Poll: A little math problem
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hmmm... well... i think the 1/3 chance actually represents the fact that we now know that we can't have a FF pair of puppies.Cheeze_Pavilion post=18.73797.823377 said:
You're right that if the question was "here's a male dog, what's the chance that the next dog i give you is male" the chance would be 1/2 or 50% male.
The difference here is that we know that this is a finite set of two dogs. By learning that one of the dogs must be male, it informs us that the chance of getting an all female set is 0% which we didn't know until we were informed that one of the dogs is male. See? If we claimed it was still a 50%
to get a male puppy then we'd be saying that the information we recieved about the male puppy was immaterial and didn't matter. In actuality it informs us that we can't have a set of two females forcing us to update our probabilities.
I saw it, answer the question. Here, I'll repeat it.Cheeze_Pavilion post=18.73797.823812 said:
You flip 4 coins and ask your friend if two of them are heads. He says "yes." What are the chances that the other two are heads compared to the chances that the other two are comprised of 1 head and 1 tails.
By your logic:
KKHH, KHKH, KHHK, HKKH, HKHK, HHKK = 6 possibilities
KKHT, KHKT, HKKT = 3 possibilities
Perhaps not over internet forums in the future, but I'll try a little longer. Returning to the issue at hand.Cheeze_Pavilion post=18.73797.823815 said:
What, are you saying there will be no female female pairs in the problem that Werepossum described? He just has random puppies, therefore there will probably be female female pairs. Why shouldn't they exist?
But this problem should be applicable to all pairs that DO contain at least one male. Meaning the other 150 (about). So, if we fit them to each of our respective probabilities, 50 percent results in significantly unequal quantities of male and female puppies across the board, which is not natural. 33 percent, however, results in the correct 1 to 1 ratio of male puppies to female puppies.
Whatever you choose to believe, this ends soon, for me. Either you go away believing what is right, or you go away believing the opposite. If it is the latter, I hope you get some second opinions before preserving your certainty for life.
On the other hand, what does it matter?
On the other hand, what does it matter?
@samirat
ugh... sorry.
My point was MF == FM
only because she selected the puppy at random and we aren't dealing with ordered pairs.
I should have instead said that you have the possibility of 1 set all male, 1 set all female, and one set with 1 a male and a female. I hope that makes it clear.
The primary reason i'd argue that we shouldn't accept 50% is because the information that one of the dogs is male actually is material and relevant. It forces us to realize that we can't get a female pair.
Accepting 50% indicates that we have no more material and relevant information about the set then we did before she told us that one of them was male. Does that makes sense?
@CheesePavillion
ugh... sorry.
My point was MF == FM
only because she selected the puppy at random and we aren't dealing with ordered pairs.
I should have instead said that you have the possibility of 1 set all male, 1 set all female, and one set with 1 a male and a female. I hope that makes it clear.
The primary reason i'd argue that we shouldn't accept 50% is because the information that one of the dogs is male actually is material and relevant. It forces us to realize that we can't get a female pair.
Accepting 50% indicates that we have no more material and relevant information about the set then we did before she told us that one of them was male. Does that makes sense?
@CheesePavillion
QFTSamirat post=18.73797.823821 said:
After reading the problem on the original wikipedia page, I think I've worked it out
According to Marilyn vos Savant, the starting scenario has 4 possible results with an equal probably:
Male Male Pmm = 1/4
Male Female Pmf = 1/4
Female Male Pfm = 1/4
Female Female Pff = 1/4
but order isn't important, so fm = mf - so, lets ADD THE PROBABLY of the two:
2xMale Pmm = 1/4
1xMale 1xFemale Pmf = 2/4
2xFemale Pff = 1/4
We know at least 1 puppy is male, so:
2xMale Pmm = 1/4
1xMale 1xFemale Pmf = 2/4
2xFemale Pff = 1/4
Therefore, its 2/3 likely that its a male and female puppy...
Yes, I'm surprised too!
Edit:
Putting a sample population to it of 1000 people with dog pups, we get:
2xMale, 250 people
1xMale 1xFemale 500 people
2xFemale 250 people
Hence, the probably of the other dog being male is 250/(250+500)=250/750=1/3=0.33 probably, or 33%
According to Marilyn vos Savant, the starting scenario has 4 possible results with an equal probably:
Male Male Pmm = 1/4
Male Female Pmf = 1/4
Female Male Pfm = 1/4
Female Female Pff = 1/4
but order isn't important, so fm = mf - so, lets ADD THE PROBABLY of the two:
2xMale Pmm = 1/4
1xMale 1xFemale Pmf = 2/4
2xFemale Pff = 1/4
We know at least 1 puppy is male, so:
2xMale Pmm = 1/4
1xMale 1xFemale Pmf = 2/4
Therefore, its 2/3 likely that its a male and female puppy...
