Poll: A little math problem

[Lukeje]

As far as I can tell, this question is more unambiguous, and should give the correct answer.

A man walks into a very strange pet shop. The strange thing is that all pets are picked at random from the type of animal you've selected (so say you ask for a cat, a random cat is selected for you). You are allowed to ask one question to the shopkeeper about the range of animals you are looking for.
The man asks for a puppy. The teller tells him that there are two in stock, and asks what his question is. He says, 'Well I really want a dog (as opposed to a *****); is there at least one dog?' the teller checks the computer, and says 'Yes'.
What are the odds that he walked out of the shop happy?

Edit: wait, that doesn't work... hmmm... can anyone think of an unambiguous wording?

You want two male puppies because you're a radical male gay separatist. You call up one shop and ask and they say "we've got two puppies, but we can't tell if they are male or female." You call up another shop and they say "we've got two puppies, we know one of them was male because the breeder told us, but the tag fell off and now we can't tell which is which."

If you want two male puppies, which shop's pair should you purchase?

That's the version that should be on wikipedia... the question is how long before they took it down?

 

[Ancalagon]

Do you really think you've got a 50% chance of winning if you pick the set that Jesse didn't investigate, but only a 33% chance of winning if you pick the set that he did?

No, you've got a 33% chance of winning if you pick the set he's informed you has a head, but a 25% chance with the one he hasn't, since it could be HH, HT, TH, or TT.

 

[werepossum]

This is just sad. More than sad, it's depressing. This is what happens when logic is no longer taught.

There is no need to keep trying to explain it. Remember that the question is "What is the probability that the other one is a male?" Now look at Saskwach's explanation in post #97 on page 3. Study it. The question's solution is laid out extremely well, walking you through each step. It can't be explained any more clearly. Anyone with a decent mind should be able to understand this. Admittedly if you haven't been taught set probability you would probably get it wrong initially unless you are extremely intelligent and logical, but with this explanation anyone should be able to understand it. The fact that wrong answers out-number the right answer by more than four to one is highly disappointing.

 

[Ancalagon]

Making one coin heads all the time not only makes it physically impossible to get two tails, it makes it physically impossible for that coin to ever be anything but heads. Therefore, you have to get rid of either HT or TH because including them both is only accurate if the facts are that two coins are being flipped.

Two coins are being flipped. The phrase 'making one coin heads all the time', is ambiguous. It could mean 'make one coin in particular heads all the time', in which case you would be right. But if it means 'make it so that at all times, at least one of the two coins will land on heads.', then that bears out my maths, and I think that's the statement which models the situation properly.

With your thing where Jesse looks at one set of coins, but not the other, would I be right in saying you make it 50% chance of being a pair of heads in the set Jesse has looked at, but 25% in the set that he hasn't?

 

[huntedannoyed]

The only options that the puppies sex could be are the following: 1. (Male and Female) 2. (Female and Female) 3. (Male and Male). There is a 30.3% percent chance that the other one is male while the other is female.

 

[Ancalagon]

So why doesn't the 'set' change like I described in comment #175

from:

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

to:

Referred To Puppy/ Other Puppy
M M
M F

Okay, so lets take your set:

Referred to Puppy/ Other Puppy
M M
M F

I agree that's a valid way of looking at it, but that the probabilities of these two instances are not equally likely.

Is puppy 1 or puppy 2 in the first table 'Referred to puppy'? In the instance M F, puppy 1 must be the referred to puppy. In the instance F M, puppy 2 must be the referred to puppy. But in the instance M M, half the time it will be puppy 1, half the time it will be puppy 2. The instance F F can not have occurred, so its probability is zero.

