Challenge your brains! Can you solve this problem?

[Bahku]

Your question is asking for the probability of the "other" dog not the combined entity of the chances of both dogs being the same thing. Two completely different questions. If that was the question, you are clearly correct but the wording throws out that possibility.

Nick you're wrong on what the wording says. It clearly is asking if both dogs are male. One arbitrary dog is given a gender, male. The other is unspecified, the only case in which it can be male both dogs must be male.

Conditional probability formula.

Things get confusing with it. Another famous example the "Monty Hall Problem".

The Monty Hall thing doesn't make much sense either, but its true. We actually ran a in person Monte Carlo simulation for are class once just to prove the result. I just kept mixing up flash cards and having a classmate pick a card.

Yes!! The Monty Hall problem ... that was one referred to in our class ... thanks, Fafnir! (TP) And I'll take door number 4 ... night all!

[Fafnir]

The coolest part of that problem is all of the other studies they've done on the problem. Everything form psychology to the different variants of the problem.

[Bahku]

The coolest part of that problem is all of the other studies they've done on the problem. Everything form psychology to the different variants of the problem.

Yeah ... it's seriously been discussed and analyzed and studied endlessly, more for the study of the elements involved than for getting a correct answer. Pretty cool stuff, actually. Demonstrates well that things aren't always as they appear to be.

[fairweatherfan]

Ok, since there's no end in sight to rewriting this manuscript, I'm taking a break and going nuclear by looking up the JESP article cited in the wiki.  The authors test this exact situation:

Half the participants read “Mr. Smith says: ‘I have two children and at least one of them is a boy.’ Given this information, what is the probability that the other child is a boy?”"  The other half got the same question except, "...the probability that both children are boys?"

Referring to the first phrase: "This facilitates a twofold, naive partition because the “other” child can be either a boy or a girl, and the sex of successive children is independent....We therefore predicted that the original wording would give rise to a higher proportion of incorrect responses of 1/2 and that the revised wording would give rise to a higher proportion of correct responses of 1/3."

Only 3.3% of people got the right answer with the original wording.  So it's definitely something people interpret in a misleading way, but the correct answer is 1/3.

[mgent]

Nick, remember that the possibilities are random.

If the guy on the phone picked a dog and told us the gender of the dog (and it happened to be male), the probability that the other one is male WOULD be 1/2.

But since the shopkeeper asked the guy, "Is at least one of the dogs male?" and he responded yes, you can't assume that.  You have to assume that either dog could be male (or both).

Bahku, this particular question tells you what to assume, so it shouldn't be interpreted in different ways.

[Bahku]

Nick, remember that the possibilities are random.

If the guy on the phone picked a dog and told us the gender of the dog (and it happened to be male), the probability that the other one is male WOULD be 1/2.

But since the shopkeeper asked the guy, "Is at least one of the dogs male?" and he responded yes, you can't assume that.  You have to assume that either dog could be male (or both).

Bahku, this particular question tells you what to assume, so it shouldn't be interpreted in different ways.

See, that's because you're starting the equation after the answer is given ... I'm saying that it should or could start before the question is aksed. Sorry, guys ... I'm asleep. Goodnight and TP's for all!

[mgent]

Your question is asking for the probability of the "other" dog not the combined entity of the chances of both dogs being the same thing. Two completely different questions. If that was the question, you are clearly correct but the wording throws out that possibility.

Nick you're wrong on what the wording says. It clearly is asking if both dogs are male. One arbitrary dog is given a gender, male. The other is unspecified, the only case in which it can be male both dogs must be male.

Conditional probability formula.

Things get confusing with it. Another famous example the "Monty Hall Problem".

The Monty Hall thing doesn't make much sense either, but its true. We actually ran a in person Monte Carlo simulation for are class once just to prove the result. I just kept mixing up flash cards and having a classmate pick a card.

Somebody already asked the Monty Hall Problem, and I think it was cdif who brought up the fact that Marilyn vos Savant (who has a column in Parade magazine and is listed in the Guinness Book of World Records as having the highest IQ) answered the question correctly and everybody gave her crap because they thought she was wrong.

I happened to know that she also answered the dog question correctly and again received lots of controversy.

I got the question straight off Wiki if anyone wants to read the article (go down to famous columns).
http://en.wikipedia.org/wiki/Marilyn_vos_Savant
Both problems give links to more detailed explanations if you're really courageous.

[nickagneta]

Your question is asking for the probability of the "other" dog not the combined entity of the chances of both dogs being the same thing. Two completely different questions. If that was the question, you are clearly correct but the wording throws out that possibility.

Nick you're wrong on what the wording says. It clearly is asking if both dogs are male. One arbitrary dog is given a gender, male. The other is unspecified, the only case in which it can be male both dogs must be male.

Conditional probability formula.

Things get confusing with it. Another famous example the "Monty Hall Problem".

The Monty Hall thing doesn't make much sense either, but its true. We actually ran a in person Monte Carlo simulation for are class once just to prove the result. I just kept mixing up flash cards and having a classmate pick a card.

