Poll: A little math problem

[Ancalagon]

No, we are reading your explanations. Mathematically, you're right. Logically, you're not. To use the tired analogy of a coin toss, if you flip a coin twice, the the first flip has no bearing on the second. That is this exact situation, seriously, you could have replaced "Male puppy" with "Heads coin" in the original problem.

You're right that the situation is exactly the same if you use coins. And if you flip a coin, and it's heads, then the probability that the other coin will be heads is 50%. But:
If someone flips two coins and keeps the result hidden from you, and you ask him if there's a 'heads', and he says yes, then the probability that they're both heads is one-third.

 

[crepesack]

25% ok heres how sex chromosones are xx & xy therefore through the use of a punnet square you can resolve this
X X therefore i could randomly select another male at a 50%*50% chance in other words 25%
X|XX XX
Y|XY XY

 

[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.

 

[Samirat]

25% ok heres how sex chromosones are xx & xy therefore through the use of a punnet square you can resolve this
X X therefore i could randomly select another male at a 50%*50% chance in other words 25%
X|XX XX
Y|XY XY

This works, except that it doesn't eliminate the situation which is eliminated in the problem. There can't be two females. This looks at the complete set:MF, FM, MM, FF. But there can't be two females. Therefore, there are only three outcomes, and only in 1 is the other dog a male.

 

[Saskwach]

Stuff I'd love to quote but feel would add too much length to this post.

I'd already put some thought into that and had a hefty piece of doubt, but I think I have the answer.
So we're talking, really, about the ordering of these dogs. When we say that M/F is a different outcome to F/M we're saying that one dog will be checked first and the other second: this is a permutative question, not a combinative one. So let's say we painted those numbers on the dogs' sides so no one could miss it.
This man is asked, is dog 1 and/or dog 2 a male, not, is dog 1 a male? This is crucial.
If the question were "Is dog 1 a male?" then you would be right; we'd have to drop F/M as a possibility and the answer would become 50%. But these dogs are being ordered, or to be more correct - and a bit more whimsical - one is an individual dog named Jesse and the other is also an individual dog named Other Sexually Ambiguous Name - so if we called out Jesse, OSAN wouldn't come running.
Instead, the man (or woman?) was asked: "Is Jesse and/or OSAN a male?", meaning that, conceivably, either M/F or F/M are still on the table because if Jesse (Dog 1) is male then M/F or M/M would be the result and we'd still have to check OSAN's nether regions, but if OSAN (Dog 2) were male then F/M or M/M would be the result and we'd be looking between Jesse's legs. This man has said, "Maybe Jesse's a boy, maybe OSAN's a boy, but I ain't saying which groin I checked."
To use coin tosses as the example, a man has flipped two coins. You asked him whether one or both came up tails and he said "Yes." What he didn't say was that his first toss came up tails, which is the same mistake people made in the dog example.

Edit: I don't think I summed up too well so I'll make clear wherein lies the rub.
If the question were: "Is Jesse/Dog 1 (ie, a specific dog) a male?" the probability for the next dog being male would be 50%, because OSAN's machismo is totally independent of Jesse's masculinity.
But the question is: "Is Jesse/Dog 1 and/or OSAN/Dog 2 (ie, at least one of the two) a male?" which leads to 33%, because OSAN and Jesse and their genitalia haven't been made independent of each other (ok, I've gone too far this time. I'll stop now.).

 

[Ancalagon]

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%.

Okay, here's the problem. We're all starting with:

Dog A/Dog B
M,M
F,M
M,F
F,F
each outcome is equally likely.

but when the information that one of the dogs is male is added, the 33%-ers are moving to:
Dog A/Dog B
M,M
F,M
M,F
each outcome is equally likely.

and the 50%-ers are moving to:
known dog/unknown dog
M,F
M,M

But by changing the heading, the two options are no longer equally likely.

If you toss two coins a hundred times, you'll a pair of heads about 25 times, a pair of tails about 25 times, and one of each about 50 times. But if you disregard any pairs of tails, you're left with 25 pairs of heads, and 50 doubles.

 

[Saskwach]

From the OP:

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?

The question wasn't, "Tell me when you find a male," but rather, "Is at least one a male?"
I think what you're saying is that any normal person would say "Yes," as soon as she found a pecker (yes, apparently it's a Peeping Tina), but this isn't a normal person - this is a literal person (we're assuming).
Edit: Actually, even then it'd be 33%.
So Tina checks both these dogs and if Jesse is a male she says yes. If OSAN is a male she says yes. If Jesse and OSAN are males she says yes. There are three possibilities here, all equally likely, so the possibility of both being male is 1/3.

No, we didn't ask him if either of two things are true, those two things being 'at least one is tails' OR 'both are tails'. We only asked "Is at least one a male?" not "Is at least one a male OR are both male."

Those last two questions, when added together, amount to the same question as just the first.

That's what I think the issue is--saying 33% is not taking into account the fact that once an event actually occurs, the probability of it occurring goes up to 100%. It's like the lottery--the chances of winning a future lottery may be .8353498574357%, but if you *actually do win the lottery* your chances of winning go up to 100%.