Okay, so a third of the time, it's M F, so puppy 1 is referred to puppy, and puppy two is female, so a third of the time, referred to puppy is M and other Puppy is F. Another third of the time it's F M, so puppy 2 is referred to puppy, and puppy 1 is other puppy. So for that third, referred to puppy is M and other puppy is F. In the third of the probability that M M occurs in, we split it in to two sixths. In one, puppy a is referred to puppy, puppy b is other, in the other, it's the other way round. But both sixths go for Referred to Puppy is M, and Other Puppy is M. Add it up and you've got:

Referred to Puppy/Other Puppy
M M 33 1/3% likely
M F 66 2/3% likely

 

[Solo508]

I think its 25%... if you start of with the basic 50/50 chance of it being either male or female then its 50%, but if you take into account that she asked the guy "is ATLEAST ONE male?" does that add another 50/50 chance to it (making it 25%) because he could answer that in 2 ways, "yes" or "yes, they both are". This is confusing the fuck out of me lol!

 

[Solo508]

Sorry if somebody posted this before but I just found it on wikipedia:

"Two boys" problem

Like the Monty Hall problem, the "two boys" or "second-sibling" problem predates Ask Marilyn, but generated controversy in the column,[13] first appearing there in 1991-92 in the context of baby beagles:

A shopkeeper says she has two new baby beagles to show you, but she doesn't know whether they're male, female, or a pair. You tell her that you want only a male, and she telephones the fellow who's giving them a bath. "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?
?Stephen I. Geller, Pasadena, California

When vos Savant replied "One out of three" readers[citation needed] wrote to argue that the odds were fifty-fifty. In a follow-up, she defended her answer, observing that "If we could shake a pair of puppies out of a cup the way we do dice, there are four ways they could land", in three of which at least one is male, but in only one of which both are male. See Boy or Girl paradox for solution details.

The problem re-emerged in 1996-97 with two cases juxtaposed:

Say that a woman and a man (who are unrelated) each has two children. We know that at least one of the woman's children is a boy and that the man's oldest child is a boy. Can you explain why the chances that the woman has two boys do not equal the chances that the man has two boys? My algebra teacher insists that the probability is greater that the man has two boys, but I think the chances may be the same. What do you think?

Vos Savant agreed with the algebra teacher, writing that the chances are only 1 out of 3 that the woman has two boys, but 1 out of 2 that the man has two boys. Readers argued for 1 out of 2 in both cases, prompting multiple follow-ups. Finally vos Savant started a survey, calling on women readers with exactly two children and at least one boy to tell her the sex of both children. With almost eighteen thousand responses, the results showed 35.9% (a little over 1 in 3) with two boys.

 

[werepossum]

SNIP
Okay, so I've read it a couple of times, and here's the question I keep coming back to: why does knowing one puppy is male only strike out one possibility, the possibility of F/F? Why don't you also have to chop in half the probability of one male and one female because *either* F/M *or* M/F can be possibilities, but not both?

I mean, puppies aren't quantum events--their sex exists before people look at them, right? So why doesn't the 'set' change like I described in comment #175

from:

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

to:

Referred To Puppy/ Other Puppy
M M
M F

The things you know are that there are two pups, that at least one pup is male, and that the chances of any pup being male are 50% - you can make no other statements of knowledge. That eliminates only one possible distribution out of four, that both are female. Ergo there are three remaining possible distributions. Of those three remaining possible distributions, only one results in two males; thus the percentage of two males is 1 in 3 or, expressed as an integer percentage, 33%.

This is where you go wrong.

Is "Excluding the case of two girls, what is the probability that two random children are of different gender?" *REALLY* an equivalent way of saying "A random two-child family with at least one boy is chosen. What is the probability that it has a girl?" What about "If we know one child is a boy, what is the probability that the other child is not?" Is that also equivalent?

Your example specifies that the family must have at least one boy; therefore one possible distribution of four in Saskwach's explanation is removed IN THE SELECTION OF THE SET.

