Poll: A little math problem

[Saskwach]

SNIP
I've read it, and you're missing the point. The question "What is the probability that the other one is a male?" makes no sense unless we can distinguish between puppies.

Based on the information in the question, in what way can you distinguish between the puppies other than by calling one 'the male' and the other 'the one we don't know about'?

That's at the heart of the problem, I think. Clearly, the question should have read "What is the probability that both pups are male." That I think would have been clear to everyone

-given that at least one s male. Let's not open up 25% as an option for debate.
And besides, where's the fun in making things obvious? That's just not how maths is done.

 

[Samirat]

Congrats, Cheeze. I was actually starting to think you might be carrying on intentionally. It's nice to know that sometimes 19 pages of argument isn't wasted time.

It certainly isn't--now I know exactly why the answer is 50% in the most sensible reading of the question: anyone who thinks it is 33% didn't really read the question that was asked: what is the sex of the *other* puppy, meaning the information in the problem wasn't about the pair, it was about *one* of the puppies.

Once I saw Doug's answer, I saw a situation where it would be 33%, because 10=/=01. On the other hand, male/female=female/male where the extra information *isn't* referring to the pair of puppies, but to one of the puppies--otherwise the phrase "the other one" makes no sense.

You know

My opinion is #3, that the final question does not contain any additional information, but rather, clues us into what the question-asker meant when they were talking about the Puppy Washing Man, that the question asker meant that he checked the pups and was thinking either

A) yeah, the first one I checked was male

or

B) yeah, the second one I checked was male, after the first turned out to be female

rather than responding on a basis of knowledge that they came from Female Pair Screening Puppy Breeder.

Yes, and if the puppy-washer looked at one or both of the puppies to find a male, that gives you 33%.

50% only makes sense if "at least one" actually means "I only have this one puppy to look at and don't know anything about the other one."

Or if it means "I am answering yes on the basis of one of the puppies I looked at that turned out to be male."

Sigh. Back to square one...

Okay, so, one column is "the other one." If you call it anything else, you're misunderstanding the language of the problem. How would you label the other column? What would you put in its rows?

The crux of the problem is that you can't tell which one the dog washer is referring to when she says "yes." If it's the first one, the chances that the other one is male are indeed 50 percent. If it's the second one, the first is female, since if the first one were male it would fall under one of the previous two cases.

We've done our job. Now let's all pass the champagne.

 

[Alex_P]

"Reflipping" anything is unnecessary, and it's a bad idea as far as producing a simple and functional model is concerned. (It is, however, correct to say that we don't really care about what happens to those FFs -- the calculations basically end up discarding the data for us when we make them.) You should just keep all the results you get.

How is discarding those results different from the Breeder holding back all the FF pairs before they get to the shop?

No, no, I'm saying that those results are physically present, but it's possible to do the math without having to keep track of what they were. (Therefore, yes, the math can work out the same if you filter them by "reflipping." I just think that makes the model poorer and more confusing.)

-- Alex

 

[Doug]

The question states that at least one pup is male (because the person doing to phoning asked and the answer was 'Yes')

 

[Alex_P]

Here's a Bayes' Theorem approach. Bayes' Theorem kicks ass because it makes it very easy to start with an initial probability distribution and then pile on additional information.

...
Bayes' Theorem:

P(A|B) = P(B|A) * P(A) / P(B)

(Notation: P(X) is the probability of an event X and P(X|Y) is the probability of event X if event Y is a given.)
...

I'm going to use "2" to refer to "two males" and "1+" to refer to "at least one male."

So, everyone who's still doubtful, feel free to play along at home -- calculation below the "spoiler":
What is P("2")? (In other words, what is the probability of a random combination of two puppies yielding two males?)
What is P("1+")? (In other words, what is the probability of a random combination of two puppies yielding at least one male?)
What is P("1+"|"2")? (In other words, what is the probability that a set of two male puppies is also a set with "at least one male" in it?)

So, given all that, what is P("2"|"1+")?

Spoiler: Answers

P("2") = 1/4 (set MM out of sets {MM, MF, FM, FF})
P("1+") = 3/4 (sets MM, MF, FM out of sets {MM, MF, FM, FF})
P("1+"|"2") = 1 (because any set of two males necessarily contains "at least one male")

Thus, P("2"|"1+") = P("1+"|"2") * P("2") / P ("1+") = 1 * (1/4) / (3/4) = 1/3

Voila!

