Is 1 = 0.99999999999...(recurring)

[LtMarvel]

Make him prove that he is not immune to the poison by drinking both glasses.

[LtMarvel]

Math hero of the week (http://www.charlotte.com/mld/observer/news/local/16649149.htm):
Part of the new exhibit on candy has been removed from Discovery Place. And you can thank Parker Garrison.

When the 8-year-old math whiz at Charlotte Christian visited the "Jelly Belly Presents Candy Unwrapped" display, he accepted the challenge: Use equations to calculate how many jelly beans were in a pyramid and other containers.

His mother, Donna, didn't want to wait while he tried all the formulas. So she copied the numbers he needed, and he took the problems home.

That's when he realized something was wrong.

For years, Parker has amazed his parents and teachers with his math skills. When he was 3, his parents could tell him which coins they had in their pocket, and he'd add the total in his head.

And when he was in first grade, he competed against third-graders in the math olympics. He would have finished second -- except he wasn't old enough to officially participate.

But that day in the family's south Charlotte home, his mom still wasn't sure what to think when Parker told her there was an error in the pyramid formula.

You made a mistake, she remembers saying.

No, I did it three times, he replied. And I know why it doesn't work out.

The equation mistakenly called for dividing the correct answer in half. Parker's father, Jim, called Discovery Place.

After the museum figured out who should handle the call, Dean Briere, a shocked vice president, decided to investigate.

An hour later, Briere called back.

The exhibit had traveled to eight cities in four years.

"And no one found this mistake -- I just couldn't believe it," Briere said.

<click on link for more>

[dingo]

Now, here's a tough one:

Suppose you and another person are engaged in a battle of wits. He shows you two cups of wine and a vial of poisonous iocane powder, then turns his back and does something you can't see. Then he puts one cup in front of him and the other in front of you, and tells you to drink from one cup, and he'll drink from the other. Should you choose the cup in front of him, or the one in front of you?

It's a battle of wits, so what I would do is say "Get Bent".

Why would I want to play Russian Roulette?

[Chris N]

It's a battle of wits, so what I would do is say "Get Bent".

Why would I want to play Russian Roulette?

That is inconceivable!

[Pól Rua]

1=1

http://homepage.mac.com/jjbeach/einheri/micros/Mr.A2.gif

[Night]

Now, here's a tough one:

Suppose you and another person are engaged in a battle of wits. He shows you two cups of wine and a vial of poisonous iocane powder, then turns his back and does something you can't see. Then he puts one cup in front of him and the other in front of you, and tells you to drink from one cup, and he'll drink from the other. Should you choose the cup in front of him, or the one in front of you? Offer a change of rules... Since you want the princess dead... anyway, offer that she drinks in his stead... (or better yet if you can get an agreement your stead)and watch the sweat as he finds a new game... Ie in the battle of wits don't let your opponent choose the game.

[MWGallaher]

Actually, mathmatically, it does.

(1) x = 0.99999...
(2) 10x = 9.9999...
(3) 10x − x = 9.9999... − 0.99999...
(4) 9x = 9
(5) x = 1

See here: http://polymathematics.typepad.com/p...sorry_it_.html


You're absolutely correct. And this is the cleanest way to see it.


Whoa...this proposed proof looks like mathematical slight of hand to me.
The only way that 9.9999... - 0.99999... can equal 9 is if there is one more instance of the digit 9 in the larger term. Where'd the extra '9' come from?
In other words, 0.99999... has (infinity) 9's. Now we multiply by 10 and somehow the result has (infinity + 1) 9's ?!?! How does the recurrance of 9's justify our padding the least significant digit of "10 * 0.9999...." with another 9 rather than a 0?
Every time I multiple a conventional number by 10, I put a zero on the end of the result. In this proof, they put a 9 on the end of the result. That's changing the rules, as far as I can tell.

[dingo]

Whoa...this proposed proof looks like mathematical slight of hand to me.
The only way that 9.9999... - 0.99999... can equal 9 is if there is one more instance of the digit 9 in the larger term. Where'd the extra '9' come from?
In other words, 0.99999... has (infinity) 9's. Now we multiply by 10 and somehow the result has (infinity + 1) 9's ?!?! How does the recurrance of 9's justify our padding the least significant digit of "10 * 0.9999...." with another 9 rather than a 0?
Every time I multiple a conventional number by 10, I put a zero on the end of the result. In this proof, they put a 9 on the end of the result. That's changing the rules, as far as I can tell.

