repeatedly. A 66% chance of winning is just that. The expected value for you of playing it dread's way, and not walking way, is to win $666K. Practically, that means you may get a million, you may get a goat.

Take the $625K and spend it on hookers.

[Huh?]

Really, you should take the second choice into account though.

You have 6 possibilites. Door A then switch, Door A then stay, Door B then switch, etc.

Without loss of generality, assume A has the prize.

You win if you:

pick A and stay
pick B and switch
pick C and switch

You lose if you

pick A and switch
pick B and stay
pick C and stay.

From that POV, it looks like staying causes you to lose 2 out of 3 times. And switching causes you to win 2/3 times.Ok, I choose C. He will show me B.

Now here are the odds:

You win if you:
Choose C and switch.
Choose A and stay.

You Loose if you:
Choose C and stay.
Choose A and switch.

Seems like even odds to me!!!!!

[Chris Nowlin]

This is absolutely correct, no matter how you want to complicate it.

Oh, I'll complicate it as much as I want to.

[Matt Algren]

No it is so much more than that, the banker makes the show, because now lets say the same scenario as the monty hall thing, only the prize has $1 million, but still 2 goats. Now he shows me a goat, so I either have a goat or $ 1 million left, with the option to switch what I chose. But, my choices in Deal or No Deal land, I have to choose between 2 cases left, switch my choice or take $625K and walk away without revealing either case.I'd be the type of person to walk away. Someone put together an Excel sheet that calculates expected return, given the pre-revealed amounts. Link here. (http://www.javajargon.com/deal.html)

[Paragon Kobold]

The odds of picking the right door in the first place is 1/3. Nothing that happens after that has the ability to go backwards in time and change those odds to anything other than 1/3.

[Huh?]

The odds of picking the right door in the first place is 1/3. Nothing that happens after that has the ability to go backwards in time and change those odds to anything other than 1/3.Thank You! I whole heartedly agree.

[Matt Algren]

Ok, I choose C. He will show me B.

Now here are the odds:

You win if you:
Choose C and switch.
Choose A and stay.

You Loose if you:
Choose C and stay.
Choose A and switch.

Seems like even odds to me!!!!!No. You're leaving out the first pick. Again, you always have three choices, even after the first reveal.

Seriously, try it.

[Chris Nowlin]

The odds of picking the right door in the first place is 1/3. Nothing that happens after that has the ability to go backwards in time and change those odds to anything other than 1/3.

You choose the door.

Huh? is watching. I ask Huh? if he wants to bet that you're right or you're wrong. Which should he choose?

[Chris Nowlin]

Seriously, try it.

Some of us are too busy posting on forums to play cards!

[Dreadstar]

AGAIN, forget the switch choice. Forget it for a second

3 doors, choose one. Stick with it.

What is your percentage chance?

What are Monty's?

Showing you a loser does not change this percentage, because he ALWAYS has a loser.

Monty's 2 in 3 is better.

[Dreadstar]

The odds of picking the right door in the first place is 1/3. Nothing that happens after that has the ability to go backwards in time and change those odds to anything other than 1/3.

YES! You're on the verge:

"The odds of Monty having the right door in the first place is 2/3. Nothing that happens after that has the ability to go backwards in time and change those odds to anything other than 2/3."

EXACTLY!

[Gilda Dent]

I wish I had made that switch from law to elementary education and reading. Not too late, I suppose.

And I never could have majored in math. I just took an advanced calculus for non-math majors my senior year to fill out the course load and because it only met once a week (plus I think it was pass/fail). So the pressure was definitely off.

I enjoyed it. How's it go....nice place to visit, but I wouldn't want to live there. Admire the thinking processes, though.

It's still very helpful if you're going to do any advanced study because you need a basic understanding of statistical analysis, which requires a pretty good grasp of algebra.

[Chris Nowlin]

Forget the first choice! Forget it. You pick anything. He shows you something. Who cares?

Now the game is on. There are 2 doors. One is right. One is wrong. 50%

[Huh?]

No. You're leaving out the first pick. Again, you always have three choices, even after the first reveal.