-- Alex

 

[Ancalagon]

The question states that at least one pup is male (because the person doing to phoning asked and the answer was 'Yes')

If that's what it states, on what basis is the end question referring to "the other one"? What information do we have that allows us to identify one puppy as "the other one" if not information that one specific pup is male, and "the other one" is the one we don't know anything about?

It's interesting to do this while replaying God of War. You all probably are thinking of cute little puppies. I'm thinking of little hell hounds that I want to pound into the ground for the extra health.

I think that the issue has moved from one of mathematics to one of semantics. If I'm understanding you right, Cheeze, then you'd agree with the following:

If the guy looks at both dogs before making his statement, and upon checking the dogs says that there is a male, either on the basis that one of the dogs is a male or both are, then the probability that they're both male is 33%.

Or:

The guy picks up a dog, it's male, he may look at the other, or he may leave it unchecked, either way, the chance of them both being male is 50%.

And that what we're really discussing is whether the language in the question (in particular, the phrase 'the other dog') implies the first or the second; or indeed that the language of the question is ambiguous about what he's done, and therefore how do we calculate the probability with that ambiguity.

Or am I wrong, and you'd still say that in the first instance, i.e. he checks both and then reports that there is a male, the chance is 50%?

 

[Ancalagon]

SNIP
I've read it, and you're missing the point. The question "What is the probability that the other one is a male?" makes no sense unless we can distinguish between puppies.

Based on the information in the question, in what way can you distinguish between the puppies other than by calling one 'the male' and the other 'the one we don't know about'?

That's at the heart of the problem, I think. Clearly, the question should have read "What is the probability that both pups are male." That I think would have been clear to everyone

-given that at least one s male. Let's not open up 25% as an option for debate.
And besides, where's the fun in making things obvious? That's just not how maths is done.

What we need is language that somehow screens out the FF pups from the predicted 25/50/25 breakdown, without making it obvious that they have been screened.

Yeah, I don't know how to do that either!

If the question were phrased: "There are two puppies. We are informed that at least one of them is male. What is the probability that they are both male?" Would we all agree the answer is 33%?

 

[Ancalagon]

I'd still say 50% in both instances and here's why--you listed the two possiblites as in your 33% scenario as:

1) on the basis that both of the dogs are male

OR

2) the basis that one of the dogs is a male.

So wouldn't (1) look like:

Pair of Dogs
M

And (2) would look like

Basis Dog/Non-Basis Dog
M/M (if he liked Basis Dog better and decided to talk about that dog and not the other or about the pair)
M/F (if he had to talk about basis dog to say "Yes!")

Right? Because putting anything but an M under Basis Dog...doesn't make any sense, because otherwise, it wouldn't be a sufficient basis for saying "Yes!" to the question.

I phrased it poorly. What I meant for 2) is: the basis that only one of the dogs is male. So 1) would look like:

Pair of Dogs
MM

and 2) would look like:

Basis Dog/Non Basis Dog
MF

So the question is: how likely is option 1, and how likely is option 2? If we go back to having just two dogs, before the dogs have been examined by the dog-sexer, we have:

Dog 1/Dog 2 (all options are equally likely)
MM
MF
FM
FF

The first leads to option 1. The second leads to option 2, with Dog 1 being the basis dog. The third leads to option 2, with Dog 2 being the basis dog. These three possibilities are equally likely. The fourth is ruled out by the sexing. So one third of the time, it's option 1, and two-thirds of the time it's option 2. That how moving from Dog 1/Dog 2 to Basis Dog/Other Dog affects the probabilities.

 

[Ancalagon]

Dog 1/Dog 2 (all options are equally likely)
MM
MF
FM
FF

The first leads to option 1. The second leads to option 2, with Dog 1 being the basis dog. The third leads to option 2, with Dog 2 being the basis dog.

No, the options rule out anything that doesn't look like them, including their converses. Otherwise, the options do not cover all possible outcomes.

So say Dog 1 is an English Beagle, and Dog 2 is an American Beagle, then the American Beagle couldn't be male while the English one is female, since that configuration (FM) doesn't look like any option available in the new set of configurations?

