Poll: A little math problem

[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

 

[Alex_P]

Cheeze,

I want to see where we diverge:

Assume I have 2.00e50 actual puppies -- not probabilistic puppies with Schrodinger's Nards, just a metric fuckton of actual puppies.
Half of the puppies are female and half are male. (There are no intersexed puppies. )
I group them into 1.00e50 pairs of puppies.

2. How many groups would you expect contain...
a. two females?
b. one female and one male?
c. two males?

3. Of all of the puppy groups, how many contain at least one male? (Number, not percentage.)

4. Of all of the puppy groups, how many contain exactly two males? (Number, not percentage.)

5. Of only the groups that contain at least one male, how many contain exactly two males? (Number, not percentage.)

-- Alex

 

[Ancalagon]

3. Of all of the puppy groups, how many contain at least one male? (Number, not percentage.)

4. Of all of the puppy groups, how many contain exactly two males? (Number, not percentage.)

5. Of only the groups that contain at least one male, how many contain exactly two males? (Number, not percentage.)

3. .75e50 (2b + 2c, right?)

4. .25e50 (same as 2c, right?)

5. .25e50 (why would this be different from 4 or 2c, if you're asking for numbers and not percentages? How could a group that contains two males NOT be of the groups that contain at least one male? Isn't containing at least one male a necessary condition for containing two males? Or is that the point of the question?)

Still going, huh? I thought maybe I'd wake up to find everyone agreeing that there were no dogs, or that the dogs had drowned or something. Oh well. Here's the thing. What you know is that the pair of dogs are in set 3. That is all you know. You do not know the order of the dogs. You are looking for a pair of dogs that is in set 4. What odds are you going to give me that the first pair you pick out is going to be two males?

 

[Samirat]

The other beagle is irrelevant it could have been a crocodile or Cthulu aswell.

There is 50% chance that the other dog is male (if thats how dogs breed)

It would be 25% if both genders had to be male and both were unknown, thats why you fools were mistaken.

No, again, you're missing the fact that either dog, which I'm going to call Dog 1 and Dog 2, could be male. If Dog 1 is male, there is a 50 percent chance either way that dog 2 could be male or female. But if the first dog the dog washer looked at was female, than the other dog has to be male, since the premise of the problem guarantees at least 1 male.

By the way, Cheese, the woman you were talking about is Marilyn vos Savant, who actually originated, and solved, the Monty Hall problem.

In the case of coins, I'm not sure why you think a reflip is inappropriate, in the situation of two tails. This is just like if you asked the washer the question, and she said, Nope, both are female. But since she didn't say that, you can reflip until at least one is heads. It's just like you knock of the 25 percent probability of two tails, and magnify the chances of the other situations to sum to 1. That's pretty much all you're doing here. Placing one coin on heads at the beginning of the problem is equivalent to assuming that Dog 1 is a male, when it could also be Dog 2 that's a male.

All right, relating to you problem with Alex_P. To solve this problem, you simply place the number of pairs with 2 males over the number of pairs with at least 1 male.

Out of 100: 25/75

The sample space of the problem are the 75 pairs that contain at least 1 male, this fraction represents the number of pairs of dogs out of that where the other dog is a male. You made the fractions yourself.

 

[Samirat]

Check the question--you're not actually checking the dogs, you're relying on Puppy Washing Man to check them, then answer the questions 'is at least one of them male' with a yes or no answer.

What kind of an argument is this? Are you saying that the problem differs significantly based on who's doing the checking? The method is the same. If the first dog is male, she doesn't check the other one, before telling you "yes." But if the first dog is female, she checks the other one, and, since she has to find it male, in the problem's premise, she calls and answers "Yes." So in the first case, the next dog could be either male or female, in the second, it is definitely a male female pair. Therefore, two male female pairs, one male male.

 

[Alex_P]

I want to see where we diverge:

Assume I have 2.00e50 actual puppies -- not probabilistic puppies with Schrodinger's Nards, just a metric fuckton of actual puppies.
Half of the puppies are female and half are male. (There are no intersexed puppies. )
I group them into 1.00e50 pairs of puppies.

2. How many groups would you expect contain...
a. two females?
b. one female and one male?
c. two males?

3. Of all of the puppy groups, how many contain at least one male? (Number, not percentage.)

4. Of all of the puppy groups, how many contain exactly two males? (Number, not percentage.)

5. Of only the groups that contain at least one male, how many contain exactly two males? (Number, not percentage.)

2a. .25e50
2b. .50e50
2c. .25e50

3. .75e50 (2b + 2c, right?)

4. .25e50 (same as 2c, right?)

5. .25e50 (why would this be different from 4 or 2c, if you're asking for numbers and not percentages? How could a group that contains two males NOT be of the groups that contain at least one male? Isn't containing at least one male a necessary condition for containing two males? Or is that the point of the question?)

Correct.

So, of the groups that contain at least one male, what percentage contain two males? .25e50 / .75e50, which is 33%, right?

Does putting it this way help anything?