Seriously, try it.What moron is going to switch his choice to B!? Seriously?

[Paragon Kobold]

As the odds of 'my' door being the right one is 1/3, the odds of one of the two other doors being right is 2/3.

Nothing ever changes these odds either.

[Chris Nowlin]

What moron is going to switch his choice to B!? Seriously?

He meant you could have picked B the first time, before it was lifted. You didn't account for that.

[Matthew E]

I took this in grad school math. It's game theory. You should always switch.

Here's another way of looking at it. When Monty opens his door and reveals the goat, he's really showing you which one of his doors has the prize, if any of his doors have the prize. Now, in this example, it so happens that he only has two doors and you have the other one, but imagine:

There are a hundred doors.

You pick a door.

Monty opens 98 of the remaining 99 doors, revealing goats behind them all.

Do you stick with your door or switch to Monty's last door?

Obviously, you would switch to Monty's last door.

The case with three doors is the same as that, but with 2/3 odds in favour of switching instead of 99/100 in favour of switching.

[Chris Nowlin]

As the odds of 'my' door being the right one is 1/3, the odds of one of the two other doors being right is 2/3.

Nothing ever changes these odds either.

True. You are then asked if you want to bet you were right to begin with, or wrong to begin with. What do you answer?

[Matt Algren]

All right, let's do this.

I have three post-its in front of me. Two are marked Queen, while the third is marked Ace. They are not necessarily in that order. You are trying to get the Ace.

Pick one, two, or three.

[jessecuster3]

AGAIN, forget the switch choice. Forget it for a second

3 doors, choose one. Stick with it.

What is your percentage chance?

What are Monty's?

Showing you a loser does not change this percentage, because he ALWAYS has a loser.

Monty's 2 in 3 is better.

Ok this makes sense, I get this, so forget Monty, and lets play 3 Card Monty. Totally on the up and up, I swear. Each time I win you owe me a dollar and each time you win I owe you a dollar. I am the dealer. I deal 1,000 hands to you I should have 666 dollars right ?

Now we play another 1,000 hands, same rules, except this time after you choose I give you the option to switch which card you chose, how much would you win in that scenario ?

[Chris Nowlin]

I took this in grad school math. It's game theory. You should always switch.


What did you study in grad school?

[Chris Nowlin]

All right, let's do this.

I have three post-its in front of me. Two are marked Queen, while the third is marked Ace. They are not necessarily in that order. You are trying to get the Ace.

Pick one, two, or three.

I'll play. But if I win, you owe me a million dollars.

One. 1. Uno.

[Matthew E]

What did you study in grad school?

I got an M.A. in math, focusing on graph theory. This particular topic came up in... I think it was my combinatorics class.

[Dreadstar]

I took this in grad school math. It's game theory. You should always switch.

Here's another way of looking at it. When Monty opens his door and reveals the goat, he's really showing you which one of his doors has the prize, if any of his doors have the prize. Now, in this example, it so happens that he only has two doors and you have the other one, but imagine:

There are a hundred doors.

You pick a door.

Monty opens 98 of the remaining 99 doors, revealing goats behind them all.

Do you stick with your door or switch to Monty's last door?

Obviously, you would switch to Monty's last door.

The case with three doors is the same as that, but with 2/3 odds in favour of switching instead of 99/100 in favour of switching.



A more visual way to see it is this way.


I let you choose a card out of a deck randomly. NO PEEKING! If you have the Ace of Spades, I pay you 1 million dollars. Further, after you choose, I will reveal to you 50 cards that are NOT the Ace of Spades and keep only one card for myself. Which card would you rather have? Mine or yours?

[Matt Algren]

I'm revealing #3 as a queen.

Stay with your choice, or switch to #2?

[Chris Nowlin]

I got an M.A. in math, focusing on graph theory. This particular topic came up in... I think it was my combinatorics class.

Awesome. I have a masters in math. Working on my PhD now. This problem came up once upon a time for me, in undergrad though.

[jessecuster3]

I'm revealing #3 as a queen.

Which do you want, 1 or 2?

Thats 50/50 then.

