Poll: A little math problem

[Ancalagon]

Puppy 1/Puppy 2
M/M
M/F
F/M
F/F

the same as this

Puppy 2/Puppy 1
M/M
M/F
F/M
F/F

Yes, it's totally the same. But it's not the same as:

Referred to Puppy/Not referred to Puppy
M/M
M/F
F/M
F/F

While the objects have not been interacted with, you can swap them if you like. But once observations have been made, you can't swap them, unless you can prove that the observations were entirely inconsequential.

What about my post no. 363? Do you disagree that the situation is as I describe in the first of my statements, i.e. "you could mean that "I've won either bet 1 or bet 2" and maintain that the real situation is one of the other two?

 

[Samirat]

Once you have an order, though, you have to stick with it. You can't change it halfway through, like you're doing. Essentially what you're going in the second one is just relabeling them. But once the problem's started, they can't be reassigned on a whim. For instance, MF is not the same thing as FM. Even when assured that one is heads, they aren't the same.

Uhh, I'm not the one doing changing halfway through--you are: "So, what if he uses the second puppy as the warrant for his response." [http://www.escapistmagazine.com/forums/jump/18.73797.815993]

How is that changing orders? Which one is the warrant for his response is irrelevant, it's still the second dog and the first dog. What you are doing is assigning them an order based on knowledge which you do not, in fact, have, namely, which puppy is the male that the washer was basing his claims on.

By the way, could you check out my question about the information difference in Post 357. It might give you a little uneasiness, which would be helpful.

 

[Jumplion]

What the hell, this topic's still going?!?!?

Havn't we found a solution yet people?

 

[Samirat]

It would appear that the only way to recreate the problem is to flip both coins at the same time. If they're both tails, it's irrelevant to the problem.

Dude, you CAN'T make up a model and then say 'oh, never mind the nonsensical results we get-just ignore them'. If you're getting nonsensical results and you have no explanation for them, then you can't trust your model yet.

They're not nonsensical, they're something which did not, in fact happen. Just like in the question at the beginning, there is a chance that the washer woman will say that both are female. In this case, both could be tails. The fact that one was a male in that particular case doesn't matter. It's like if you asked, out of a group of 100 dogs, if 50 were male. They could say yes, but they could just as easily say no. If I propose a problem where the answer happens to be "yes," it doesn't mean that the answer is always "yes." So if I recreate the experiment, and the answer is "no," it's merely a separate case, and irrelevant to the problem.

EDIT: If you recreated the original problem, with two random dogs of unknown gender, as the problem specifies, would the answer always be "yes?"

 

[Samirat]

I just noticed something, anytime in an argument when someone pulls "The majority believes its right" as evidence to the validity of something; I'm going to send them here to look at that poll.

Hehe, true dat.

 

[Alex_P]

Compare these two scenarios:
A. Your friend looks under one of the cups and says you have won at least one of your bets.
B. Your friend looks under both cups and says you have won at least one of your bets.

Do you see why A actually provides you with more information than B?

Yes, but I fail to see the relevance to the problem--check the OP: the word problem doesn't give us any information by which to figure out whether the Puppy Washing Man checked one puppy and then said yes, or both puppies and then said yes.

I agree totally--it's just that your example gives us more information about the situation than we have in the question under discussion, and therefore, isn't a good fit.

What are the probability matrices of these two scenarios?

A.
Investigated and Referred to Cup/Uninvestigated, Other Cup
Even/Even
Even/Odd



B.

Investigated and Referred to Cup/Investigated and Other Cup
Even/Even
Even/Odd

This business with referents is messing it up.

Here's a simpler one to consider:

You have a number-generating machine. It generates numbers based on the following known probability distribution:
25% 2
50% 1
25% 0

You press the button on the machine. Then your friend looks at the output and says "well, it's a positive number." What's the probability that machine output is 2?

-- Alex

 

[Samirat]

I don't understand. The problem I get. What I don't understand is how this has eleven pages when the answer and a good explanation were stated multiple times on the first page.

Actually, I don't think that many people even got it on the first page. Mostly it was just a torrent of 50 percents. That's the intuitive answer. If there's one thing to understand about problems like these, it's that they're non intuitive. So if the question seems ridiculously simple, take a closer look.

 

[Samirat]

Here's a simpler one to consider:

You have a number-generating machine. It generates numbers based on the following known probability distribution:
25% 2
50% 1
25% 0

You press the button on the machine. Then your friend looks at the output and says "well, it's a positive number." What's the probability that machine output is 2?