Actually, I am not wrong because in all those cases the questions are giving something arbitrary not defined. In the paradoxes discussed by mgent and you there is always a definitive given doubt about which dog or whatever is one specific thing. "At least one dog", or "one of the children" leaving doubt as to which of the dogs or children are which. In the given question that is removed because the person doing the washing clearly knows which dog is which and can clearly identify them in giving the answer.

Hence, the questionable, arbitrary identification of the original dog is removed from the equation. You people that are steeped in statistics are being tricked by the wording of the question thinking it is a statistics question, which this question is clearly doing with you. There is no question in identifying the dogs in this case, The person doing the washing clearly can identify the dogs apart and so removes the statistical end of things out.

[Roy Hobbs]

I see it exactly like nick does.  I think it's meant to be a trick question, but I think the questioner outsmarted him/herself, if they're looking for 1/3 as the answer.

[fairweatherfan]

The cite I posted above directly addresses the difference in wording.  Using "the other dog is male" instead of "both dogs are male" definitely causes a lot more people to pick 1/2, but it doesn't change the actual right answer, which is 1/3.  Remember that in every case where the other dog is male, both dogs are male, so the two statements are interchangeable in terms of probability.  But I get the feeling most who disagree have their minds set at this point. 

[Master Po]

“Not everything that counts can be counted, and not everything that can be counted counts.”

Albert Einstein

[nickagneta]

“Not everything that counts can be counted, and not everything that can be counted counts.”

Albert Einstein

We bow to the wisdom of the wise one.

[Fafnir]

The person doing the washing clearly can identify the dogs apart and so removes the statistical end of things out.

The person washing the dog has all the information. The question is framed from the perspective of the customer.

The washer can identify them, but hasn't. He has supplied you with incomplete information. The arbitrary nature remains intact from the perspective of the customer. You know the minimum number of male dogs hence the conditional probability.

I can see why you think "other" is ambigious enough to imply that is independent though.

[Fan from VT]

I was a 50% responder, but I'm converted after reading the wiki logic.


Basically, there's a guy in a store. In a separate situation are two puppies. They already exist, and their sex is determined. At this point in time, they could be:

BB BG GB GG


Then there's a phone call. A question: "is at least one of the puppies male?" An answer that yes, at least one is male. So from our original, set situation, we eliminate GG, and are left with BB, BG, GB. And yes, GB is still an option because the man on the phone arbitrarily picked a puppy; he did not necessarily pick "the first" puppy.


The phone call basically confirms that 1/4 of the results (GG) have been eliminated.





There are 2 very similar situations I see in which the answer would certainly be 50%.

1. A man adopts a male puppy. The puppy is lonely. He wants to adopt the puppy's younger sibling, but only if it's a boy. What are the odds that this will work out the way he wants it to?

Answer: 50%. A mother dog can have 4 combinations of 2 puppies, in order of birth:

BG BB GB GG.  The man has adopted the elder puppy, which is a Boy. Thus BOTH GB and GG can be eliminated. Therefore, the puppy in utero could be the 2nd puppy of either the BG or BB situation; thus 50%. 

***there must be a time sequence in the question for the answer to be 50%!


2. A man is running a big puppy adoption site. He is super organized, so he has his puppies in different warehouses. Warehouse 1 is for single puppy children, warehouse 2 is for puppies of 2-puppy litters, warehouse 3 is for puppies born from a litter of 3 puppies, etc. (Additionally, each puppy has a detailed name tag so that his or her sibling(s) can be located later).  Furthermore, obviously we don't want puppies breeding here, so within each warehouse there's a boy side and a girl side.

A woman comes in and wants to adopt 2 puppies. She wants both to be male. AND she wants them to be brothers. So they head over to Warehouse 2. They walk over to the Boy side. The man picks up a random boy puppy. What are the odds that this puppy's sibling is a boy?

Answer: 50%.

Here's why: Each puppy on this man's puppy ranch can be from a BG, BB, GB, GG litter. By going to the boy side, there cannot be a puppy from the GG litter picked. eliminate that. We're down to BG, BB, or GB litters (the only ones that could contribute a puppy to the boy side of the 2-litter warehouse).
Now lets label each of the puppy possibilities left:

G1B1 B2B3 B4G2

So these litters have contributed B1, B2, B3, B4 to the boy puppy pool. One of these has been picked by the man at random. It could have been B1, B2, B3, or B4. B1 has a brother, B2 has a brother, B3 has a sister, B4 has a sister. Thus, when you pick a random puppy from B1-4, you have a 50% chance of picking a puppy with a brother.

***The key difference here is that we have Pre-screened for boys, by going to the boy puppy pen!


*******In the original question, at the time of the phone call there has been no pre-selection and there is no sequential timing in play; there is a static, unchangeable situation that exists and you are learning about.
BG, BB, GG, GB exists. Now you are learning about it. First info learned is that it's NOT GG. Now you are trying to figure if it's GB, BG, or BB. 1/3 chance that the boy has a brother.

[Redz]

Hey...Just wanted to thanks to everyone who responded to my request for some math problems, and for all of the other problems and discussion on this thread.  I used the 40 pound brick problem as my problem to present to the class, and used a bunch of the thought processes laid out in this thread for a Problem Solving Reflection paper (fun stuff, huh)

Gotta love being able use this place for a resource!