"Simple" Math Problem

[Charla]

A balloon with just enough helium to offset its weight (so it floats in place without rising or falling) is placed in an elevator. Will the balloon move with the elevator or hit the ceiling/floor, assuming it doesn't accelerate too quickly?

What about a balloon on a non-enclosed platform open to the air?

A bird in an enclosed elevator?

A bird on a non-enclosed platform?

The balloon will move when the elevator accelerates and only when it accelerates. If the balloon is held in place while the elevator is accelerating and released only while the elevator's velocity is constant, it will not move.

This should go for a hummingbird as well (other types of birds don't or can't hover in place I think, and in any event a hummingbird is the easiest to imagine doing so).

As for a non-enclosed elevator it's more complicated. I'm assuming for simplicity that the top and bottom of the platform are covered, and the sides consist of either a grating or just four poles at the corners. In either case, I think some sort of air current would be set up which would cause the balloon to move. A bird could probably remain motionless against a slight air current, but it would have to do so intentionally. And once you start wondering whether the bird would try to remain motionless, you have to wonder why it was hovering in an elevator in the first place, and that doesn't work.

[Charla]

we are left with 2 possibilities:

You are holding Bag A.
You are holding Bag B.

If you are holding Bag A, the coins will match.
If you are holding Bag B, the coins will not match.

You are no more likely to be holding Bag A than you are to be holding Bag B.

This is wrong and I'll show you why

It's true that there are now two possible states, but either of those states must have been reached by two random events:
You picked a bag, and
You picked a coin.

If the bag was originally the bag with one of each coin, then there was also a chance that it was discarded when you picked the silver coin. Therefore, the state in which you are holding bag B is less likely than that in which you are holding bag A. If you picked the silver coin from bag B, you discarded it, and didn't reach this point in the algorithm (so to speak).

Yes, this does in fact contradict my earlier reasoning using pseudocode. It appears that I was wrong then.

[Jotaru]

You are no more likely to be holding Bag A than you are to be holding Bag B.

This is where the other 'theory' breaks away from your reasoning. The idea is that the act of pulling out a gold coin has given you information about what bag you have; its more likely to pull a gold coin out of one bag than the other, so that bag has a higher chance of being the one you have.. um.. kinda hard to explain.. perhaps watch that video if you haven't.

Basically, without ANY information (i.e. you just take two of the bags, don't pull any out, and then say, ok, what're my chances of pulling TWO gold coins out of the bag) its 50/50- it could be, or it could not be. You now have more information about the bag though; its more likely to be bag A than bag B, thus, its more likely you'll pull out a gold coin than a silver coin.

[Amele]

A balloon with just enough helium to offset its weight (so it floats in place without rising or falling) is placed in an elevator. Will the balloon move with the elevator or hit the ceiling/floor, assuming it doesn't accelerate too quickly?

What about a balloon on a non-enclosed platform open to the air?

A bird in an enclosed elevator?

A bird on a non-enclosed platform?

the enclosed cases, it stays put (edit: relative the elevator I mean). (if you go from standing to immediate free fall it might float up slightly)

not sure about the other case; I think in this case it still stays put relative to the platform, more or less. (buoyancy being what it is and assuming there's no breeze or anything.)

[Niiro]

This is wrong and I'll show you why

It's true that there are now two possible states, but either of those states must have been reached by two random events:
You picked a bag, and
You picked a coin.

This is answering a different question.

It's not asking what the chances are that you picked a gold/silver coin from the mixed bag or the 2 gold or 2 silver bag, it's simply asking what the chances are that the coins you have will match.

No matter which bag you choose the chances are still the same.

The kind of coin being gold or silver is irrelevant, the point is we know what coin we are holding, and can infer what the chances are that the other coin is the same.

You pick a coin.

You know it can't be the bag with 2 unlike coins, so that's out.

You have 2 possibilites:

You chose the coin from the mixed bag, they don't match.
You chose the coin from a homogeneous bag, they do match.

Neither is more likely than the other.

[Amele]

This is answering a different question.

It's not asking what the chances are that you picked a gold/silver coin from the mixed bag or the 2 gold or 2 silver bag, it's simply asking what the chances are that the coins you have will match.

No matter which bag you choose the chances are still the same.

The kind of coin being gold or silver is irrelevant, the point is we know what coin we are holding, and can infer what the chances are that the other coin is the same.

You pick a coin.

You know it can't be the bag with 2 unlike coins, so that's out.

You have 2 possibilites:

You chose the coin from the mixed bag, they don't match.
You chose the coin from a homogeneous bag, they do match.

Neither is more likely than the other.

the homogenous bag is twice as likely... (because half the time you have the mixed bag, you would draw dead on a silver) being told you have a gold coin just tells you which step in the logic process you're at - prior to that your chances of having any given bag are 1/3. drawing gold makes the homogenous gold bag twice as likely; drawing silver makes the homogenous silver bag twice as likely; etc.

[archibaldcrane]

This is wrong and I'll show you why.