Had the two pups been selected to have at least one male, you would be correct. But we know the pups were selected without regard to their sexes, else the woman would not have had to make a phone call to determine that there was indeed at least one male pup in the two-pup set. In the case of the pups, the "Referred To Puppy" can be in the first position OR in the second position as long as it is male. Thus the distribution in your analysis should be:
Referred To Puppy/Other Puppy
M/M
M/F
Other Puppy/Referred To Puppy
F/M
There are still three possible distributions out of a set of two binary members if one possible distribution is known.

In Solo's two examples, the woman's children are expressed exactly like the puppy example; you know there is at least one boy, nothing else. Therefore if you know that one child is male, the chance that both are male is 33% because we have eliminated only one possible distribution out of four, namely the female/female one.

For the man's children, however, you also know WHICH child is male - it's the oldest. That eliminates two of the four possible distributions, female/male and female/female. Thus the chance of the second child being male is 50%. Going back to the pup example, if there was a brown pup and a black pup and the groomer told us the brown pup was male, that would eliminate two of the four possible distributions. The chance that the black pup is also male would be 50% because we have already established that the brown pup is not female, eliminating both possible distributions requiring a female brown pup. Thus,
Black/Brown
M/M
M/F
F/M
F/F
becomes
Black/Brown
M/M
F/M
for a 50% chance the black pup is also male. This is a different possible distribution because we have been given additional information, allowing us to eliminate more possible distributions.

 

[Wormthong]

Ok its very simple.
it IS 50 % chance if you deny that you fail(exept when you count in the chance of a hermaphrodite but lets just ignore that for now because we will probably all get a headache from that).

The thing is i haven't seen it being explained properly yet (i haven't read the last page so i am sorry if its in there but its late and i am tired so i don't want to read the rest of all the silly 33.333endless theory's)i will explain it as best i can.

ok lets begin by calling one dog sparky and the other bill

so now comes the hard part

these are the things that can happen

male:bill male:sparky 25%
male:bill female:sparky 25%
male:sparky female:bill 25%
female:sparky female:bill 25%

ok now we count in the fact that one is male (see how im not only ruling out the fact that they can both be female because that is what all the 33% guys do wrong)

so now the chances are

male: bill male: sparky 25%
male: sparky male: bill 25%
female: sparky male: bill 25%
female: bill male: sparky 25%
see its that simple now the only option that remains is the the other one has a 50 % chance of being male.

 

[Samirat]

All right, let's try this. If you have two dogs, the chances that at least 1 will be male are 75 percent, which is the probability of both being male added to the probability of only one being a male. The probability that both will be a male is 25 percent, while the probability that only 1 will be a male is 50 percent.

Since the probability of only 1 being a male (the male the dog washer saw) is twice that of both being males, the odds are, respectively, 66 percent and 33 percent.

The problem is though that we get new information--we get the Bath Giving Man searching among the puppies for a male. The 75 percent you're talking about is not only M/M and M/F, but also F/M. Once the Bath Giving Man goes searching and finds a male, we can eliminate not only F/F but also F/M because the questions changes from "Two Unknown Puppies" to "One Known Puppy and One Unknown Puppy."

There's a difference between 'what's the probability of at least one male among a set of two unknown puppies' and 'what's the probability of at least one male among a set of two puppies, one known to be male'. The former is 75%, but the latter is 100%.

EDIT: I think the crux of this is that you're talking about two *unknown* dogs while the problem is talking about one known dog and one unknown dog. There's where the issue is.

No. The order of the dogs remains constant. When you find out one's gender, its place in the problem doesn't change.

I'm confused. Why don't you just do this problem, with coins or something. Rule out all the tails tails outcomes, so that there is always at least one head. And count the number of times the other coin is also heads. I don't think you'd actually have to do this. Just look at it, and puzzle it for a moment. It should make sense. If you still don't get it, I'm sorry for you.

Oh, and by the way. The answer is 33 percent. That's set in stone. This problem was devised by Marilyn vos Savant, and she proved the answer to be 1 third. All we're trying to do is explain it to you.