Poll: A little math problem

[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?"