I think that's exactly the issue the 33% percent crowd is taking into account - but I can't understand your logic here, so feel free to explain it again.

 

[Lukeje]

Ahh... now I understand. There are two questions.
1.If the person sexes one of the dogs, finds it to be male, and then puts it down, what is the probability, that if the this person then picks up a dog at random, the dog is male? The answer being 2/3.
2.If the person sexes one of the dogs, finds it to be male, and then keeps hold of it, what is the probability that the the other dog is male? The answer in this case being 1/2.
Edited, as 1. =2/3, not 1/3.

 

[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?

(2/3... I think).
Edit: wait, that doesn't work... hmmm... can anyone think of an unambiguous wording?

 

[Saskwach]

However, what if we don't count T/T because we put one coin Heads up and only flip the other coin? Then we can't just stop at pulling T/T out of the matrix. We ALSO have to pull out *either* H/T *or* T/H, and that's an exclusive 'or' there. If we leave both H/T and T/H in, then our matrix no longer reflects the reality that we're only flipping one coin. It *would* reflect the reality if we eliminated F/F because the rules were that we flip two coins and T/T is considered a reflip, but if the rules of the game are that one coin is placed Heads up before any flip, then you can't just stop at eliminating T/T--you also have to eliminate, like I said, one and only one of T/H or H/T.

The connection to the puppy question is that once Bath Giving Man says "Yes!" to the question "Is at least one a male?" that is the equivalent of changing the game from flipping two coins to flipping one and leaving one Heads up. The 33%/redistribute .25 is not accurate under the 'lay one down Heads up and flip one' rules, but only under the 'reflip any T/T result' rules.

And it's the Puppy Bathing Man, not the Puppy Sex Reassignment Surgery Man, so no reflipping! ;-D

Yes, I think I have it (funny how we each have the cure for the others' ills...)
The thing is we're NOT placing one head on the table: we're getting a shady dealer to look at both without showing us and then telling us if one or both are heads. These are different operations.
If we put a head on the table, that would be like throwing it away and saying "new toss - will this be heads?" (I was shocked to learn some people play this game at casinos.) After all, that's the very definition of independent events.
But, and this is unarguable, we are NOT doing this in the dog question. We're flipping both coins, getting this shady dealer, who, nonetheless, has a heart of gold, not a tongue of silver, clauses(!), to check both coins together and tell us not just if the first coin came up a heads (in which case we'd throw that coin away and be back to 50/50 for the next), but rather whether one or other or both are heads. These coins are already flipped; they're heads or tails and we aren't flipping them again. All we know is that Coin 1 is a head or Coin 2 is a head, but we aren't sure which, and we aren't sure what the other is at all.
Probabilities!
T/H
T/T
H/T
H/H
If we're told coin 1 is a head then we get:
H/T
H/H
We may as well throw coin 1 away and get this:
T
H
So a 50% chance.
But, again, we're just not doing that. We're stating that coin 1, or coin 2, or both, are heads, but we aren't stating which:
H/T
T/H
H/H
In which case we can't throw away coin 1 because, in this case, if coin 1 were a head, then it actually does affect the likelihood of coin 2 being a head: coin 2 is not independent of coin 1. The important part, the really important part, is that the first coin - the one we're told to put face up in your example - doesn't actually have to be heads for the other to be heads, and thus the answer to still be "yes, we have at least one head."

The real lesson we should take from this thread, though, is: when you don't want to deal with probabilities you should ask a better question.

 

[Ancalagon]

True, but in this case, we're no longer flipping two coins. We're only flipping one--the Unknown Dog. The Known Dog is like a coin left on the table facing heads up while we flip the other one.

That's the thing though, you don't re-flip the coin. The genders of the dogs remain the same throughout the exercise. I know it's stating the obvious, but bear with me. Actually, let's go back to the coins. The probability that a coin lands on heads is 50%. You toss two coins. You make it physically impossible to get two tails, all other options retain their probability. So you've got HH, HT, TH, representing the fact you know one is a head. What I think you're doing is seeing the guy holding a head in his hands (like I think people on both sides of the discussion are beginning to...), and saying that the other coin is still unknown. Which it is. But then you're saying that because it's unknown, it must have a 50% chance of being a head, as stated above.

That seems logical, since it could be either, and had an equal chance of being either when it was tossed. But what you're doing is mentally re-flipping the coin. By which I mean your looking at the situation as it stands, with one coin being heads, and the other coin being unknown, and taking the fact that a coin has a 50/50 chance of being heads, so saying that the unknown coin has a 50/50 chance of being heads. But assuming it still has a 50/50 chance ignores the fact that while it is unknown to you, it was not unknown to the person tossing the coins, and by looking at both coins, and choosing one which was a head, the probabilities from the onlooker's point of view have changed. Apologies if I've misunderstood what your saying though.

EDIT: It's the dogs I feel sorry for. They always get caught in the middle of these things.

 

[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.