Poll: A little math problem

[FrcknFrckn]

Well, good luck guys - I'm done with this.

 

[Alex_P]

Our problem isn't with the initial matrix of permutations. We've all agreed on that. M/M, M/F, F/M, F/F. The question isn't about permutations, it's about trying to figure out the probability of three mutually exclusive sets of combinations.

Well, this is separate from that. It speaks to how to test the problem empirically if you want.

Here's what I'm driving at...

The problem scenario goes like this:
1. Someone makes a pair out of two random puppies.
2. You ask a question about that pair and find out the answer is yes, allowing you to use this new knowledge to adjust your probabilistic guess.
3. You are asked to make a guess about the pair.

In order to "test" something like this, you should copy that general flow. First, make a pair of random puppies. Then ask "Is at least one male?" Then look at whether there are two males in this set.

If you just try any old way to create a pair of puppies, then you're not effectively simulating the problem anymore.

-- Alex

 

[Samirat]

One puppy was picked from an infinite number of puppies, then another puppy was picked from an infinite number of puppies.

So, that's analogous to "two random puppies," right?

So, what's the matrix of permutations?

Our problem isn't with the initial matrix of permutations. We've all agreed on that. M/M, M/F, F/M, F/F. The question isn't about permutations, it's about trying to figure out the probability of three mutually exclusive sets of combinations.

Not quite. Natural numbers vs. real numbers is countable infinity vs. uncountable infinity. In informal language, one of those infinities is like "infinity times infinity" (basically the idea is that you can squeeze infinite real numbers between any two rational numbers).

This would be more like just taking a limit.

Ahh, okay. My bad.

You said it yourself. There are three mutually exclusive situations. They are all equally probable, are they not?

MM, MF, and FM

 

[geddydisciple]

now my head hurts, i would guess 50%

 

[Alex_P]

Our problem isn't with the initial matrix of permutations. We've all agreed on that. M/M, M/F, F/M, F/F. The question isn't about permutations, it's about trying to figure out the probability of three mutually exclusive sets of combinations.

Well, this is separate from that. It speaks to how to test the problem empirically if you want.

Here's what I'm driving at...

The problem scenario goes like this:
1. Someone makes a pair out of two random puppies.
2. You ask a question about that pair and find out the answer is yes, allowing you to use this new knowledge to adjust your probabilistic guess.
3. You are asked to make a guess about the pair.

In order to "test" something like this, you should copy that general flow. First, make a pair of random puppies. Then ask "Is at least one male?" Then look at whether there are two males in this set.

If you just try any old way to create a pair of puppies, then you're not effectively simulating the problem anymore.

-- Alex

"1. Someone makes a pair out of two random puppies."

Two unknown puppies put in a pair.

"2. You ask a question about that pair and find out the answer is yes, allowing you to use this new knowledge to adjust your probabilistic guess."

One puppy is male, the other, we don't know.


"3. You are asked to make a guess about the pair."

Well, the male puppy is guaranteed to be male, and the unknown puppy is either male or female. And puppies are as likely to be male as they are to be female, so there's a .5 chance the unknown puppy is male, and a .5 chance the unknown puppy is female. So the chances of the pair being two males is 50%, and one of each, 50%

Now where, as the LOLCats would say, am I doin' it wrong?

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."

What you should be doing is starting with a random pair -- because that's how the pair in the problem is actually constructed. Then for each pair you say "Does this pair contain one or more male puppies?" and "Does this pair contain two male puppies?" and you look at how the distributions are aligned.

Otherwise any "experiment" is flawed because you're replacing the original probability distribution of the sets with one of your own devising.

-- Alex

 

[Samirat]

Our problem isn't with the initial matrix of permutations. We've all agreed on that. M/M, M/F, F/M, F/F. The question isn't about permutations, it's about trying to figure out the probability of three mutually exclusive sets of combinations.

Well, this is separate from that. It speaks to how to test the problem empirically if you want.

Here's what I'm driving at...

The problem scenario goes like this:
1. Someone makes a pair out of two random puppies.
2. You ask a question about that pair and find out the answer is yes, allowing you to use this new knowledge to adjust your probabilistic guess.
3. You are asked to make a guess about the pair.

In order to "test" something like this, you should copy that general flow. First, make a pair of random puppies. Then ask "Is at least one male?" Then look at whether there are two males in this set.

If you just try any old way to create a pair of puppies, then you're not effectively simulating the problem anymore.

-- Alex

"1. Someone makes a pair out of two random puppies."

Two unknown puppies put in a pair.

"2. You ask a question about that pair and find out the answer is yes, allowing you to use this new knowledge to adjust your probabilistic guess."

One puppy is male, the other, we don't know.


"3. You are asked to make a guess about the pair."

Well, the male puppy is guaranteed to be male, and the unknown puppy is either male or female. And puppies are as likely to be male as they are to be female, so there's a .5 chance the unknown puppy is male, and a .5 chance the unknown puppy is female. So the chances of the pair being two males is 50%, and one of each, 50%

Now where, as the LOLCats would say, am I doin' it wrong?

You are, indeed, doin' it wrong. You are assuming that you know which puppy is the known puppy, and which is unknown. Indeed, if the washer told you, "yes, Sparky's a male," the answer would be 50 percent.

However, since you don't know, these three possibilities are all equally likely;

Sparky is a male, Othello is a male
Sparky is a male, Othello is a female
Sparky is a female, Othello is a male

You see how knowing one is male, and knowing which one is male are different? These three solutions are all exclusive possibilities, and all are equally likely. With what do you disagree?

 

[Samirat]

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.

You can't describe order from information that we don't have. Saying "the next one" implies that you know the male dog is first. You can't place him first because you don't know who he is. I don't even think the rules of probability problems allow you to change ordering after a problem is already set up. Sparky and Othello have one set order. It doesn't matter if it's Sparky and Othello or Othello and Sparky. Consistency is important.

 

[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!)