[Chris Nowlin]

I'm revealing #3 as a queen.

Which do you want, 1 or 2?

2. Two. Dos. Make the switch!

[Dreadstar]

Now we play another 1,000 hands, same rules, except this time after you choose I give you the option to switch which card you chose, how much would you win in that scenario ?

Are you going to reveal a queen to me before allowing me to re-choose?

[Chris Nowlin]

Thats 50/50 then.

Let's see how this plays out.

[Matt Algren]

Thats 50/50 then.
Read the edit for clarification. It does make a difference.

[jessecuster3]

Are you going to reveal a queen to me before allowing me to re-choose?

yeah sorry thats what I meant.

[Matt Algren]

2. Two. Dos. Make the switch!
2 is the Ace.

Now, Huh?, jesse, or whoever else, find someone to do this with a hundred times.

[Chris Nowlin]

A more visual way to see it is this way.


I let you choose a card out of a deck randomly. NO PEEKING! If you have the Ace of Spades, I pay you 1 million dollars. Further, after you choose, I will reveal to you 50 cards that are NOT the Ace of Spades and keep only one card for myself. Which card would you rather have? Mine or yours?

How about this way. You have a deck. I choose one card. Then I have to bet if the ace of spades is in my hand or still in the deck.

[jessecuster3]

Read the edit for clarification. It does make a difference.

No thats semantics and language should not play a factor.

[Chris Nowlin]

2 is the Ace.

Now, Huh?, jesse, or whoever else, find someone to do this with a hundred times.

Dread is proven right!

Dogs marry cats!

The universe implodes!

[Dreadstar]

How about this way. You have a deck. I choose one card. Then I have to bet if the ace of spades is in my hand or still in the deck.

Exactly.

Turning over 50 non ace of spades do NOT affect that percentage.

[jessecuster3]

2 is the Ace.

Now, Huh?, jesse, or whoever else, find someone to do this with a hundred times.

Thats so dumb(nothing personal) I have a coin at my desk I am going to flip it 100 times, you jot down right now, what those hunderd times will work out to. The catch is you never see me flip the coin or know the results. And to prove my point tails is going to come up 75% of the time.

[MWGallaher]

The Monte Hall problem is easier to grasp if you express the rules a little differently:
Monte shows you n doors. There is a prize behind only one of them and a goat behind all the rest of them. Monte knows which single door opens on a prize.
You now select a door. Monte now reveals the contents of all but one of the remaining n-1 doors. If the prize is behind one of those doors you didn't pick, that's the one he'll leave closed. If the prize is behind the door you did pick, Monte can leave any single one of his doors closed, since he knows that all of his n-1 doors open to goats. Then you're offered the chance to pick.
In the typical case, there are 3 doors (n=3). But suppose there were 1000 doors? You pick door #1, now Monte opens all the doors except yours (#1) and door #730, and then offers you the chance to switch. Do you switch? Of course you do! The odds were only 1:1000 that you were right, and there's obviously a 999:1000 chance that the prize is behind #730. Similar (although increasingly lesser) advantages will apply if there are 100 doors (Do you want #1 or #80?), or 10 doors, all the way down to 3 doors (the game can't be played with only 2 doors). At the smaller scale game, psychology masks the advantage.

[Chris Nowlin]

Matt, you now owe me a million dollars. See my stipulation is post 183

[Dreadstar]

yeah sorry thats what I meant.

Then I'm a gonna switch, and win 67% of the time.

[Shellhead]

Thinking it over, I get it.

Your first pick is more likely to be wrong, because your initial odds are 1 in 3 of picking the right card.

Monte shows you a queen, which is irrelevant, except that it may give you greater confidence in your first pick, even though your odds were bad.

Now you have the option to switch. It may seem like it doesn't matter, but because your initial odds were so bad, it makes sense to switch.

It's easier to understand if you picture Monte using a full deck of cards, and you only win if you pick the Ace of Spades. You pick your card. Then Monte shows you 50 cards that are not the Ace of Spades. You can stay with your original pick, which was probably (51 out of 52 times) the wrong card, or you can switch to the other card.

[Matt Algren]

No thats semantics and language should not play a factor.It is semantics, but I think the wording makes a difference in how you visualize this. You aren't making a fresh choice between one and two, you're betting on my odds on the first choice versus your odds on the first choice.

Here's a site (http://explorepdx.com/montyhallpuz.html) that may explain it better. They did the experiment all day at a fair and posted the results. Like they say there, the distinction is subtle, but important.

[Matthew E]

I think the thing that people trip over here when they say it's 50-50 is that this is not a random situation. It's not raw probability at work here. Because Monty Hall, or whoever's doing the revealing, *knows* where the prize is and is using that knowledge when he chooses which doors to reveal. If Monty *didn't* know, and opened a door at random, then it *would* be 50-50... except there would be a chance that he'd open the door that the prize was behind (thus collapsing all the probabilities nicely), and from the point of view of the show that's suboptimal, so they don't do it that way.

[Chris Nowlin]

So the pack of cards.

At days end, there are 2 cards in front of me. One is Ace of Spades. One is not. But there is not equal weight with each.

It's not like flipping a coin. It's like flipping a rigged coin that is top-heavy.

One of those cards was chosen randomly from a deck of 52.

The other has a 1/52 chance of being randomly chose from a deck of 51 and has a 51/52 chance of being the ace of spades.

EDIT: Matthew beat me to it.

[Matt Algren]

Thats so dumb(nothing personal) I have a coin at my desk I am going to flip it 100 times, you jot down right now, what those hunderd times will work out to. The catch is you never see me flip the coin or know the results. And to prove my point tails is going to come up 75% of the time.Seriously, that was the real outcome. I originally had them AQQ, but changed it to QAQ before we started.

One time proves nothing, though. It's results over a higher number of experiments that shows the odds.

[Chris Nowlin]

Seriously, that was the real outcome. I originally had them AQQ, but changed it to QAQ before we started.

One time proves nothing, though. It's results over a higher number of experiments that shows the odds.

One time proves you owe me a million dollars.

[Matt Algren]

Matt, you now owe me a million dollars. See my stipulation is post 183
Now, would you like that check, or do you want to switch your answer?

(Nice edit, BTW.)

[Dreadstar]

One time proves you owe me a million dollars.


If he had a million dollars, he wouldn't be living in Dayton.


duh.

[jessecuster3]

Ok higher numbers are where it makes lense sense. If I pull a card from a deck at random, and I am trying to pick the Ace of Spades. You throw away every other card in the deck except my card and the Ace Of Spades. Now, my card's odds are 1 in 50 and the Ace of Spades odds are 1 in 50.

[Chris Nowlin]

Now, would you like that check, or do you want to switch your answer?

(Nice edit, BTW.)

Edit?

Dread said always switch answer. So I'll switch back to 1!

(As a sometimes math teacher, I can say with certainty that is how all students think)

[Chris Nowlin]

Ok higher numbers are where it makes lense sense. If I pull a card from a deck at random, and I am trying to pick the Ace of Spades. You throw away every other card in the deck except my card and the Ace Of Spades. Now, my card's odds are 1 in 50 and the Ace of Spades odds are 1 in 50.

I don't understand.

[Matt Algren]

If he had a million dollars, he wouldn't be living in Dayton.


duh.
True. .

[Huh?]

Ok higher numbers are where it makes lense sense. If I pull a card from a deck at random, and I am trying to pick the Ace of Spades. You throw away every other card in the deck except my card and the Ace Of Spades. Now, my card's odds are 1 in 50 and the Ace of Spades odds are 1 in 50.Nope, your card's odds are 1 in 52, and the other card's odds are 51 in 52.

BTW, I now see how this works....:)

[Dreadstar]

Hmmmm... let me try something.

Monty: "You have chosen door number one! Now let's see---"

Dread: "I changed my mind. I don't WANT door number 1. I want both door number 2 AND door number 3!"

Monty: "Waitasecond, I have to show you what's behind---"

Dread: "Don't CARE what you have to show me. At all. I want both door 2 and door 3. You can keep door 1."