-- Alex

 

[AuntyEthel]

Ow. My brain hurts.

 

[Uncompetative]

50%

 

[Saskwach]

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?

That two out of the three possible outcomes is female AFTER we've figured out that one of the puppies is male--that's the premise I'm not accepting.

I see now. Thanks for that. Ok, well as I laid out, the reason is because of how the woman has been asked to check the dogs and then how we've been asked dto check the dogs. She wasn't asked "Is Jesse a male?" This is crucial because such a question would actually give us two pieces of information: the gender of one dog; and which dog we're referring to. It would make the gender of the other completely independent and thus 50/50.
We can both agree that Jesse and OSAN, no matter their gender, will not suddenly change it depending on the order we check them. For example, if Jesse were male and OSAN female, to check J-O would give use M then F, and to check O-J would give us F then M. Therefore, the order in which we check these dogs would matter, if the question asked us to consider checking order. In other words, it would be a permutation question (order of the outcomes is important), not a combination question (the sum of each outcome is all). This is because Jesse being male and OSAN female is very different to OSAN being male and Jesse female.
This is why there are still two 1-female 1-male options instead of one. What the question asks is a bit tricky, but what happened was the bathing woman was asked to consider the combination of the dogs (how many were male and how many female, not which is what) and then tell us that - in this case, for the woman, it doesn't matter whether Jesse is the male or OSAN is the male, only that there was 1+ males. We, however, are asked to consider a permutation problem: assuming we've found an male first, what is the sex of the other? In other words, we're being given an order of checking but, importantly, we aren't told for sure the order - which dog will be considered checked first (the male). If we were told which dog was checked first we could know precisely which dog to check next and the gender of the second dog checked would exist independently of the other for the purposes of our check. Unfortunately, the order of checking isn't independent of circumstances so neither is the gender of the dog we're checking second.
This is why we now have to consider F/M and M/F as two distinct possibilities: because one represents Jesse being the male dog referred to and the other represents OSAN being the male. These are two distinct possibilities, which would affect the order in which we check the dogs - it's a permutation problem.
I'll bring out my Jesse and OSAN thingammies again because my explanations are getting tangled.

J O
- -
M M
M F
F M
These are the three gender combinations that could come about after the woman has said yes. As you can see, Jesse being male and Osan being female is different to Jesse being female and Osan being male. The question that you've posed is: why can't we just say it's 1-male and 1-female all the same?
We can't do that because we aren't sure which dog is the male - it could be Jesse or it could be Osan, and we have to look at this problem permutatively - in the order of the dogs that have been checked, which we just aren't told. If Jesse turns out to be the male dog then he is the one we attach a hypothetical "checked first" card to, and we then check Osan. If Osan was the assured male, he is now the "checked first" dog and Jesse is checked second - the problem has become one of permutation and not combination. But we don't know which dog has been checked first (is the male). We haven't been shown one "heads" on the table, so we can't consider the other as an independent event: that other coin could just as easily be the "heads" coin as well.
I really hope this post makes more sense that it's seeming to.

 

[Geoffrey42]

Another approach to why you can't eliminate either of the M/F combinations:

Let's say we had 4 sample setups. 8 beagles, 4 immediate observers (the sex-checkers), 4 questioners, and you, the omnipotent observer. You are made privy to the following information at the beginning: setup A has 2 male beagles, setup B has 1 male and 1 female beagle, setup C has 1 female and 1 male beagle, and setup D has 2 female beagles.

Each of the questioners (A-D) asks each of the sex-checkers (A-D) the question: "Is at least one a male?"

Sex-checkers A-C respond: "Yes!", sex-checker D responds: "No!"

When you evaluate that situation, as the omnipotent observer, what probability should questioners A-D (assuming rational questioners) assign to their chances that both dogs are male?

++++

Another question to consider: It seems that you are making the assumption that the sex-checker has only checked one of the dogs. What if you assume that the sex-checker knows the sex of both dogs?

++++

Lastly, yet another scenario, but, one that I think SUPPORTS the idea of removing one of the split (F/M, M/F) options.

The store has two dogs, which they have labeled Dog 1, and Dog 2. The questioner asks: "Is Dog 1 male?". Sex-checker says: "Yes!". What are the odds that Dog 2 is male?

Original distribution:
A. 1-M 2-M (.25)
B. 1-M 2-F (.25)
C. 1-F 2-M (.25)
D. 1 F 2-F (.25)

Since I know that Dog 1 is male, I can eliminate C AND D. My odds that Dog 2 is male are 50%. The key difference between this scenario and the other is the layer of obfuscation in the question "Is one of the dogs male?" as opposed to "Is the first dog male?"

 

[Geoffrey42]

My sticking point is that I don't accept that we don't change the numbers once one of the events goes from a probability to a certainty.

As you said before, since this is math, isn't half the point knowing whether we got to the answer the right way? The only reason we care about that is for the sake of reproducibility. When you have two competing methods/models, that both accurately predict the outcome of a given scenario, the best way to find which one is right is to provide an example where one of them fails, and the other succeeds. Can you think of an example where our model fails? Up to that point, it is absolutely semantics; two ways of getting to the same place, reliably and speedily, and nothing but a pillow fight over "my way being better".

