Poll: A little math problem

[Doug]

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

 

[Doug]

http://en.wikipedia.org/wiki/Conditional_probability

because the two dogs being male are independent events:
P(2nd is male given 1st is male)=P(2nd is male)=0.5

CONFIRMATION:
P(2nd is male given 1st is male)=P(1st and 2nd are male)/P(1st is male)=(0.25)/(0.5)=0.5

simple high school maths

Wrong, the events are NOT indepent WITHIN THE SCENARIO

 

[Doug]

http://en.wikipedia.org/wiki/Conditional_probability

because the two dogs being male are independent events:
P(2nd is male given 1st is male)=P(2nd is male)=0.5

CONFIRMATION:
P(2nd is male given 1st is male)=P(1st and 2nd are male)/P(1st is male)=(0.25)/(0.5)=0.5

simple high school maths

Wrong, the events are NOT indepent WITHIN THE SCENARIO

Explain why they are not independent

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

Bias population, assuming the answer was YES

 

[Doug]

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

 

[Doug]

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.

I see, she has not specified which dog is male so there are three possibilities of which only one is (male-male)
If she did specify which one was male it would be 50%
wow, confusing

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.

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

 

[Bakery]

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.

 

[Lukeje]

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.

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'.

 

[Doug]

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.

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'.

I think he was being funny...I think

 

[Bakery]

'two options' does not mean that each has the same probability.

Order makes no difference in this question. If there's a boy _and_ a girl, then theres a boy and a girl, not a boy _then_ a girl/girl _then_ a boy.

To look at it any other way is to look at it too deeply and miss the point of the question.

EDIT:

I think he was being funny...I think

I was in the ending sentance but alas, no, my answer of 50% is dead serious.

 

[Doug]

'two options' does not mean that each has the same probability.

Order makes no difference in this question. If there's a boy _and_ a girl, then theres a boy and a girl, not a boy _then_ a girl/girl _then_ a boy.

To look at it any other way is to look at it too deeply and miss the point of the question.

EDIT:

I think he was being funny...I think

I was in the ending sentance but alas, no, my answer of 50% is dead serious.

I was talking about you! Look at my previous answers.

EDIT:

Misread, my bad. You are reading the scenario incorrectly though.

 

[Doug]

Order makes no difference in this question. If there's a boy _and_ a girl, then theres a boy and a girl, not a boy _then_ a girl/girl _then_ a boy.

To look at it any other way is to look at it too deeply and miss the point of the question.

To look at it your way is to misread the situation and the maths of the scenario all together, I'm afraid.

So, the initial 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?

So, breaking it down, we have

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.

So, assuming P(male)=0.5, P(female)=0.5, we get:

Hence:
P(Male and Male) = 0.25
P(Male and Female) = 0.25
P(Female and Male) = 0.25
P(Female and Female) = 0.25

But if we assume the order isn't important, then we have:
P(2 Males) = 0.25
P(1 Male and 1 Female) = (0.25 + 0.25) = 0.5
P(2 Female) = 0.25

Next:

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.

Thefore, we can exclude P(2 Females) as that case has been removed, so we can multiply:
P(2 Female) = 0.25
P(2 Males) = 0.25/(0.25+0.5) = 0.25/0.75 =(approx) 0.3333 (or 1/3 exactly)
P(1 Male and 1 Female) = 0.5/(0.25+0.5) =(approx) 0.6667 (or 2/3 exactly)

What is the probability that the other one is a male?

P(2 Males) = 0.33333 or 33.33% or 1/3

EDIT:
For example, what are the odds of rolling 2 dice, one after the other, or at the same time, and getting a total of 5

36 (i.e. 6 * 6) possible combinations, of which the following match:
1+4=5
2+3=5
3+2=5
4+1=5

P(rolling a total of 5)=4/36

if you assumed order wasn't important, you'd get:
1+4=5
2+3=5
P(rolling a total of 5)=2/36

Which would be wrong

 

[Jimmydanger]

Scenario #2: Toss a fair coin ten times in a row without looking at the results. Now begin looking at the results for the first nine. They are all heads. What is the chance of the tenth being heads? About one in a thousand. This is set probability.

