Poll: A little math problem

[Samirat]

since fm == mf
if you place fm AND mf in the set you have to repeat your other sets too.

ugh. take your coin tossing and simplify it say you have 1 coin.
since, trivially, H == H and T == T. you can say your list of possible sets

is
H
or
T

which condenses

H
H
H
H

T
T
T
T

with an equal number of H and T's to however much your patience takes you.

Well, this actually isn't important, because the ratios remain the same, between heads and tails. With two coins, though, you suddenly think you have an equal chance of getting two heads as getting one heads and one tails.

 

[Samirat]

And so would you also say that out of a random collection of dogs, you would have this set:

MM
MM
MF
FM
FF
FF

.....

Is HHHH the same as HHHH, and HHHH as well? What about HHHH

Nope--because while MM=MM, KM=/=MK,

and

while HHHH=HHHH=HHHH=HHHH, KHHH=/=HKHH=/=HHKH=/=HHHK

You're confusing the possible combinations of a permutation in a set of possible permutations with the possible permutations of a permutation in a set of possible permutations.

All right, I said that you had at least two heads, so:

KKHH, KHKH, KHHK, HKKH, HKHK, HHKK
There are 6 possibiliites, just as I said. So, given that there are two heads, are you saying that this is actually more likely than 3 heads, 1 tail?
KKHT, KHKT, HKKT

And you're using the word "permutation" randomly.

 

[Samirat]

You can't deny that out of a random selection of 200 puppies, you will end up with, on average, 50 female female pairs.

I don't.

I deny that we can act as if we *did* wind up with 50 F/F pairs.

We took the 50 female pairs and placed the aside, separately.

How do you place aside something that doesn't exist?

What, are you saying there will be no female female pairs in the problem that Werepossum described?

Nope. I'm saying there won't *necessarily* be any female female pairs in the problem that Werepossum described.

He just has random puppies, therefore there will be female female pairs.

Think about that one for a while, and the differences between the words 'will' 'might' 'should' and 'could'.

Not relevant. He probably will have female female pairs. Do you minor moral victories over semantics feed your resolve?

 

[mnimmny]

In the puppy problem, eliminating the possibility of the all-female pair doesn't raise the probability of the One of Each pair from 1/2 to 2/3, it stays at 1/2

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.

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.

 

[Samirat]

All right, I said that you had at least two heads, so:

If you think that's an issue, you don't understand what I'm saying.

Go back and get the logic of the Three Card Problem down--it *really* helps here, I just realized.

I saw it, answer the question. Here, I'll repeat it.

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

 

[Samirat]

He just has random puppies, therefore there will be female female pairs.

Think about that one for a while, and the differences between the words 'will' 'might' 'should' and 'could'.

Not relevant. He probably will have female female pairs. Do you minor moral victories over semantics feed your resolve?

If you think the difference between 'will' and 'might/could' and 'should' in the context of trying to explain probability to someone is just 'semantics' and 'not relevant', you really shouldn't try to explain probability to people.

Perhaps not over internet forums in the future, but I'll try a little longer. Returning to the issue at hand.

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.

 

[Samirat]

In the puppy problem, eliminating the possibility of the all-female pair doesn't raise the probability of the One of Each pair from 1/2 to 2/3, it stays at 1/2

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.

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.

Check this post for a longer, more detailed explanation:

http://www.escapistmagazine.com/forums/jump/18.73797.823693

Hehe, you think that makes sense?

 

[Samirat]

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?

 

[mnimmny]

@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

Hehe, you think that makes sense?

QFT

 

[Samirat]

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

I'm sorry but, it's not working out between us. You just don't get me, respond to quickly, or something, but we just can't get through to each other.

I mean, if you think my logic would be:

KKHT, KHKT, HKKT

and not something more like (just doing this fast now without double checking)

KKHT, KHKT, HKKT, HKTK, HTKK, KHTK, KTHK, HTKK, THKK, KKTH

You either don't understand me, how to do permutations, or both.

Sorry I couldn't be of more help.

Is it just me, or does that look like, I don't know, a listing of Soviet agencies?

Yeah, you're right, there should be 12 of the things. 4 spaces for the T, times 3 arrangements for the other three. Which still isn't correct. The chances are

6 2 heads 2 tails
4 3 heads 1 tails
1 4 heads

 

[Samirat]

I think we both know that there is only one way for us to settle this now.

Duel to the death!!!

Or Starcraft. 1v1 Python, right now.

Or chess. I call white.

 

[Doug]

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

fixed: that says the same as the one above it so it's irrelevant

Ah, but there's a reason it was like that- it was the order.

I came back for three mins. after leaving to have fun so im not getting into this cirle logic again

I agree with (ZHU) Michael - order is wholy irrevelant in this scenario.

 

[Doug]

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%

 

[Doug]

double post