Heh, this is fun.

6 possibilities. 2 are that you pick a silver coin from the silver-only bag. 2 are that you pick a gold coin from the gold-only bag. 1 is that you pick a silver coin from the mixed bag. 1 is that you pick a gold coin from the mixed bag.

I know it sounds counter-intuitive, and it took me a long time to wrap my head around the first time I heard it too. But, if you pulled a gold coin, 2 out of the 3 possibilities that happens because you grabbed the gold-only bag. The other one of the 3 is when you grab the mixed bag.

Therefore 2 out of the 3 times this happens, you will pull a second gold coin. This probability problem has it's own "paradox" named after it for a reason, it's very counter-intuitive. But be that as it may, I'm right and you 50%ers are wrong. Very very wrong.

[Niiro]

1/3. drawing gold makes the homogenous gold bag twice as likely; drawing silver makes the homogenous silver bag twice as likely; etc.

I'm not seeing it like this

"You pick one bag, and toss the other two out. You pick out a coin; its gold. What are the chances the other coin in the bag is gold?"

The parameters are that you know what is in each individual bag and you know what coin you are holding. No matter which bag you choose you can always eliminate one bag.

What I'm trying to say is when you draw the coin you know it's not the unlike homogeneous bag, so you're left with 2 potential bags and neither is more likely than the other.

I think the difference is coming from this:

Ignoring what kind of coin you have: you are going to have the different coins 1/3 of the time and the same coin 2/3 of the time.

Considering coin type:
Drawn coin/Bag coin
G/G = 33%
S/G = 16%
G/S = 16%
S/S = 33%

If you repeated this a million times you would get these results.

This is merely the luck of the draw, my argument was that if you break it down into smaller pieces you can infer certain things about which bag you have and which you don't.

[Amele]

Heh, this is fun.

6 possibilities. 2 are that you pick a silver coin from the silver-only bag. 2 are that you pick a gold coin from the gold-only bag. 1 is that you pick a silver coin from the mixed bag. 1 is that you pick a gold coin from the mixed bag.

I know it sounds counter-intuitive, and it took me a long time to wrap my head around the first time I heard it too. But, if you pulled a gold coin, 2 out of the 3 possibilities that happens because you grabbed the gold-only bag. The other one of the 3 is when you grab the mixed bag.

Therefore 2 out of the 3 times this happens, you will pull a second gold coin. This probability problem has it's own "paradox" named after it for a reason, it's very counter-intuitive. But be that as it may, I'm right and you 50%ers are wrong. Very very wrong.

you got it right but for the wrong reasons.

it doesn't actually matter how many coins there are in any given bag, just the ratio.

the likelihood is still two thirds, even if the all gold bag has 300 gold coins and the mixed bag is 1 silver and 1 gold.

edit: the right way to think about it is in my post above.

[Amele]

I'm not seeing it like this

"You pick one bag, and toss the other two out. You pick out a coin; its gold. What are the chances the other coin in the bag is gold?"

The parameters are that you know what is in each individual bag and you know what coin you are holding. No matter which bag you choose you can always eliminate one bag.

What I'm trying to say is when you draw the coin you know it's not the unlike homogeneous bag, so you're left with 2 potential bags and neither is more likely than the other.

I think the difference is coming from this:

Ignoring what kind of coin you have: you are going to have the different coins 1/3 of the time and the same coin 2/3 of the time.

Considering coin type:
Drawn coin/Bag coin
G/G = 33%
S/G = 16%
G/S = 16%
S/S = 33%

If you repeated this a million times you would get these results.

are you agreeing with me? that's pretty much what I said.

edit: oh weird. did you mean to contradict yourself? your truth table (while accurate) shows that the homogenous bag is twice as likely as the heterogenous bag, given the a priori of a particular coin color. but you had earlier stated:

What I'm trying to say is when you draw the coin you know it's not the unlike homogeneous bag, so you're left with 2 potential bags and neither is more likely than the other.

which isn't borne out by your truth table...

[Niiro]

Read my edit at the bottom, I think that's where my difference in logic came from.

[Amele]

Read my edit at the bottom, I think that's where my difference in logic came from.

ok yeah, the belief that you can make an inference is wrong; that's the paradox.

it's very similar to the monty hall problem. (or the boy girl problem for that matter)

[archibaldcrane]

you got it right but for the wrong reasons.

it doesn't actually matter how many coins there are in any given bag, just the ratio.

Yes, whether the gold-only bag has 2 coins or 2000 it's the same.

[Niiro]

ok yeah, the belief that you can make an inference is wrong; that's the paradox.

it's very similar to the monty hall problem. (or the boy girl problem for that matter)

I'm not seeing how this is a paradox then.

Without trying to figure out which bag is which, the chance of the other being gold is flatly 33%.

Where's the problem?

[aurik]

While we're on the topic, I'd like to bring up something I saw on lol4chan last night, and really got me thinking, much in the same fashion as the 'Will the plane on a treadmill fly?' question...