-- Alex

So it creates two 0/1's? That's genius. I love it. Saying it is a positive number guarantees at least 1 "1," so it fits perfectly.

Yeah. Sweet. Chomp on that one.

 

[Samirat]

You never responded to my information question on post 357. I would still like your thoughts.

 

[Ancalagon]

What about my post no. 363? Do you disagree that the situation is as I describe in the first of my statements, i.e. "you could mean that "I've won either bet 1 or bet 2" and maintain that the real situation is one of the other two?

Glancing back on it quickly, I guess--what's your point?[/quote]

That if you follow the logic that follows from situation A, i.e. that "I've won either bet 1 or bet 2"

bet1/bet2
M/M
M/F
F/M

then the chance that there are two males is 33%, whereas if you follow your original logic, that you've either won bet 1(situation B), or you've won bet 2(situation C)(both of which you've since stated you can't prove):

Then either

bet1/bet2
M/M
M/F

or:

bet1/bet2
M/M
F/M

 

[Samirat]

It would appear that the only way to recreate the problem is to flip both coins at the same time. If they're both tails, it's irrelevant to the problem.

Dude, you CAN'T make up a model and then say 'oh, never mind the nonsensical results we get-just ignore them'. If you're getting nonsensical results and you have no explanation for them, then you can't trust your model yet.

They're not nonsensical, they're something which did not, in fact happen. [/quote]

If your model is supposed to recreate what happened, and you get results that did not happen, then yes--the results are nonsensical and they need to be explained.[/quote]

The question assumes a random pair of dogs. This alone ensures that sometimes, the washer woman will say "no." In this particular instance, the washer woman said "yes." Therefore, you base your probabilities off of other particular instances, where the washer woman says "yes." They aren't "results" when the situation is different. They just aren't the same problem. They aren't nonsensical, they're just a different case from the premise stated in the problem. You can't have "results," since the problem doesn't even apply to situations where the washer woman says "no."

Again, back to the 100 dogs, magnified example. If I have 100 dogs, and ask if 50 of them are males, and the answer is "yes," it doesn't mean the answer will always be "yes," for a random set of 100 dogs. I could still propose a problem based on the premise that in one instance, the answer was "yes."

 

[Samirat]

You never responded to my information question on post 357. I would still like your thoughts.

You're hung up on the first element on a matrix representing the first element in time. That's where you're misunderstanding me.

You're not answering my question. I'm asking how you can justify the probability being the same, when in one case you're receiving more information than in the other.

[
It is; unfortunately any problem without that step of knowing one result doesn't capture what's being expressed in the original problem, so a machine that only spits out one number isn't an equivalent for the issue raised by the original problem.

No, it's equivalent. Say that it instead spits out two pairs of binary toggles: 0 and 1. So saying it is positive is the same thing as guaranteeing at least one "1."

 

[Samirat]

The question assumes a random pair of dogs.

No, it really doesn't. It assumes one male dog and one random dog. It starts out talking about a random pair of dogs, but by the time it asks you to come up with any probabilities, it assumes one male dog.

I can't make that any simpler--just read the word problem over until you get it.

Yes, you find out that one of the dogs is male. Not knowing which one, though, is important. You don't have information as to which was the male dog, so you can't assign it an identity or a placement in your solution set. Which you do. You can't assign an order based on which the washer saw to be male, because you don't know which.

Again, you have three solutions.

Sparky is Male, Othello is Female.
Sparky is Female, Othello is Male.
Sparky is Male, Othello is Male.

You see how the order remains constant. That's key.

 

[Samirat]

You never responded to my information question on post 357. I would still like your thoughts.

You're hung up on the first element on a matrix representing the first element in time. That's where you're misunderstanding me.

You're not answering my question. I'm asking how you can justify the probability being the same, when in one case you're receiving more information than in the other.

What information is that, again? If it's that important and you think it'll really change my mind, I'd appreciate it if you summed up this train of thought into one single piece of text.

No, it's equivalent. Say that it instead spits out two pairs of binary toggles: 0 and 1. So saying it is positive is the same thing as guaranteeing at least one "1."

Is zero odd or even or neither according to you?

Where do odd and even come in? He says that it's positive. Zero is not positive. You are guaranteed one "1."

 

[Alex_P]

This business with referents is messing it up.

It is; unfortunately any problem without that step of knowing one result doesn't capture what's being expressed in the original problem, so a machine that only spits out one number isn't an equivalent for the issue raised by the original problem.

