Poll: A little math problem

[santaandy]

The wording of the question says "if they're male, female, or a pair." It makes it sound like a trick question. If they're male or female, then they're both the same thing.

... And if they're "a pair", that means one of each sex.

The wording sucks but it doesn't suck quite that badly.

-- Alex

Incorrect. It doesn't specify. A pair of what? A pair of males, a pair of females, or a pair of puppies of opposite gender? Context clues indicate a pair means the same gender.

 

[Alex_P]

Incorrect. It doesn't specify. A pair of what? A pair of males, a pair of females, or a pair of puppies of opposite gender? Context clues indicate a pair means the same gender.

From context, it's obvious "a pair" is supposed to mean one of each.

Context like:
1. The structure implies that they're both male, both female, OR "a pair" -- does a common-sense reading of the text really justify "both male, both female, or both male"?
2. The thread title says "math problem" instead of "solve my stupid riddle".

-- Alex

 

[Stupu]

Oh, sweet jesus why has this gone to 31 pages? It's pretty basic probability, the question is made to show how probability can give counter-intuitive, but still correct, results, much like the Monty Hall problem, which has likely been mentioned at least once per page so far. The whole thread is just a battle between people who understand it and have thought through the possibilities, and those who just go with their gut, and are willing to argue it. It's like all those thousands of threads about whether 0.999... repeating equals 1 (it does), or who'd win in a fight, Alien or Predator (Predator, obviously), or indeed the once again posted CoD vs Halo (incomparable).

just give it a rest people, if you dont get why the answer is 1/3, try starting from scratch without any presumptions (ie "Male-Female is the same as Female-Male", easily disproven: if you're a guy, imagine you have a sister; that is clearly not the same as if you were female and she were male, becoming female would not make you into the female your sister would be. Just switch that round if you're a gal)

Damn, poor mathematics gets me angry...

 

[Flap Jack452]

I say 33% for reasons already explained.

 

[Alex_P]

P (both male | whatever we know) = P (1st is male | whatever we know) * P (2nd is male | whatever we know)

Not according to two of the scenarios from that textbook. I'm looking for something that explains *why* out of the three choices this is the one to go with, that makes specific reference to why it's okay to use this even in a short run situation where we know probabilities not from rules governing a pool/observations of the pool (like the example about uniforms in that link you gave me) but from knowledge of how the still-to-come events are generated.

The P's describe independent events. One "coin flip" doesn't affect the other. Therefore, you can find their union (AND) by multiplying the probabilities(*).

-- Alex
__________
* - N.B. you cannot do this directly with P (1st is male | at least one is male) and P (2nd is male | at least one is male) from my previous example because the information you've added makes these P's no longer independent of each other -- there's no way you can have the 1st and 2nd be female if "at least one is male".

 

[Alex_P]

That's what I'm talking about: I understand in other contexts how multiplying possibilities works, but in this instance, I'm not so sure.

What concern's me is that this may be like the prosecutor's fallacy.
http://en.wikipedia.org/wiki/Prosecutor%27s_fallacy

Please elaborate. I'm not clear on why you think this is an example of the prosecutor's fallacy as defined there.

-- Alex

 

[oktalist]

1/3

Choosing two puppies of indeterminate sex and then finding out that at least one of them is male, is not the same thing as choosing two puppies of which one is of indeterminate sex and the other is male. In the latter case the probability is indeed 1/2.

Before we are told that at least one of them is male: P(both male) = 1/4; P(both female) = 1/4;
probabilities must sum to 1, therefore P(one of each sex) = 1/2 = P(male, female) + P(female, male) = 1/4 + 1/4.
Then we find out that in fact, P(both female) = 0.

[small]Thanks to whoever bumped this thread or I never would've found this awesome problem [/small]

the question is made to show how probability can give counter-intuitive, but still correct, results, much like the Monty Hall problem,

If you think this is anything like the Monty Hall problem, you don't understand the Monty Hall problem.

It is only inasmuch as it is a probability problem and the result is counter-intuitive.

 

[DemetriTheGreek]

the answer is 50%, either they are or they aren't.

 

[Bat Vader]

It should be 50%. Although knowing that math is my worst subject and that algebra messes everything up I already know I am most likely wrong.