Poll: A little math problem

[Samirat]

EDIT: you know what I think is causing difficulties? That we've got Mendel's genetics in the back of our heads. In the case of the offspring of a Tt and a Tt, where, say, tt is a fatal birth defect so there are no tt's, there's a difference between Tt and tT--in one case the gene is coming from the father while in the other it's coming from the mother. This is more like a case where you've got two Tt parents, but let's say we've done genetic screening on the sperm and all the t sperm have been taken out before insemination so that it's human selection and not birth defects making tt offspring impossible. In the case of human selection, you could never get a tT offspring but you *could* get a Tt.

Hey, nice. If this makes sense to you, I think I can use it. In the situation that you described, say that it is impossible to have a TT offspring. There is at least one recessive gene. What are the odds that the offspring would have the recessive trait, v. the odds of having the dominant trait. There are two situations on the Punnet square that result in a manifest dominant trait, and only one that will result in the recessive trait. Therefore, 66 percent chance of Tt, meaning one dominant gene, equivalent to females in the dog problem, and only one chance of the recessive trait, which is tt, equivalent to two males.

The fact that both situations with the dominant trait are expressed as Tt is irrelevant.

Oh, and if you actually look this problem up, you'll find that the answer is 33 percent. It is the correct answer. People just need to understand it.

 

[Geoffrey42]

I say we only have 1xM/F and 1xM/M once we learn that one of the dogs is male:

Order (in this case/perspective) is irrelevant, and using it to filter out 1/4 of the results is a mistake. The thing about Mendel and the square is that it is relevant insofar as it informs us that there are 2 possible outcomes that are equivalent, because whether 't' came from the mother or the father, is irrelevant. tT and Tt are equal, in how they affect the offspring, and how they affect any future offspring. They still make up 2x the amount of the probability (when combined) than either tt or TT alone. If I'm not trying to determine which parent it came from, I can consider tT and Tt to just be 2xtT. If I then learned that "one of the pairs of relevant chromosomes has a 't'", I would not be able to rule out either tT OR Tt, because no one told me which one it was. There's still a 33% chance that the kid is tt, and will die.

M/F and F/M are indistinguishable (in the contexts of this question), and together they account for 50% of the original, un-informed, potential result set. Knowing that "one of the dogs is male" only removes the possibility of the F/F outcome. Nothing in the original problem provides an order, nor does it ask for anything related to order, and thus your attempt to factor in order is misguided.

Let's say that there was an order. I have put one beagle in a container labeled '1', and one in a container labeled '2'. Now they have an order. My potential result set is thus:
A. 1=M, 2=F
B. 1=M, 2=M
C. 1=F, 2=F
D. 1=F, 2=M

Now, the person on the phone asks the question, "Is at least one a male?" The sex-checker says "Yes!" Everyone seems to be in agreement that we eliminate option C. You seem to be arguing for eliminating D, because 1=F, and the sex-checker said "Yes!". How did you deduce that the sex-checker was saying "Yes!" in reference to the beagle in the container labeled '1'?

EDIT:

Oh, and if you actually look this problem up, you'll find that the answer is 33 percent. It is the correct answer. People just need to understand it.

Judging by popular opinion, the answer is 50%. I guess this is why "public opinion" has never been seriously considered as the criteria for getting something on a science curriculum... oh, wait... DOH!

I'm doubly entertained by the ~3 posts that appeared in between the time I hit "quote", and the time I hit "submit", with everyone basically saying the exact same thing. Go us!

 

[Captain Wes]

the main problem I see with the male/female and female/male problem is the fact that we aren't looking for the gender of the set. we're looking for the gender of the other dog, it can be male or female, the other dog isn't important. it breaks down like this,

we have onle dog and we need to know if it's male (the other one is compleately out of the scenario)
male 49.8%
female 49.8%
hermaphrodite 00.4%
using the magic or rounding 50% of the other dog being male

 

[werepossum]

EDIT: you know what I think is causing difficulties? That we've got Mendel's genetics in the back of our heads. In the case of the offspring of a Tt and a Tt, where, say, tt is a fatal birth defect so there are no tt's, there's a difference between Tt and tT--in one case the gene is coming from the father while in the other it's coming from the mother. This is more like a case where you've got two Tt parents, but let's say we've done genetic screening on the sperm and all the t sperm have been taken out before insemination so that it's human selection and not birth defects making tt offspring impossible. In the case of human selection, you could never get a tT offspring but you *could* get a Tt.

Hey, nice. If this makes sense to you, I think I can use it. In the situation that you described, say that it is impossible to have a TT offspring. There is at least one recessive gene. What are the odds that the offspring would have the recessive trait, v. the odds of having the dominant trait. There are two situations on the Punnet square that result in a manifest dominant trait, and only one that will result in the recessive trait. Therefore, 66 percent chance of Tt, meaning one dominant gene, equivalent to females in the dog problem, and only one chance of the recessive trait, which is tt, equivalent to two males.

The fact that both situations with the dominant trait are expressed as Tt is irrelevant.

Oh, and if you actually look this problem up, you'll find that the answer is 33 percent. It is the correct answer. People just need to understand it.

You - are the man. Or woman. Or rat. Whatever, that explanation was excellent.