Yes, I'm surprised too!
Edit:
Putting a sample population to it of 1000 people with dog pups, we get:
2xMale, 250 people
1xMale 1xFemale 500 people
Hence, the probably of the other dog being male is 250/(250+500)=250/750=1/3=0.33 probably, or 33%
Incorrect - we aren't assuming they choose from a random list of people with at least one male.
They selected from a random list of people, INCLUDING people with FF pairings.
Putting a sample population to it of 1000 people with dog pups, we get:
2xMale, 250 people
Answered: YES
1xMale 1xFemale 500 people
Answered: YES
2xFemale 250 people
Answered: NO
Hence, the resulting population is bias towards MF's
250 MM's vs 500 FM's
They selected from a random list of people, INCLUDING people with FF pairings.
Putting a sample population to it of 1000 people with dog pups, we get:
2xMale, 250 people
Answered: YES
1xMale 1xFemale 500 people
Answered: YES
2xFemale 250 people
Answered: NO
Hence, the resulting population is bias towards MF's
250 MM's vs 500 FM's
If the question was:
X phones up people who have at least 1 male puppy out of 2, then it would be 0.5 and 0.5 probablities.
In this scenario, the probably is skewed by the filter of NO answers.
Edit:
In other words, Male-Male pairs are half as likely as Male-Female/Female-Male pairs, regardless of scenario
X phones up people who have at least 1 male puppy out of 2, then it would be 0.5 and 0.5 probablities.
In this scenario, the probably is skewed by the filter of NO answers.
Edit:
In other words, Male-Male pairs are half as likely as Male-Female/Female-Male pairs, regardless of scenario
Well, the original statement was:
A shopkeeper says she has two new baby beagles to show you, but she doesn't know whether they're male, female, or a pair. You tell her that you want only a male, and she telephones the fellow who's giving them a bath. "Is at least one a male?" she asks him. "Yes!" she informs you with a smile. What is the probability that the other one is a male?
?Stephen I. Geller, Pasadena, California
She was asked that.
Its a trick on your elementary maths
If it makes you feel better, I answered 50% at first too!
http://en.wikipedia.org/wiki/Marilyn_vos_Savant
A shopkeeper says she has two new baby beagles to show you, but she doesn't know whether they're male, female, or a pair. You tell her that you want only a male, and she telephones the fellow who's giving them a bath. "Is at least one a male?" she asks him. "Yes!" she informs you with a smile. What is the probability that the other one is a male?
?Stephen I. Geller, Pasadena, California
She was asked that.
Well... not really - what she meant was the first answer did matter overall. Because 2 Male pairs are half as likely as having 1 male and 1 female.elitheiceman post=18.73797.823978 said:
Its a trick on your elementary maths
If it makes you feel better, I answered 50% at first too!http://en.wikipedia.org/wiki/Marilyn_vos_Savant
The probability of the other puppy being a boy is one half, 0.5, 1/2, 50%. Nothing else.
With two puppies, there are four possible gender combinations.
G,G
G,B
B,G
B,B
Because the question states that _at least one_ is a male (as opposed to 'the first one' or 'the second one') then the order doesn't make a difference. BG and GB are the same thing. If _at least one_ is a boy, then the options are like this:
B,B
B,G/G,B
That's only 2 options. It's 50:50.
I disagree with the geniuses and the schoolteachers and the mathemeticians because I know I'm right. I know I'm right about this, the Kennedy Assassination, the Moon Landing, 9/11, any moral or ethical decision and everything else in the universe.
Goodnight.
With two puppies, there are four possible gender combinations.
G,G
G,B
B,G
B,B
Because the question states that _at least one_ is a male (as opposed to 'the first one' or 'the second one') then the order doesn't make a difference. BG and GB are the same thing. If _at least one_ is a boy, then the options are like this:
B,B
B,G/G,B
That's only 2 options. It's 50:50.
I disagree with the geniuses and the schoolteachers and the mathemeticians because I know I'm right. I know I'm right about this, the Kennedy Assassination, the Moon Landing, 9/11, any moral or ethical decision and everything else in the universe.
Goodnight.
There is a way to prove its 1/2; but that's not it. You've made the same assumptions the 1/3ers have, but then forgotten that 'two options' does not mean that each has the same probability. If there are twice as many microstates for a particular outcome, then it will have twice as much probability, hence your answer should read '1/3'.Bakery post=18.73797.824151 said: