Poll: A little math problem

[HSIAMetalKing]

This is such a long thread... the answer is so simple.

Probability doesn't exist. Fact.

 

[Doug]

EDIT:
Think about what P(1 Male and 1 Female) *really* is, and don't just fill in values from above, and remember: you don't add probabilites of two events together to find out the probability of them both occuring, you multiply them, otherwise the chances of two male dogs would be P(1 Male and 1 Other Male)=P(.5 + .5)=P(1) which is obviously wrong, you agree?


What are the chances of a male? .5 right? What are the chances of a female? .5 again, right? So shouldn't P(1 Male and 1 Female) actually be:

P(1 Male and 1 Female)=P(.5 x .5)=.25

and from there I think you'll see why the chances are actually 50/50.

No, I don't agree.

P(Anything) == 1 (i.e. One of the three has to be true, agreed?)
P(2 Males) = P(2 Females) = 0.25 (we all agree that its 0.5x0.5, correct?)

P(Anything) = P(2 Males) + P(2 Females) + P(1 Male AND 1 Female)
Hence:
1 = 0.25 + 0.25 + P(1 Male AND 1 Female) = 0.5 + P(1 Male AND 1 Female)

Therefore, P(1 Male AND 1 Female) = 1 - 0.5 = 0.5

IF P(1 Male AND 1 Female) was 0.25, the total would be 0.75

 

[Doug]

Let me put in another way... binary counting.

So, for two bits, we have possibly:

00 (P of 0.25)
01 (P of 0.25)
10 (P of 0.25)
11 (P of 0.25)

What is the probably of a randomly selected number containing a one and a zero is:
P(01) + P(10) = P(one 1 and one 0) = 0.25 + 0.25 = 0.5

So, if we assume order isn't important, we get:
P( two 0's ) = 0.25
P( one 1 and one 0 ) = 0.5
P( two 1's ) = 0.25

The maths isn't lying. You've just based your probablities on false assumptions.

 

[Doug]

Let me put in another way... binary counting.

So, for two bits, we have possibly:

00 (P of 0.25)
01 (P of 0.25)
10 (P of 0.25)
11 (P of 0.25)

What is the probably of a randomly selected number containing a one and a zero is:
P(01) + P(10) = P(one 1 and one 0) = 0.25 + 0.25 = 0.5

So, if we assume order isn't important, we get:
P( two 0's ) = 0.25
P( one 1 and one 0 ) = 0.5
P( two 1's ) = 0.25

The maths isn't lying. You've just based your probabilities on false assumptions.

Ahh, I get it now--coins and dogs and children were confusing. The two mixed pairs really are two mixed pairs, not just the same way of expressing the same mixed pair possibility--that was the false assumption.

Now I get it! Thanks!

*smiles* Excellent - and no worries, its a confusing question at first.

 

[Vallen00]

What I'm about to say has no bearing on the actual math and may have already been mentioned, but, does the gender of the other dog matter? The guy said he wanted a male, and it was confirmed that at least one of them was a male, so why does he need to know if they both are? The only way it could possibly matter is if the shopkeeper had already promised a male to another person.

 

[Doug]

Ah well, its because he a wierd mathematican, collecting results for a survey

 

[KittywifaMohawk]

Okay. I feel stupid, but the 75% was jumping out at me.

Just a little flag going up in my brain somewhere. Either that, or early onset dementia.

That's exactly what happened to me, I had a feeling it was going to be right, but I have yet to find out.

 

[Dark Crusader]

Well, the second gender does not rely on the first gender(of the dogs), so it's nothing more than a 50% chance.
I also don't know what you mean by the pair.
Also, this is not like the monty hall problem at all.
The monty hall problem is conditional probablity.
The probablity of one door depends on how many other doors, there are.
The monty hall problem removes a door, so therefore changes the probability. They rely on each other, they are dependant to use the proper term.
Meanwhile the gender of the two dogs is independent, I am ignoring the pair at the moment, because I don't know what you mean.

Then again, I'm bloody tired, and I may be remembering wrong.

 

[Alex_P]

2 is gambler's fallacy. You're saying "this particular pair of puppies contained a male, so I must create an arbitrary system for generating puppies that puts a male in every pair."

No it isn't--gamblers fallacy is that luck going one way will be balanced out by luck going the other way very soon. Gamblers fallacy would be that because we got one male, we need to construct a system that makes it more likely that we get a female for the next result--which is what you guys are going--rather than a system where we've got an equal shot at getting a male or a female next.

I'm not doing anything to make the "second female" somehow more likely than it should be, though. All I'm saying is that you should start with each one being "fair" -- random with an equal probability of being male or female -- and construct sets from that. Otherwise you're not creating a probability distribution that's representative of the initial setup of the problem.

It's a mistake to say "well, in this problem there was a male so I will just make one a male every time" because what happened is that some fair coin ended up heads this particular time.

-- Alex

 

[geizr]

This thread proves how much people actually do misunderstand probabilities and how they can change with change in knowledge. Of the respondents, 65.5% got the wrong answer. Let me make an attempt to explain why 33% is the correct answer.

At the start of the problem, we have only know that there are two puppies, but we have no idea what their sexes are. So, the possible outcomes are the following:
dog1 dog2
M M
M F
F M
F F

There are 4 possible outcomes. This means that the probability of each outcome is 1/4 or 25%. If we ask what is the probability that both dogs are male, the answer is 25%. This is because only one of the four outcomes is M/M. But, this is the case that we don't know anything at all about the dogs other than the fact we have 2 of them.

Now, let's suppose we are told that dog 1 is a male. This means that we now know the F/M and F/F outcomes did not occur. So now we ask what are the possible outcomes. The possible outcomes are
dog1 dog2
M M
M F

There are only 2 possible outcomes. So the probability of each outcome is 1/2 or 50%. If we are told that dog2 is the one that is male, then we have again 2 possible outcomes:
dog1 dog2
M M
F M

But, look closely at what happened as a result of our change in knowledge. While the first possible outcome remains the same, the second possible outcome is a different configuration from the case when we knew dog1 was the male. Regardless, because there are only 2 possible outcomes now, the probability of each outcome is 50%.

Now, here is where things change. In the problem, we are told that at least one of the dogs is male. But, we don't know which one. Because we don't know which dog is male, we must consider all possible outcomes in which there is a male dog. There are 3 such outcomes:
dog1 dog2
M M
M F
F M

Notice that the number of outcomes has changed to 3 because we don't know which dog is male, only that at least one is male. Because there are now only 3 possible outcomes, thanks to us knowing now that at least one dog is male, the probability of each outcome is 1/3 or 33%. The problem is asking us what is the probability that both dogs turn out to be male when we know that at least one dog is male. We see that only 1 of the three outcomes have both dogs as male. So, the probability that both dogs are male is 33%.

The KEY of this problem is realizing that the probabilities change as a result of our change in knowledge about the situation. Notice carefully how the probability of having 2 male dogs changed as a result in our change in knowledge about the situation. When we don't know anything at all about the two dogs, the probability of 2 males is 25%. When we know that dog1 or dog2 is the one that male, then the probability of 2 males is 50%. When we only know that at least one dog is male(but not which one), then the probability of 2 males is 33%.

One last try to give a thorough explanation why the correct answer is 33%(I quote my own post because I don't want to type all that again). Someone mentioned about conditional probabilities; that's exactly what's going on in this problem. The problem is asking what is the probability of 2 males given that you know at least 1 dog is male.

 

[werepossum]

Let me put in another way... binary counting.

So, for two bits, we have possibly:

00 (P of 0.25)
01 (P of 0.25)
10 (P of 0.25)
11 (P of 0.25)

What is the probably of a randomly selected number containing a one and a zero is:
P(01) + P(10) = P(one 1 and one 0) = 0.25 + 0.25 = 0.5

So, if we assume order isn't important, we get:
P( two 0's ) = 0.25
P( one 1 and one 0 ) = 0.5
P( two 1's ) = 0.25

The maths isn't lying. You've just based your probabilities on false assumptions.

Ahh, I get it now--coins and dogs and children were confusing. The two mixed pairs really are two mixed pairs, not just the same way of expressing the same mixed pair possibility--that was the false assumption.

Now I get it! Thanks!

*smiles* Excellent - and no worries, its a confusing question at first.

Cheeze gets it? Yay! All hail Doug, master of logic and explanation! Sir, I bow before your superior teaching ability.

 

[Geoffrey42]

Now I get it! Thanks!

*pokes head out from behind a rock*

*whispers*Is it safe to come out now?

I thought this was never going to end. Since I bid this "adieu", I've been coming back every day just to see how many pages there were. Each day, I was increasingly perplexed that it was still going so strong. In fact, I was increasingly suspicious that Cheeze_Pavilion was taking advantage of [a href=http://xkcd.com/386/]this[/a] and doing an incredible job of [a href=http://i19.photobucket.com/albums/b161/P3Shinobi/1192392913251.jpg]that[/a]. Honestly Cheeze, I mean that with the utmost respect; I couldn't fathom the purported explanation that you sincerely didn't get it, and it seemed to make much more sense if you were doing it on purpose.

Still, I have a mild suspicion that this is all just one of [a href=http://photo.gangus.com/d/26788-2/ackbar.jpg]these[/a]...

EDIT: For the grammar!

 

[Lukeje]

What I'm about to say has no bearing on the actual math and may have already been mentioned, but, does the gender of the other dog matter? The guy said he wanted a male, and it was confirmed that at least one of them was a male, so why does he need to know if they both are? The only way it could possibly matter is if the shopkeeper had already promised a male to another person.

I've pointed this out numerous times, but nobody seems to care. (ps., the answers zero!)

 

[werepossum]

SNIP
You know what it is? What I think confuses people is not this particular problem. I think what confuses people is how counter-intuitive probability itself is, that it's natural to have "based your probabilities on false assumptions" as Doug pointed out.

I think what makes it confusing is that it doesn't seem like permutations should have anything to do with the probability of different results in a two-event, on/off outcome in a system without memory. What's counter intuitive isn't what happens when the new information comes along, it's the initial probabilities. What's fishy is that the probabilities involved in two independent random trials is the exact same as two picks from a system with memory.

As I said long ago in this thread, I think the difficult part is that your mind seizes on what you know and stops. If the question had been the chance that, say, three other dogs out of six were male, you'd work the problem.

 

[gerrymander61]

It's 50%, or slightly higher given that the dogs could be identical twins. Anyone who says otherwise should take a class in basic stats or probability to find out just how STUPID they are.

EDIT: Wow, just checked the poll results. 34.6% of people here are stupid.

Take discrete math, please. It's actually like the best don't-have-to-be-a-math-major college math class: useful and light on memorization.

-- Alex

Um, excuse me? The fact that one dog is male has NO BEARING WHATSOEVER on the other dog's gender, if anything, it makes it slightly more likely that it is male due to the possibility, albeit unlikely, of the dogs being twins, putting the second puppy's chance at being male slightly over 50%. What you're arguing for is what we Statistics and probability people call "The Gambler's Fallacy." See? It's even got a name because there are so many idiots like YOU out there in the world.

http://en.wikipedia.org/wiki/Gambler's_fallacy

Do a little reading, and then maybe instead of taking a probability course, pay a-f**king-ttention in YOUR discrete course. I've taken it.