 

[Ancalagon]

In the case of coins, I'm not sure why you think a reflip is inappropriate, in the situation of two tails.

Because you can't change the sex of the puppies or pick a new set of puppies every time you get two females. So if we're using coins as a stand in for puppies, they have to obey the same rules as puppies.

The reflip wouldn't be modelling a hasty gender reassignment, it just models the fact that we know that two tails cannot have happened, or we cannot have had two female dogs to start with, so if we get it, we reflip. If there were a way of making it so that two flipped coins turning up tails simply couldn't happen, then we could do that instead.

All right, relating to you problem with Alex_P. To solve this problem, you simply place the number of pairs with 2 males over the number of pairs with at least 1 male.

Out of 100: 25/75

The sample space of the problem are the 75 pairs that contain at least 1 male, this fraction represents the number of pairs of dogs out of that where the other dog is a male. You made the fractions yourself.

Everything changes once we know one of the events--which are independent--is no longer in the realm of probability, but has become a certainty.

Absolutely. The only thing that changes the probabilities throughout the whole exercise is that we find out that at least one of the two dogs is definitely a male.

We start off with:
Dog 1/Dog 2
M/M
M/F
F/M
F/F
all being 25% likely. Where we go our different ways is when we are told that at least one of the two dogs is definitely a male. How does that affect the probability that each of these was the starting position? F/F cannot have been the starting position, since we know that at least one of the two dogs is definitely a male. So we change it to:

Dog 1/Dog 2
M/M 33% likely
M/F 33% likely
F/M 33% likely
F/F 0% likely

If on the other hand we were told that Dog 1 was a male, then F/M could not have happened either, so we would have:

Dog 1/Dog 2
M/M 50% likely
M/F 50% likely
F/M 0% likely
F/F 0% likely

But that isn't what we're told.

Okay, from first principles:

First, what can we agree on:

1.) There are two puppies, which we can call Dog A and Dog B.

2.) There are four equally likely combinations of sex that the two puppies can have: MM, MF, FM, FF, where the first letter indicates the sex of dog A, and the second letter indicates the sex of dog B.

3.) That if you remove one equally likely combination from consideration, the rest remain equally likely to one another.

Okay, I don't think anyone is claiming that any of these is false.

So when you remove FF from consideration, you still have three equally likely starting positions that are valid. Dog A could be male, and Dog B could be male. Dog A could be male, Dog B could be female. Dog A could be female, Dog B could be male.

Now what most of the 50%-ers have been saying is that you don't have three equally likely outcomes, you have two. The dog in your hand is male, the dog not in your hand is either male or female. So you have MM and MF, where the first letter denotes the dog in your hand, the second letter denotes the one that isn't. The problem is that these situations are not equally likely. Why are they not equally likely? Because the dog-bather chose a dog. He may have looked at both dogs, found a male and a female, and picked up the male. If, on the other hand, he had picked up a dog at random, not looked at the other, and the one he picked up was male, then there would be a 50% chance that the other is male.

Another way I've heard it put is that if you're saying that the starting situation FF must be removed, so must either MF or FM. Why? Which one would you remove? If you remove the first, what you're saying is 'Dog A can't be a male, while Dog B is a female'. If you say the other, then you're saying the opposite. You cannot make either of these assumptions based on what you know.

How do I know that my three combinations MM, FM, and MF are equally likely? because they always have been. We agreed that they were equally likely in the beginning. Nothing has happened to make any of these situations any more or less likely. So we have three equally likely starting situations, two of which lead to the other dog being female, one of which leads to the other dog being male. The problem comes from people thinking that given the set:

Dog A/Dog B
M/M
M/F
F/M
F/F

as equally likely initial starting points, it can be changed to:

Dog in Hand/Dog not in Hand
M/M
M/F
and that these two situations are equally likely. They aren't.

 

[Geoffrey42]

Thing is, we're not getting to the same place. Let's say this is a bet on picking a pair of male puppies. Even money when we don't know either puppy's sex is 4-1 odds, right? I'm saying that once we know that one puppy is male, that even money goes down to 2-1 odds. Everyone else is saying that the odds only go down to 3-1 for even money, if I'm doing the math right.

I misunderstood your earlier comment then. I thought we were all in agreement about the 33%, and the sticking point was how we get to that number (whether it be a recalculation of the odds, a reduction of the result set, whatever).

Let's try: puppies are paired at the puppy farm randomly. It turns out that for this particular breed of dog, pairs of females always instantly kill one another. Therefore, the universe of possible puppy pairs only consists of MM and MF pairs. Due to the nature of the universe (this is not an assumption of the problem, just a statement about the nature of the universe when sample sizes are sufficiently high), there are twice as many MF as MM pairs. What are your break-even betting odds of getting a MM pair?

How does the above differ from the original question?