Actually, you aren't putting a zero at the end, you are moving the decimal place one place to the right.

I still think, you are right in as far as it is sleight of hand.

I was thinking about this today.
The problem is the mathematical axiom that anything (other than infinity) divided by infinity is equal to zero.

If this was not true then I would say that...


0.999999.... = 1 - (1/infinity)

[Tadhg]

Every time I multiple a conventional number by 10, I put a zero on the end of the result. In this proof, they put a 9 on the end of the result. That's changing the rules, as far as I can tell.

No. Everytime you multiply a number by 10, you move the decimal place over one space.

If you multiply 3.3 by 10 you get 33 not 30.

Edit: Dingo beat me to that point.

[Matthew E]

It's a perfectly respectable proof. All of the steps are commonplace in math.

[jessecuster3]

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.

Thank you, I get it. That is all.

[Loren]

Whoa...this proposed proof looks like mathematical slight of hand to me.
The only way that 9.9999... - 0.99999... can equal 9 is if there is one more instance of the digit 9 in the larger term. Where'd the extra '9' come from?
In other words, 0.99999... has (infinity) 9's. Now we multiply by 10 and somehow the result has (infinity + 1) 9's ?!?!

There's no such unique thing as (infinity + 1). Infinity = (Infinity + 1). The whole point of an infinite sequence of numbers is that it has no end, so it makes no sense to say that it loses a decimal place when multiplied by 10.

As I pointed out earlier, Infinity also equals (Infinity x 2).

[dingo]

There's no such unique thing as (infinity + 1). Infinity = (Infinity + 1). The whole point of an infinite sequence of numbers is that it has no end, so it makes no sense to say that it loses a decimal place when multiplied by 10.

As I pointed out earlier, Infinity also equals (Infinity x 2).

I'm sorry Loren, I don't think you are right.

Don't think of infinity as a magical concept, think of it as a number that can be compared.

The whole one-to-one pairing again.

All even integers is a lower order of infinity than all integers, because there are some integers that cannot be paired up.

[Tadhg]

There's no such unique thing as (infinity + 1). Infinity = (Infinity + 1). The whole point of an infinite sequence of numbers is that it has no end, so it makes no sense to say that it loses a decimal place when multiplied by 10.

As I pointed out earlier, Infinity also equals (Infinity x 2).

Your rationale would have πr² = π then.

[Loren]

I'm sorry Loren, I don't think you are right.

Don't think of infinity as a magical concept, think of it as a number that can be compared.

The whole one-to-one pairing again.

All even integers is a lower order of infinity than all integers, because there are some integers that cannot be paired up.

Except they *can* be paired up. See Cantor's Set Theory (http://www.mathacademy.com/pr/minitext/infinity/).

[Matt Algren]

Here's another fun (that is, headache inducing) statistics problem. How many people must be viewing the Community Board to make the probability of finding two people with the same birthday at least 50&#37;?

Limitations:

1) Looking for month and day, but not year.

2) Forget leap year. There is no February 29th.

3) Assume that a person has an equal chance of being born on any day of the year, even though some birthdays may be slightly more likely than others.

(Doesn't look like Dread's around, and I'm not up to explaining it, so here's a link (http://mathforum.org/dr.math/faq/faq.birthdayprob.html) to the answer.) This is a great party trick.

[Matthew E]

I'm sorry Loren, I don't think you are right.

Don't think of infinity as a magical concept, think of it as a number that can be compared.

The whole one-to-one pairing again.

All even integers is a lower order of infinity than all integers, because there are some integers that cannot be paired up.

Loren is right. Infinity is not a number, because a number is finite. It's infinity. Add to it, multiply it, whatever; it's still infinity.

[MWGallaher]

No. Everytime you multiply a number by 10, you move the decimal place over one space.

If you multiply 3.3 by 10 you get 33 not 30.

Edit: Dingo beat me to that point.
Obviously so. I guess I didn't express myself clearly enough.