Monty: "Fine, you can have both door 2 and door 3. By the way, there's a goat behind door--- oh, it doesn't matter."

Dread: "WooHoo!"

[Dreadstar]

Nope, your card's odds are 1 in 52, and the other card's odds are 51 in 52.

BTW, I now see how this works....:)
WHOO HOO!


Now go scam someone in a bar!

[Gilda Dent]

Here (http://www.stat.sc.edu/~west/javahtml/LetsMakeaDeal.html) is an online simulator. Play it a few dozen times and see what results you get.

[Huh?]

WHOO HOO!


Now go scam someone in a bar!Yeah, expanding the number of intitial choices makes it crystal clear.

The initial 1 in 3 made it too difficult for me to get.

[Chris Nowlin]

Thanks, Dread. I've had fun. But I must go.

Remember, kids. The real lesson here is that in gambling, there are no winners, only losers. If you want money, don't look for an easy solution. Get a job. Work hard and contribute to society. You can realise your dreams. But you have to earn it. Work hard enough and this will truly be, a land of opportunity.

[Dreadstar]

Here (http://www.stat.sc.edu/~west/javahtml/LetsMakeaDeal.html) is an online simulator. Play it a few dozen times and see what results you get.

I could not find that, for the life of me. I think my internet is broken.

[MWGallaher]

You're absolutely correct. And this is the cleanest way to see it. The fact that 1/3=.333333... more properly follows from this observation, rather than the reverse.

The other way to see this is, if you've taken any calculus, is to think of the number .999999 as the limit of the sequence:

9/10,99/100,999/1000,...

As you go to infinity, the limit is .99999..., by definition. That's the only thing the crazy ellipse can mean.

But, one can prove mathematically that the limit is 1. The point being numbers like 99999999/10000000 (pretend the number of zeroes and nines match) are damn close to one and getting arbitrarily closer.

One sequence can only have one limit point. So 1=.999999.... Equality is correct.

The biggest problem I have with ".9999... = 1" is that it seems to leave mathematics with no means of representing the concept of "the largest real number that is less than 1". Math has no shortages of ways to represent "1" itself--in fact, it has an infinite ways of representing that value!

[Dreadstar]

Thanks, Dread. I've had fun. But I must go.

Remember, kids. The real lesson here is that in gambling, there are no winners, only losers. If you want money, don't look for an easy solution. Get a job. Work hard and contribute to society. You can realise your dreams. But you have to earn it. Work hard enough and this will truly be, a land of opportunity.

What he said. Now who's up for blackjack?

[Matt Algren]

I could not find that, for the life of me. I think my internet is broken.It's java. I had to give it a few seconds to load up.

[jessecuster3]

Forget it.

[jessecuster3]

What he said. Now who's up for blackjack?

Forget blackjack, I will play poker against any of you any day of the week and twice on Sunday.

[dingo]

I don't know. I go to sleep and look what I miss.

[Chris Nowlin]

I don't know. I go to sleep and look what I miss.

That'll learn ya' to sleep. You began all this while I was asleep, so I needed to catch up too. The summary was: current mathematical thinking is that you're wrong, but you can always fight the power

[dingo]

The other way to see this is, if you've taken any calculus, is to think of the number .999999 as the limit of the sequence:

9/10,99/100,999/1000,...

As you go to infinity, the limit is .99999..., by definition. That's the only thing the crazy ellipse can mean.

But, one can prove mathematically that the limit is 1. The point being numbers like 99999999/10000000 (pretend the number of zeroes and nines match) are damn close to one and getting arbitrarily closer.

One sequence can only have one limit point. So 1=.999999.... Equality is correct.

I have kind of given up this, but you logic seems to be predicated by the fact that approaching a limit point is the same as equality, which is what is being disputed.

This makes this particular "proof" pointless in this context. That is why I was ignoring it.

[Chris Nowlin]

I have kind of given up this, but you logic seems to be predicated by the fact that approaching a limit point is the same as equality, which is what is being disputed.

This makes this particular "proof" pointless in this context. That is why I was ignoring it.

Never give up.
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