You don't know "one result." You know a single new fact about the whole set together. If I pick a single element from that set, you still can't say anything definite about it.

-- Alex

 

[Alex_P]

The question assumes a random pair of dogs.

No, it really doesn't. It assumes one male dog and one random dog. It starts out talking about a random pair of dogs, but by the time it asks you to come up with any probabilities, it assumes one male dog.

I can't make that any simpler--just read the word problem over until you get it.

A fair coin is still a fair coin even if you already know its current value.

-- Alex

 

[Saskwach]

Two out of the three possible outcomes is female. The other is male - hence 1/3. Let's not talk about matrices, or definitions or applicability - instead, tell me where the logic is refutable. Where is this premise that you must accept?

That two out of the three possible outcomes is female AFTER we've figured out that one of the puppies is male--that's the premise I'm not accepting.

I see now. Thanks for that. Ok, well as I laid out, the reason is because of how the woman has been asked to check the dogs and then how we've been asked dto check the dogs. She wasn't asked "Is Jesse a male?" This is crucial because such a question would actually give us two pieces of information: the gender of one dog; and which dog we're referring to. It would make the gender of the other completely independent and thus 50/50.

But the gender of the other dog is *always* completely independent. We know that from the initial conditions, that each and every dog has a 50/50 shot at being male or female.

The gender of which dog? That is the crux of the matter - we don't know, so we have to consider the possibility that one or the other dog might be the one male.

 

[Samirat]

]You don't know "one result." You know a single new fact about the whole set together. If I pick a single element from that set, you still can't say anything definite about it.

The single new fact is that one of the results will be male. So I do know a result--I might not know which element that fact describes, but I know it will describe one of those elements.

Yes, and in this case, you know 1 of them will be a "1." It's the same problem. Exactly. It's just like if you turned the gender of the dogs into a numerical value, and then added them together, with males being 1 and females being 0. Saying that it's positive is equivalent to saying that there is at least one "1," which correlates to saying at least one male.

 

[Saskwach]

[
The gender of which dog? That is the crux of the matter - we don't know, so we have to consider the possibility that one or the other dog might be the one male.

The dog who's gender we don't know.

I know we have to consider the possibility, but you're over-considering it. You're reifying a set of symbols representing abstractions in a matrix into real things.

But they are real things; each is a dog with a name and an individual identity. You are instead forgetting that and attaching the name "Jesse" to whichever dog is the male we first discard. This is why I was so obsessive in my naming: because to forget that M/F is different to F/M is the same as declaring that Jesse is whichever damn dog you say he is.

 

[Samirat]

It's been fun everybody, but we're going around in circles. We all start out with the same set of probabilities, we all (well, me at least) accept each other's 50%/33% if we grant for the sake of argument that the other has incorporated the "Yes!" from the Puppy Washing Man correctly, and we seem to be left with just the issue of how to correctly incorporate that information.

So let me try one last time to express it in a slightly different way. Now, we can all agree on this, I think, if we were setting up the probabilities based on the information in the word problem before we deal with the Puppy Washing Man (XOR is the exclusive or [http://en.wikipedia.org/wiki/Exclusive_or#Exclusive_.E2.80.9Cor.E2.80.9D_in_natural_language]):


Both male = .25 (.5 x .5)

One of each = .50 (.5 x .5 + .5 x .5 -XOR- .5 x .5 + .5 x .5)

Both female = .25 (.5 x .5)


Now, what actually happens when we get that information from the Puppy Washing Man? We know at least one dog is male. Now, what happens to the probability of one dog being male when we know one dog is male: it becomes 1, and therefore the probability of one dog being female becomes 0. Well, here's what I think happens: I've put the changes that I think are correct in bold.


Both male = .5 (1 x .5)

One of each = .5 (1 x .5 + 0 x .5 -XOR- 0 x .5 + 1 x .5)

Both female = 0 (0 x .5)

That's...about as well as I think I'm going to be able to explain myself.

Nice try. All you need to recognize is that having two dogs, and knowing one is male, doesn't have the same probabilities as just having one unknown dog. There are two unknowns. One of the dogs has an unknown gender, and which dog is a male is unknown. You must see that there is some difference between the situation where you know Dog 1 is male, and where you know one of the dogs is a male. On the problem that Alex_P proposed, which is an exact replica of this one, but in simpler terms, do you think the chances of the result being 2 are equal to the chances of a 1?