Poll: A little math problem

[thedoclc]

Saph, you've either run into a troll or someone too smug to listen to any ideas outside his/her own ungrounded and undefended certainty that their intuitions are right. Might as well just let it go.

 

[Avatar Roku]

Guys, it would be 33.3(repeating)% if the question was asking for a set, but it's not. It's asking for the other puppy, not how they are together. As such, the first puppy might as well not even be there, as it has no bearing on the gender of the second. Therefore, it's 50%.

No, because it doesn't say the "first" puppy is male. It could be either one. So this means that instead of just having two choices: Male and Female, and Male and Male, you have three: Male and Female, Male and Male, and Female and Male. In only one of these is the other dog male. Hence, 1 out of 3.

In hindsight, I don't know why I'm bothering to write this, since those that say 50 percent aren't even reading the explanations.

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. The main discrepancy is cause by the fact that the problem is worded vaguely enough that the answer could be either 50% 0r 33%.

EDIT: To address the other thing you brought up, it doesn't matter whether the known puppy is first or second, we just assign it the title "first" to make it quicker to refer to one or the other.

 

[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.