10 * 0.9 = 9.0 (one 9 on either side of the equation)
10 * 0.99 = 9.90 (two 9's on each side of the equation)
10 * 0.999 = 9.990 (three 9's on each side of the equation)
10 * 0.9999 = 9.9990 (four 9's on each side of the equation)
As I continue this pattern, one thing should remain the same: there will always be the same number of instances of the digit 9 on each side of the equations. If I want the same number of digits on the right hand side of the decimal point (after I perform the multiplication), I must pad it with a zero, as in:
10 * 0.1345 = 1.3450 ; that's the "zero" I'm talking about, the one at the end of 1.3450.
But somehow, this "proof" requires that
10 * 0.9999... = 9.9999... (an infinite number of instances of the digit 9 on both sides of the equation...but, somehow, "one more" on the left hand side--that being the 9 to the left of the decimal place!?)
At an infinite sequence of 9's, I'm somehow able to grab one more from out of the blue so that my subtraction works out to the desired difference of exactly 9? I guess that's the magic of infinity. ~shrug~

[Matthew E]

That's the whole point of 'infinity': you aren't grabbing 'one more', because there are already an infinite number of them. The 9s go on forever. There's no final nine after which you'd need to conjure up one more.

[Dreadstar]

As I continue this pattern, one thing should remain the same: there will always be the same number of instances of the digit 9 on each side of the equations. If I want the same number of digits on the right hand side of the decimal point (after I perform the multiplication), I must pad it with a zero...

But somehow, this "proof" requires that
10 * 0.9999... = 9.9999... (an infinite number of instances of the digit 9 on both sides of the equation...but, somehow, "one more" on the left hand side--that being the 9 to the left of the decimal place!?)

At an infinite sequence of 9's, I'm somehow able to grab one more from out of the blue so that my subtraction works out to the desired difference of exactly 9?

This kind of thinking only works for finite numbers. In fact, the entire idea of "one more" fails when discussing the concept of infinity.


I guess that's the magic of infinity. ~shrug~

Exactly.

[Matthew E]

Here's a fun question I saw somewhere once.

You have an infinite deck of cards. Each card is numbered; there's a 1, a 2, a 3, et cetera.

I pick a card at random and look at it. Then you pick a card and look at it.

What are the odds that your card is higher than mine?

[Tadhg]

Obviously so. I guess I didn't express myself clearly enough.
10 * 0.1345 = 1.3450 ; that's the "zero" I'm talking about, the one at the end of 1.3450

Dread already covered the fact that "one more" doesn't work when talking about infinite sequences. But I'll add to it in that the above example you are not adding a zero, and that the zero already exists. 1.345 has an infinite number of zeroes after it, if you will. 1.3450 = 1.345 = 1.3450000000000000 .

[Loren]

This kind of thinking only works for finite numbers. In fact, the entire idea of "one more" fails when discussing the concept of infinity.

Such was one of the definitions of "infinity" that I found:

A set is infinite if we can remove some of its elements without reducing its size.

[MWGallaher]

That's the whole point of 'infinity': you aren't grabbing 'one more', because there are already an infinite number of them. The 9s go on forever. There's no final nine after which you'd need to conjure up one more.

Maybe that's exactly what I'm feeling is wrong with the "proof", then:
Let's try looking at it this way:
A = 9.9999... has an infinite number of instances of the digit '9'.
B = 0.9999... also has an infinite number of instances of the digit '9'.
Both values have the same number of instances of the digit '9', which is consistent with how I initially derived A (from A = 10 * B). That number of instances just happens to be infinity.
There is a step in the "proof" that says:
9.9999... - 0.9999... = 9.0000...
To confirm that result, I must perform an infinite number of subtractions (all of the form '9 - 9 = 0').
But somehow, I end up with one left-over 9! If A doesn't have one more instance of the digit '9', than B does, then I can't possibly reach the result the proof requires! For the subtraction to work, we have to somehow introduce infinity + 1, otherwise, the result of the subtraction is not, so far as I can tell, justifiable.

[Matthew E]

But 'infinity plus one' equals 'infinity'. That's a property of 'infinity'; you can add any finite number to it and still get infinity; you can multiply it by any finite number and still get infinity. I don't know how else to explain it. 'Infinity plus one'... to ask what it is if it isn't infinity is like saying, 'what happens after forever?'.

[Dreadstar]

For the subtraction to work, we have to somehow introduce infinity + 1, otherwise, the result of the subtraction is not, so far as I can tell, justifiable.