 

[jim_doki]

Must.....resist.....urge......to make........Maths debate joke

 

[Samirat]

You - are the man. Or woman. Or rat. Whatever, that explanation was excellent.

Thankee kindly, Mr. Werepossum. Or Mrs. Werepossum. Or werepossum Werepossum.

 

[Saskwach]

So to return to my naming exercise, the first dog (Dog 1) is Jesse and the second dog (Dog 2) is OSAN. There are four possible outcomes if we compare the gender of the two:
Jesse OSAN
M M
M F
F M
F F
Now we have a new piece of information: at least one dog is male. This means the first outcome is still possible, because both dogs are male. The second is still possible because Jesse is male. The third is still possible because OSAN is male. Only the fourth is now impossible.
Jesse OSAN
M M
M F
F M
We are now asked whether the dog that isn't guaranteed to be male actually is (but aren't told which dog is so guaranteed, as will be shown). So I'll take out one M from each outcome.
Jesse OSAN
M
F (This one is actually under OSAN, but for some reason the post won't accept blank space before writing on a line.)
F
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?

 

[Geoffrey42]

No, it's not irrelevant, because Tt and tT describe two different events, either sperm T and egg t OR sperm t and egg T. But that's not how the one female dog situation works--the dog being female can result from one and only one possible event--an X egg and an X sperm. An X egg and a Y sperm is a boy, and unless there's some weird genetic disease or the Holy Spirit involved and this is the Jesus Puppy, you can't get Y from an egg.

Samirat is equating t to M and T to F (or vice versa, it doesn't matter) for the purposes of comparing the question at hand with the results of a traditional genetic Punnett square, and you are confusing the issue by bringing up the genetics of gender. Nothing relevant to this problem, or what Samirat is talking about, has anything to do with sperm and eggs.

tT and Tt describe two events that are differentiable in the grand scheme of things, but within the scope of what we care about, they are undifferentiable. It does not matter which dog is female, but whether one of them is female. The only thing worth mentioning about M/F, or Tt, is that they make up twice as much of the universe of possible results as M/M, F/F, TT, or tt alone. The difference between M/F, F/M, Tt, and tT exists, but doesn't matter to us. See Saskwach's naming of the dogs, and my containering of the dogs, to see why.

EDIT:

Ahh--this is perfect for me to explain myself. I'm not arguing for eliminating D; I'm arguing for eliminating *one and only one* from these choices: D and A (edit).

I'm saying we have to deduce that the sex checker was saying "Yes!" in reference to *either* the beagle in the container labeled '1' OR the beagle in the container labeled '2'. Isn't that both the correct and the necessary deduction?

It is neither the correct nor the necessary deduction. I have the sinking suspicion that even if we KNEW it was in reference to the puppy in container '1', we could not change the answer to anything other than 33%, because order still isn't part of the question.

There are two puppies. Puppies are male or female, and occur in equal numbers. There's a random pair of puppies. That means there are three possibilities: both are male, both are female, or one is female and one is male. We find out that one is male. That means there are only two possibilities. The one we don't know about is male or it's female. Since both possibilities are equally likely, that means there's a 50% chance we've got two males and a 50% chance we've got one female and one male.

You're excluding the point that "both are male" = 0.25, "both are female" = 0.25, and "one is female and one is male" = 0.5. Yes, the "distinct set of possibilities" contains only 3 entries, but those 3 entries do not have equal weights. This means that of those 3 possibilities, they are NOT "equally likely". When you remove the "both are female", the other two retain their relative weightings.

 

[Avatar Roku]

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.

Oh, now I see what you're getting at. Very badly worded problem, either of us could have been right with the information given.

 

[Alex_P]

That's my point--how are you 'ruling out' all the tails tails outcomes so there is always one head? Are you doing it by reflipping any tails tails outcome until you get at least one head? Or are you doing it by putting down one coin on the table Heads up? It makes a huge difference. You can't just say "Rule out all the tails tails outcomes, so that there is always at least one head" without telling me how you rule them out.

It's easiest if you try to emulate the problem as closely as possible. Don't think of it in terms of "reflipping" anything or "fixing" anything or whatever. Just list all the possibilities and then answer simple questions about them.

The question you know the answer to is "Is at least one of them male?"

You want a (probabilistic) answer to "Are both of them male?" that incorporates as much known information as possible.

Dog1  Dog2  1+ Ms?  2+ Ms?
M     M     yes     yes
M     F     yes     no
F     M     yes     no
F     F     no     no

All four combinations are equally likely. I specifically didn't name them because "Dog1" just represents the first dog that is sexed -- the table is symmetrical if you say that the other dog is sexed first.

You're incorrectly eliminating the "F/M" case by asking a different question ("Is Dog1 male?") instead.

-- Alex

 

[Trace2010]

Just tell the lady to send you the damn male already.

 

[Alex_P]

You're incorrectly eliminating the "F/M" case by asking a different question ("Is Dog1 male?") instead.

But I'm not doing that. I'm saying we have to eliminate *either* the "F/M" case *or* the "M/F" case (and that's an exclusive or) because the male dog is either Dog1 or Dog2.

No. If you pick one dog up and it is female, then you pick up the other dog to check its sex.

-- Alex