Bag 1 has a gold coin, and a silver coin
Bag 2 has two silver coins
Bag 3 has two gold coins.

You pick one bag, and toss the other two out. You pick out a coin; its gold. What are the chances the other coin in the bag is gold?

Here's my take on it. It's very simple.

* The question "what's the chance that the other coin is gold?" is equivalent to asking "are both coins gold" is equivalent to asking "did you choose bag A".

There are 6 situations:
Select coin A1
Select coin A2
Select coin B1
Select coin B2
Select coin C1
Select coin C2

We also know that
P(A1) + P(A2) = P(B1) + P(B2) = P(C1) + P(C2) -- each bag has the same overall probability
and
P(A1) = P(A2), P(B1) = P(B2), P(C1) = P(C2) -- each coin is equally likely as its neighbors to be selected
From here we can derive that P(A1) = P(B1) etc.

Now we can calculate P(A) given that we know B2, C1, and C2 have not occured
P(A|not B2, C1, C2) = [P(A1) + P(A2)] / [P(A1) + P(A2) + P(B1)]
= 2 / 3



Another way to think of it is a probabilistic approach:
if we were to repeat the experiment X times, we would pull out a gold coin from bag A 33% of the time and a gold coin from bag B 33% * 50% = ~16% of the time, and we would have a mis-trial the other 50% of the time.

So:
50% mistrials
33% other coin is gold
16% other coin is silver.

"The other coin is gold" is twice as likely "the other coin is silver", which means if those are the only two possibilities, gold is 66% and silver is 33%.

[Max™]

Time for depressing results of probabilistic logic.

I have in my hand a marble with your first initial on it.

I place it in a wooden box which could hold thousands of marbles this size, I assure you that your marble is the only one with your initial on it.

I ask you if you think the box holds 10 marbles, or 1000 marbles.

I then ask you to press a lever on the side of the box which will dispense a marble.

You press it three times and get your marble with your initial on it.

Naturally you conclude that the box has 10 marbles.


Now replace "press the lever til you get your marble" with "be born at a time when more human beings have ever been alive at one time than ever before".

Then tell me what the odds of finding yourself at an arbitrary point in the expansion of the human population is, versus the odds of finding yourself at a point where most of the humans who ever have, been, or will be alive are.

http://en.wikipedia.org/wiki/Carter_catastrophe

Have fun!

[Charla]

Time for depressing results of probabilistic logic.

I have in my hand a marble with your first initial on it.

I place it in a wooden box which could hold thousands of marbles this size, I assure you that your marble is the only one with your initial on it.

I ask you if you think the box holds 10 marbles, or 1000 marbles.

I then ask you to press a lever on the side of the box which will dispense a marble.

You press it three times and get your marble with your initial on it.

Naturally you conclude that the box has 10 marbles.


Now replace "press the lever til you get your marble" with "be born at a time when more human beings have ever been alive at one time than ever before".

Then tell me what the odds of finding yourself at an arbitrary point in the expansion of the human population is, versus the odds of finding yourself at a point where most of the humans who ever have, been, or will be alive are.

http://en.wikipedia.org/wiki/Carter_catastrophe

Have fun!

Logic can be a scary thing

Now it is such a bizarrely improbable coincidence that anything so mind-bogglingly useful could have evolved purely by chance that some thinkers have chosen to see it as a final and clinching proof of the non-existence of God. The argument goes something like this:

"I refuse to prove that I exist," says God, "for proof denies faith, and without faith I am nothing."

"But," says Man, "the Babel fish is a dead giveaway isn't it? It could not have evolved by chance. It proves that you exist, and so therefore, by your own arguments, you don't. Q.E.D."

"Oh dear," says God, "I hadn't thought of that," and promptly vanishes in a puff of logic.

"Oh, that was easy," says Man, and for an encore goes on to prove that black is white and gets himself killed on the next zebra crossing.

Most leading theologians claim that this argument isn't worth a pair of fetid dingo's kidneys, but that didn't stop Oolon Colluphid from making a fortune with his book Well That About Wraps It Up For God.

[Amele]

I'm not seeing how this is a paradox then.

Without trying to figure out which bag is which, the chance of the other being gold is flatly 33%.

Where's the problem?

the chance of the other being gold (given that the first is gold) is 66% not 33%. you're still in the paradox..

[Wulfgang]

Boss just received an answer from the teacher...

The answer is D: 153

Pattern: 57 is an odd number, 134 is an even number that is larger than the prior number in the series; ergo the next number should be an odd number that is larger than 134; the only applicable answer is D: 153.

Yes, it is a poorly designed question, and yes I am upset.

[Norellicus]

Boss just received an answer from the teacher...

The answer is D: 153

Pattern: 57 is an odd number, 134 is an even number that is larger than the prior number in the series; ergo the next number should be an odd number that is larger than 134; the only applicable answer is D: 153.

Yes, it is a poorly designed question, and yes I am upset.

Mother of god that's retarded, and we're supposed to pick that up from a preceding series of TWO?