Once again, the idea that Infinite Series A has more or less members than Infinite Series B is where this fails.

You're thinking of it in *finite* terms and that won't work.

[Loren]

But somehow, I end up with one left-over 9! If A doesn't have one more instance of the digit '9', than B does, then I can't possibly reach the result the proof requires! For the subtraction to work, we have to somehow introduce infinity + 1, otherwise, the result of the subtraction is not, so far as I can tell, justifiable.

Maybe it would help if you read about Hilbert's Paradox of the Grand Hotel (http://en.wikipedia.org/wiki/Hilbert's_paradox_of_the_Grand_Hotel).

Imagine a hotel with an infinite number of rooms, which has a customer in every room. Then a new customer arrives. The hotel puts him in Room 1, moves the guy in Room 1 to Room 2, the guy in Room 2 to Room 3, and so on. Even though the hotel already had an infinite number of customers, it can still give a room to the new guy.

[Chris N]

Obviously so. I guess I didn't express myself clearly enough.

10 * 0.9 = 9.0 (one 9 on either side of the equation)
10 * 0.99 = 9.90 (two 9's on each side of the equation)
10 * 0.999 = 9.990 (three 9's on each side of the equation)
10 * 0.9999 = 9.9990 (four 9's on each side of the equation)
As I continue this pattern, one thing should remain the same: there will always be the same number of instances of the digit 9 on each side of the equations. If I want the same number of digits on the right hand side of the decimal point (after I perform the multiplication), I must pad it with a zero, as in:
10 * 0.1345 = 1.3450 ; that's the "zero" I'm talking about, the one at the end of 1.3450.
But somehow, this "proof" requires that
10 * 0.9999... = 9.9999... (an infinite number of instances of the digit 9 on both sides of the equation...but, somehow, "one more" on the left hand side--that being the 9 to the left of the decimal place!?)
At an infinite sequence of 9's, I'm somehow able to grab one more from out of the blue so that my subtraction works out to the desired difference of exactly 9? I guess that's the magic of infinity. ~shrug~


Your addiing of the zeroes above is superfluous.

9.9=9.90
9.99=9.990

How decimals work. No reason to add zero. Moving decimal is all that happens.

Even given that, you seek to add infinity to the end of the list of 9's. But an infinite list has no end!



I'm sorry Loren, I don't think you are right.

Don't think of infinity as a magical concept, think of it as a number that can be compared.

The whole one-to-one pairing again.

All even integers is a lower order of infinity than all integers, because there are some integers that cannot be paired up.


It is a number and can be compared. As a cardinal number (a number that represents the size of a collection), we call it aleph-naught, as I think Loren had mentioned.

Think of what you said above. The comparing thing. A number represents the size of a collection. Two collections are the same size if we can line them up. If I have 5 comics, and 5 coke cans, even if I didn't know to ascribe the number 5 to both of them, I could see they were the same size by putting one can on top of each comic.

This works for numbers. See a post of Gilda's from a few pages ago. But I'll say something similar.

Think of the positive integers, the collection
Then think of them and 0, the collection,

But you can see how they are lining up:

1,2,3,4,5,6,7,8,9...
0,1,2,3,4,5,6,7,8,...

They are the same size! The number x in the top holds hands with the number x-1 down below, and everybody has a partner! If you want to claim the second set is infinity+1 and that this is 1 bigger, there should be somebody left without a partner.

Since there's not. They're all the same size.

Now think of the positive even integers. Same thing. They line up

1,2,3,4,...
2,4,6,8,...

1 and 2 pair up. 2 and 4. x and 2x. And nobody is left out! So infinity/2=infinity.



Maybe it would help if you read about Hilbert's Paradox of the Grand Hotel (http://en.wikipedia.org/wiki/Hilbert's_paradox_of_the_Grand_Hotel).

Imagine a hotel with an infinite number of rooms, which has a customer in every room. Then a new customer arrives. The hotel puts him in Room 1, moves the guy in Room 1 to Room 2, the guy in Room 2 to Room 3, and so on. Even though the hotel already had an infinite number of customers, it can still give a room to the new guy.

Another good example. Gilda had mentioned that earlier too, though.

[Chris N]

Here's a fun question I saw somewhere once.

You have an infinite deck of cards. Each card is numbered; there's a 1, a 2, a 3, et cetera.