 

[Alex_P]

Remember, this is under the heading you called "If the guy looks at both dogs before making his statement"; if he's looked at both dogs, F/M doesn't add a row, it just switches
Basis Dog/Non Basis Dog to Non Basis Dog/Basis Dog

But if you fold F/M into M/F that doubles the effective probability of M/F.

-- Alex

 

[Ancalagon]

Remember, this is under the heading you called "If the guy looks at both dogs before making his statement"; if he's looked at both dogs, F/M doesn't add a row, it just switches
Basis Dog/Non Basis Dog to Non Basis Dog/Basis Dog

Okay, so you've got:

Pair of Males: M/M

Basis Dog/Non Basis Dog
M/F

Non Basis Dog/Basis Dog
F/M

The last two options give the same result, but came from different equally likely options, and are still equally as likely.

EDIT: Changed to make sense(to me at least...)

 

[geizr]

Now, here is where things change. In the problem, we are told that at least one of the dogs is male. But, we don't know which one.

Actually, we do. Look at the question the word problem asks: "What is the probability that the other one is a male?"

If the only info we have for discriminating between one dog and the other is that one is male, and we're asked about the "other" dog, that means our info must pertain to not just the set but one of the dogs, the non-other one, so things don't actually change, other than our labels for the matix, maybe.

No, you really don't. Here's what the question says:

"Is at least one a male?" she asks him. "Yes!" she informs you with a smile. What is the probability that the other one is a male?

First off, the problem explicitly asks if at least one is male(which is answered in the affirmative), but there is no indication if we are talking about dog1 or dog2. Consequently, we don't really know which one is the "other" one?is it dog1 or dog2? It is only if the problem explicitly labels dog1 or dog2 as being that one that we are told is male do we get a probability of 50%. But, because we don't know which one is being referenced when we are told at least one is male, we obtain a probability of only 33%. It is invalid logic to presuppose that it MUST be dog1(or dog2) that the problem is referencing as the known male dog.

 

[werepossum]

SNIP
First off, the problem explicitly asks if at least one is male(which is answered in the affirmative), but there is no indication if we are talking about dog1 or dog2. Consequently, we don't really know which one is the "other" one?is it dog1 or dog2? It is only if the problem explicitly labels dog1 or dog2 as being that one that we are told is male do we get a probability of 50%. But, because we don't know which one is being referenced when we are told at least one is male, we obtain a probability of only 33%. It is invalid logic to presuppose that it MUST be dog1(or dog2) that the problem is referencing as the known male dog.

Quite true. "The other one" refers to the pup which was not identified as male; both the pup which has been identified as male and "the other one" have equal probability of being either pup in the set, in any position in the set, by any ordering of the set, because we have no information about any particular pup, only about the set. In other words, if we are looking down at the two pups in the pet store, we do not know if they are male and female, female and male, or male and male; only female and female can be ruled out by the information we have available. This statement is true regardless of how we order the pups within the set; we have no conclusive knowledge about either pup, only about the set of pups.

This is demonstrably true because either pup being male OR both pups being male will produce the exact same answer, "Yes", from the pet store lady to the question "Is at least one a male?" Given that we do not hear the groomer's side of the conversation, we have no idea if he said:
"I just looked at the pup in my hand and it is a male." (50% chance of two males.)
"I remember that at least one was a male, maybe both." (33% chance of two males.)
"I remember that one was a male, but not which one." (Indeterminate - could mean only one male, or that he only specifically remembers one pup.)
"There is one female and one male." (0% chance of two males.)
"They are both males." (100% chance of two males.)
"The brown one is a male, but the black one is a female." (0% chance of two males.)
"The little one is a female, but the bigger one is a male." (0% chance of two males.)

All these statements and a near-infinite number of others from the groomer would have caused the pet store lady to say "Yes". From this it should be clear that the ONLY possibility we can eliminate is that of two females. From our oft-quoted charts, we know that this possibility is 25% for a set of two pups regardless if we treat this as a permutation or combination problem, and that the possibility of two males is 25% for a set of two pups regardless if we treat this as a permutation or combination problem. No matter how you work the problem, it is inescapable that the chance of having two males is 1 in 3 or 33%.