Just back in this quagmire for a second this was explained wrong and I think it can help with the main problem.

you cannot look at the first nine coins in a set and still treat them as a set. In this example you have reduced yourself to only two options coin 10 is heads or tails leaving a 50-50 chance.

For this to remain a theoretical set no knowledge must be known about individual coins. Say if someone told you at least nine of those coins were heads. Then you would have 11 options. First all coins are heads OR coin 1 is tails OR coin is tails 2 OR coin 3 etc. Then the chance of all the coins being heads would be 1 in 11.

This is equivalent to both puppies male OR first puppy female OR second puppy female. Then the chance of both male is 1 in 3

 

[Jimmydanger]

Unless, when people have been saying that the female male combination can also be written as male female, can male male also be written as male male.

I *thought* you were on to something, and you were. You've got the solution right there.

We start with this, which everyone agrees on:

M M
M F
F M
F F

Then we get the information that at least one dog is male. Let's call this the Known Dog, or K for short. Now we have to go back to our table, and work on the line in bold, applying our new knowledge.

M M
M F
M F
F F

Well, we know Known Dog can't be female, so we have to get rid of that line completely, so we get (understanding that the "-" is just a placeholder that indicates no possibility):

M M
M F
F M
- -

Now we go to the next line, the one in bold, and once again apply the knowledge that Known Dog is Male.

M M
M F
F M

Well, Known Dog is male, there's only one M, so by process of elimination, we must have to replace the lone M with a K, so we get:

M M
M F
F K

Now we go to the middle line, the one in bold, and apply the knowledge that Known Dog is Male a third time.

M M
M F
F K

Well, same as before--one M, that must be Known Dog

M M
K F
F K

Now we move to the top line, and apply our knowledge.

M M
K F
F K

Oh Noes! How do we apply our knowledge to this line? We have to apply that knowledge, we can't just pretend we don't have it. The solution? We have to create a new line, a line that reflects the fact that in pairs where both puppies are male, the Known Dog could be either dog. So we get:

K M
M K
K F
F K

And as all are equally likely because they are all (1x.5) or (.5x1), and K=M for purposes of the question 'is this a pair of male dogs,' two out of four equally likely possibilities means a probability of 50%.

Ok this got glossed over because someone started talking about 4 coins. By doubling the MM chances you have brought order into the problem when is isn't really needed but we can go through using it anyway. This much is true you cannot ever create new options for results halfway through, all the results must be there from the beginning so by your logic the original options are.

MM
MM
MF
FM
FF
FF

which as we have pointed out is obviously wrong because then the chance of getting 2 heads or 2 tails is 1/3 instead of 1/4. To fix this you must also double FM and MF. To make this more clear I will attach numbers to the dogs 1 and 2 this looks very repetitive at this point because it is it only becomes important once you start calling one of the results K.

1M 2M
2M 1M
1F 2M
2F 1M
1M 2F
2M 1F
1F 2F
2F 1M

Your final equation removing all FF now looks like this

1K 2M
2K 1M
1K 2F
2K 1F
1F 2K
2F 1K

1F 2F
2F 1F
Returning us to our 1/3 33% probability

edit: I noticed that the MF FM looks like I'm doubling because I still have repeated pairs like 1K2F and 2F1K. It only looks more repetitive than the MM's because the K is obscuring the doubling there. I added in the FF pairs show that it is the same as the grouping above and if it didn't have those doubles then we would have the 1/3 of the time you get 2 heads problem again.

 

[MagnetoHydroDynamics]

Ok this is like the monty hall problem.
If atleast one is male, you have three possible situations
Beagle1 Beagle 2
Male Male
Male Female
Female Male

Of these three possible outcomes, only one of them results in the second beagle being male.
Thats where the 33% comes from...
That's the logic behind the problem, however there's also the idea of independent events... the mind boggles.

This guy is right:

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?

As long as you havn't seen them yet "The least one" could be any of them.

Makes perfect sense.