I pick a card at random and look at it. Then you pick a card and look at it.

What are the odds that your card is higher than mine?

I'm gonna go with 100%

[Matthew E]

That is one of three correct answers.

[Chris N]

That is one of three correct answers.

Three correct answers? That'll make my brain explode.

But you chose a card. Call it n. It was fixed first. n-1 numbers lose. infinitely many win. 0% of the cards lose.

[Matthew E]

Doesn't matter who goes first; your card has a finite number on it too, and they were drawn independently. Therefore whatever the number is on my card, it has a 100% chance of being higher than yours. So that's the second valid answer.

The third is, obviously, that we've got two cards coming out of the same deck, and one of them's got to be higher than the other, and since they're both randomly chosen, the odds are 50-50.

(Since all three of those calculations are mutually exclusive, the real conclusion here is therefore something like, 'therefore we don't have an infinite deck of cards'.)

[Gilda Dent]

Here's a fun question I saw somewhere once.

You have an infinite deck of cards. Each card is numbered; there's a 1, a 2, a 3, et cetera.

I pick a card at random and look at it. Then you pick a card and look at it.

What are the odds that your card is higher than mine?

I'm thinking it's 100&#37;.

Once you look at your card, you know the exact value of the card. Whatever that value, it is finite, meaning that there is a finite number of cards with a lower value, but an infinite number of cards with a higher value. Since the chances of my choosing an larger card are infinitely larger than that of a smaller card, the probability of my card being larger is 1.

Edit: C & C beat me too it, and said it much more concisely than I did.

[Matthew E]

I don't think it makes any difference to how the question works, but if it'll make anybody else feel better, assume that the two cards are drawn and viewed simultaneously.

[jessecuster3]

I don't think it makes any difference to how the question works, but if it'll make anybody else feel better, assume that the two cards are drawn and viewed simultaneously.

I don't care about that, how would you shuffle such a deck ?

[Chris N]

I don't think it makes any difference to how the question works, but if it'll make anybody else feel better, assume that the two cards are drawn and viewed simultaneously.

You're probably right. There is no reason that should make me feel better. But it does.

What if you drew first, and then showed me your card? Does that matter? Now I have to make a draw and I know I just have to beat 23. Seems to me the odds are good.

It must matter if I see your card mustn't it? With a finite deck, if we play the game and each draw one, we obviously have equal odds of winning. If you draw first and get the lowest card and show it to me. It's now clear I'm going to win before I draw.

[Matthew E]

I don't care about that, how would you shuffle such a deck ?

You would need to possess hands with fingers infinitely far from your thumbs.


What if you drew first, and then showed me your card? Does that matter? Now I have to make a draw and I know I just have to beat 23. Seems to me the odds are good.

Yeah. But there's still a way around it, like, I could say before drawing first, "I know this guy's going to draw a card after me, and whatever he gets is going to be finite, so I have a 100% chance of beating him."

It just illustrates that some mathematical concepts don't translate well into the physical universe. Like i.

[Dreadstar]

The infinite deck problem hinges on who views the card first. Or "fixes" the value.

I view first, your card will be bigger. You view first, my card will be bigger. We view at the same time, 50/50.

[Matthew E]

The infinite deck problem hinges on who views the card first. Or "fixes" the value.

I view first, your card will be bigger. You view first, my card will be bigger. We view at the same time, 50/50.

But these aren't quantum particles here. They're physical cards. Their values are as fixed as fixed is fixed.

[Chris N]

In the finite case, even if you take into account who draws first, either player has a 50&#37; chance of winning.

But it seems incorrect to assume the draws are indepent of each other. You need to draw one of the 52 first. Then I need to choose from the remaining 51. It so happens that I have a 50% chance of beating you.

However, if you draw and show me, it does change the odds.

Instead of me having a 50% chance of winning, based on not knowing what your card is, I now know how many cards in the deck win and how many don't.

[Gilda Dent]

Here's a fun one:

Two women meet for lunch one day. Each has been given a Valentines day present of a new purse by her girlfriend. Each woman believes that she has the more expensive purse, and they decide to bet on this. Each will contact her girlfriend and find out how much her purse cost. The woman with the more expensive purse will give it to the other woman as a consolation prize.

The first woman thinks to herself, "If I lose my purse, I'm losing the value of the less expensive purse, but if I win the other purse, I'm getting the value of the more expensive one. The risk is smaller than the reward, but the odds of winning are exactly 50/50, therefore this agreement is in my favor."

The second woman thinks the same thing, using the same reasoning.

The reasoning seems sound, but the results are contradictory. How can both have an advantage at the same time?

[Matthew E]

In the finite case, even if you take into account who draws first, either player has a 50&#37; chance of winning.

But it seems incorrect to assume the draws are indepent of each other. You need to draw one of the 52 first. Then I need to choose from the remaining 51. It so happens that I have a 50% chance of beating you.

However, if you draw and show me, it does change the odds.

Instead of me having a 50% chance of winning, based on not knowing what your card is, I now know how many cards in the deck win and how many don't.

...I wouldn't put it like that. There's a difference between the true odds and the odds as far as we know. But that doesn't make the second draw 'dependent' on the first draw, because you don't have any control over what card you draw. It's still independent. It does change the odds, yes, but that's not the same thing.

Gilda Dent: I'm not following. What are the stakes of this bet, exactly?

[Gilda Dent]

Gilda Dent: I'm not following. What are the stakes of this bet, exactly?

The two purses. Whoever has the more expensive purse gives it to the other woman. This produces a situation in which it seems that each stands to gain more than she risked.

[Chris N]

There's a difference between the true odds and the odds as far as we know.

I haven't really studied statistics, so forgive me, but does that make sense?

You lay the cards out on a table. I decide to flip over the first card. As far as I know, the odds it's a jack of clubs is 1/52. Clearly it is or it isn't a jack though, and the universe knows. The "true odds" must be either 0 or 1. It seems we only ever talk about the odds as far as we know. More knowledge on my part changes the odds, does it not?

[Chris N]

Gilda, you're hurting my brain.

[Dreadstar]

But these aren't quantum particles here. They're physical cards. Their values are as fixed as fixed is fixed.

Nevertheless, that's how it works.

c&c showed the proof earlier. The value of the known card is X. There are an infinite number of cards higher than X. There are a finite number of cards lower than X.

Always bet on the larger percentage, especially when it works out to 100%.

[Gilda Dent]

Gilda, you're hurting my brain.

Sorry. Here's an easier to understand version:

You have before you two envelopes, one with twice as much money it it as the other. You choose one envelope. You are now given the opportunity to switch to the other envelope. Should you switch?

The second envelope has either twice as much money in it or half as much; you don't know which. If you switch and the second envelope has half as much, you've lost half your original amount. If you switch and the second envelope has twice as much, you've won your original amount. Because the odds of the other envelope having more or less are the same, it's always in your favor to switch.

You are, in effect, risking half your money for twice the potential gain.

This works for both envelopes at the same time. How can switching both ways be an advantage?

[Matthew E]

Sorry. Here's an easier to understand version:

You have before you two envelopes, one with twice as much money it it as the other. You choose one envelope. You are now given the opportunity to switch to the other envelope. Should you switch?

The second envelope has either twice as much money in it or half as much; you don't know which. If you switch and the second envelope has half as much, you've lost half your original amount. If you switch and the second envelope has twice as much, you've won your original amount. Because the odds of the other envelope having more or less are the same, it's always in your favor to switch.

You are, in effect, risking half your money for twice the potential gain.

This works for both envelopes at the same time. How can switching both ways be an advantage?

It can't, and I think I've found the fallacy; I've bolded the key phrases in your premise. The question treats 'your original amount' as if it's the same in both cases, when it isn't.

Switching is no particular advantage, assuming you gain no additional knowledge.

It is a profitable game to play, though, as long as someone else is supplying the filled envelopes; you're bound to end up with *something*.

I'm not sure this is the same as the purse bet, unless the women in the purse thing were going to *switch* purses...

I'm getting lost in all these different conversations...

Dreadstar: Yeah, I understand it; it's just that I can't believe that, in the physical world, the order makes any difference. Which is where this breaks down, because you can't have an infinite deck in the physical world, which is actually the point of the whole question as far as I'm concerned.

coke & comics: You're right; I lost my way partway through my response to you and forgot that my point was the dependence/independence thing.

[Chris N]

I don't see it.

Say they're worth $10 and $25, respectively.

There's a 50&#37; chance you stand to lose $25 and a 50% chance you stand to win $25.



I'm also confused by the Valentine's day gifts coming from girlfriends. Don't women have boyfriends?

(kidding, sorry, stop hitting me)

EDIT: Oh, I missed that switching was happening. I had thought one woman walked away with both purses. Same thing, though. 50% says you lose $15. 50% says you win $15

[Dreadstar]

Dreadstar: Yeah, I understand it; it's just that I can't believe that, in the physical world, the order makes any difference. Which is where this breaks down, because you can't have an infinite deck in the physical world, which is actually the point of the whole question as far as I'm concerned.

Ah. I see.

Again, you can't move infinity from the conceptual to the real-world. It's either a conceptual problem or it's a real-world problem. If it's a real-world problem then it becomes a finite problem, and it changes. Like I said a ways back in this thread, this is where the problem comes in. The inablity to separate the concept from reality.

[Chris N]

Ah. I see.

Again, you can't move infinity from the conceptual to the real-world. It's either a conceptual problem or it's a real-world problem. If it's a real-world problem then it becomes a finite problem, and it changes. Like I said a ways back in this thread, this is where the problem comes in. The inablity to separate the concept from reality.

I can't separate anything from reality. Concept, fantasy, dreams, infinity, television, reality television. So I just pretend reality doesn't exist and assume everything is my imagination. This creates no confusion with what's real and what isn't.

[Gilda Dent]

It can't, and I think I've found the fallacy; I've bolded the key phrases in your premise. The question treats 'your original amount' as if it's the same in both cases, when it isn't.

Ah, but it is. There is a fixed amount in the envelope. For argument's sake, let's call it $10. The other envelope might contain $5, or it might contain $20. Your potential loss is $5, your potential gain is $10.

Now, substitute any other random amount for the first amount, and the general principle stays the same.

But let's say you open evelope 2 instead, and in that envelope is $5. You have no knowledge of the contents of the other envelope except that it has either twice as much or half as much. By switching, your potential gain is $5 and your potential loss is $2.50.

From each individual perspective, the potential gain is greater than the potential loss.

[Gilda Dent]

I don't see it.

Say they're worth $10 and $25, respectively.

There's a 50% chance you stand to lose $25 and a 50% chance you stand to win $25.




I'm also confused by the Valentine's day gifts coming from girlfriends. Don't women have boyfriends?

(kidding, sorry, stop hitting me)

Eh. Make it a gift from her wife instead. It doesn't affect the outcome much.

EDIT: Oh, I missed that switching was happening. I had thought one woman walked away with both purses. Same thing, though. 50% says you lose $15. 50% says you win $15

No, you had it right, one woman gets both purses, but the potential loss and potential gain are not the same.

Let's say I have a $25 purse. If the other woman has a $10 purse, I lose mine, in effect losing $25. If the other woman has a purse worth more than $25, I keep my purse and gain one worth more than $25. Regardless of the value, I more than double my money if I win.

Let's use your amounts, $10 and $25, with each woman not knowing how much the other's is worth. If I have the $10 purse, I'm risking $10 and winning $25. If I have the $25 purse, I'm risking $25 and losing $25. If I lose, I lose the amount I risked, but if I win, I win more than what I risked.

[Loren]

Ah, but it is. There is a fixed amount in the envelope. For argument's sake, let's call it $10. The other envelope might contain $5, or it might contain $20. Your potential loss is $5, your potential gain is $10.

Now, substitute any other random amount for the first amount, and the general principle stays the same.

But let's say you open evelope 2 instead, and in that envelope is $5. You have no knowledge of the contents of the other envelope except that it has either twice as much or half as much. By switching, your potential gain is $5 and your potential loss is $2.50.

From each individual perspective, the potential gain is greater than the potential loss.

And that's to be expected, because you've set up a game that has a positive expected outcome.

I'll use $100 instead of $10. There's a 50% chance the second envelope has $200, and a 50% chance it has $50. Switching envelopes might net you an additional $100, or lose you $50, so your expected outcome is (0.5*100) + (0.5*-50) = 25.

Very few real games offer positive expected outcomes. The lottery occasionally does, when the jackpot gets high enough. Real-world gambling proprietors depend on games that have negative expected outcomes. That's how they make profits.

And the expected outcome is even if the odds you'll lose a certain amount are the same as the odds you'll win that same amount. Say, betting $1 on a coin toss. If you win, you get an extra $1. If you lose, you lose the $1 you bet.

[Matthew E]

Gilda Dent: There's still something about this that I can't get my mind around. (Even after Loren's explanation.) I'm going to have to think about it.

Dreadstar: Yeah, I understand it...

[Chris N]

And that's to be expected, because you've set up a game that has a positive expected outcome.

I'll use $100 instead of $10. There's a 50% chance the second envelope has $200, and a 50% chance it has $50. Switching envelopes might net you an additional $100, or lose you $50, so your expected outcome is (0.5*100) + (0.5*-50) = 25.

Very few real games offer positive expected outcomes. The lottery occasionally does, when the jackpot gets high enough. Real-world gambling proprietors depend on games that have negative expected outcomes. That's how they make profits.

And the expected outcome is even if the odds you'll lose a certain amount are the same as the odds you'll win that same amount. Say, betting $1 on a coin toss. If you win, you get an extra $1. If you lose, you lose the $1 you bet.


I think that's the point, though. Casino games have a negative expected value for you and a postive one for the casino, who you're betting with. This seems to be strictly positive for both. That's impossible.

[Loren]

I think that's the point, though. Casino games have a negative expected value for you and a postive one for the casino, who you're betting with. This seems to be strictly positive for both. That's impossible.

But it's not positive for both. Rather, the way it's been presented, it's negative for the person offering the envelopes, because he stands to lose at least $50 no matter which envelope is finally chosen. He doesn't stand to gain anything he didn't already possess before the game began.

[Chris N]

But it's not positive for both. Rather, the way it's been presented, it's negative for the person offering the envelopes, because he stands to lose at least $50 no matter which envelope is finally chosen. He doesn't stand to gain anything he didn't already possess before the game began.

So you do trade envelopes, or do you not? Or does it not matter?

[Justin D.]

Is this an almost 320-post thread on math with the first post being about what almost equals the number 1?

You know what's coming, right?









Nerds!

[Gilda Dent]

But it's not positive for both. Rather, the way it's been presented, it's negative for the person offering the envelopes, because he stands to lose at least $50 no matter which envelope is finally chosen. He doesn't stand to gain anything he didn't already possess before the game began.

Sure, but we're looking at options once the money has already been distributed. The experimenter offers the envelopes to A and B. At this point, each has the option of keeping what is in her own envelope or switching with the other person. Whatever is in the envelopes already belongs to A and to B, yet switching still seems to be an advantage for each from her perspective.

This is, so far as I know, an unresolved paradox.

[Chris N]

Something's funny, here.

Nobody's risking $10.

Do you play the game or not? You stand to gain $25. You stand to lose $25. It depends which purse you have. Why isn't that just even?

And you know who's really a nerd: Justin Davis

[Night]

Well, on the Monty problem.... I had never really tried it experimentally so I gave a crack at it

using a billion tries each for the switch case and for the stay case (I used an old "Wargames" speedup by setting the number of players to zero)

I got 33.3351768% in favor of staying (~1/3)
and 66.6650107% in favor of switching (~2/3)

adding them together and dividing by 100 you get 1 ~ 1.000001875 not 0.999999999 :P


also thought it would be interesting to test processor speeds on this
Took the desktop 8.52 minutes
and my laptop 6.08 minutes (with expected value deviations after the 2nd decimal point)
Thought the laptop would do better since the desktop is getting old... then again it is 32-bit single-threaded code

[Chris N]

I say not.





I've changed my mind.

In standardly accepted calculus, they are equal.

But a new system is (extremely slowly) gaining acceptance, where instead of studying the classical real number line, people study hyperreals, which include infinitesimally small numbers.

One of the professors at my school is a big advocate for this.


wiki (http://en.wikipedia.org/wiki/Hyperreal_number)


It's a subject of much debate.

[Chris N]

I say not.





I've changed my mind.

In standardly accepted calculus, they are equal.

But a new system is (extremely slowly) gaining acceptance, where instead of studying the classical real number line, people study hyperreals, which include infinitesimally small numbers.

One of the professors at my school is a big advocate for this.


wiki (http://en.wikipedia.org/wiki/Hyperreal_number)


It's a subject of much debate.