Poll: A little math problem

[werepossum]

Can I chime in a quicky? I was taking the PSAT today and there was a question like this. (x-3)(x+3)=a and then it asked what would (X-6)(X+3)= too in terms of a if I am right.>.> I dunno can't remember but I would like to know how you would solve problem like this..

(X-3)(X+3) = X^2 -3X +3X -9 = X^2 -9 = a

(X-6)(X+3) = X^2 -6X +3X -18 = X^2 -3X -18 = a -3X -9

That would be one way. Am I answering what you're asking? I'm afraid I'm used to setting up my own equations now, so "in terms of a" throws me a bit. (Disclaimer: I took the PSAT thirty years ago.)

 

[werepossum]

Damn. I had abandoned this thread and then I just HAD to look at it one more time.

Samirat had the clearest explanations, I think. But I'll start with the ten coin tosses.
Scenario #1: Toss a fair coin ten times in a row. If the first nine tosses come up heads, the tenth toss still has a 50-50 chain of being heads. Each coin flip is independent. This is sequential probability.

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.

In the second scenario, I start with ten coin tosses in my set. It makes absolutely no difference whether I look at them as they are tossed OR all at once when they are all tossed SO LONG AS I GROUP THEM WITH NO KNOWLEDGE OF ANY OF THE INDIVIDUAL ITEMS' VALUES.

The means by which you pick your set in the two scenarios are:

Scenario #1: Toss a fair coin ten times in a row.

Scenario #2: Toss a fair coin ten times in a row without looking at the results.

So we've grouped both of them "with no knowledge of any of the individual items' values," right? The words in S#2--"without looking at the results"--are irrelevant because you said it "makes absolutely no difference whether I look at them as they are tossed OR all at once when they are all tossed."

So why in S#1 "the tenth toss still has a 50-50 chain of being heads" while in S#2 "the chance of the tenth being heads? About one in a thousand" is your answer.

What exactly is the difference between the scenarios besides the fact that you define them as sequential vs. set probability?

If there should be no difference in odds "SO LONG AS I GROUP THEM WITH NO KNOWLEDGE OF ANY OF THE INDIVIDUAL ITEMS' VALUES" then where's the grouping *with* knowledge of values in one of the scenarios and not the other? Don't you group both sets identically and with no knowledge of their value when you decide the set will be the next ten coin flips?

If you decide the next ten coin tosses will be your set, then yes, they are identical. It's only if you throw nine heads in a row and THEN throw the tenth coin toss that the two sets differ.

I brought that up (and apparently explained it poorly) because that's what you are doing with the pups problem. In simple combination, two pups of unknown sex will be female/female 25% of the time, female/male 50% of the time, and male/male 25% of the time. If you start your set with the pair of the pups sent to the groomer, then your distribution will look like that. Running the experiment 100 times should result in twenty-five female/female pairs, fifty female/male pairs, and twenty-five male/male pairs. This is like designating ten coin tosses without knowing their contents and guessing that all ten will be heads, except by using ten coins I exaggerate the odds for clarity.

On the other hand, claiming that there is a fifty percent chance that the second dog is male requires that you re-form your set and say THIS dog is male; what are the chances the other dog is male as well? The only way the math works is if you identify one particular dog as male, with the second dog unable to be the dog identified. Thus you eliminate not only the female/female pair, but also one male/female permutation and thus half the possible male/female combinations. An exaggerated form of this is to select a run of nine heads tosses and then calculate the probability of the tenth toss ALREADY KNOWING THE OUTCOME OF THE FIRST NINE TOSSES. Reforming the set with the fifty percent chance for a second male likewise requires knowledge you don't have.

 

[werepossum]

Set theory statistics (and common sense) tell you that if you have two hundred randomly paired pup pairs, you'll have fifty female/female pairs, fifty male/male pairs, and one hundred male/female pairs.

No it doesn't. It tells you that you *most likely* have that distribution--not that you actually have it. You can't behave as if that's an establish fact when it's just a possibility. The most likely possibility, but a possibility nonetheless.

Of those two hundred pup pairs, fifty pairs (those with two females) are clearly not the pair in the problem because they lack a male. So I set those aside. But they have to be there as possible pairs or the woman would not have needed to make the call.

She doesn't call because she doesn't know the pair isn't all-female: she calls because she doesn't know anything about the sexes of the puppies.

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?

What is in bold is the only information she has, and what is underlined is the lack of knowledge she has. The woman still has to make the call even if every single one of the 400 random puppies is female, because she doesn't know anything about the sex of any puppy OR any set.

You're confusing the knowledge the Puppy Washing Man gets when he checks, with the *lack* of knowledge the Shopkeeper Woman has that makes her call necessary.

But if you can't rely on set theory to set the number of possibilities of female/female pairs, then you can't use it to determine any percentage of male/male pairs. Either the math works, or it doesn't. If it doesn't, then the problem is indeterminate. You might just as well yell "Burma!" as throw out a percentage, which could just as easily be negative since we no longer have math capable of handling the problem.

I give up! I'm going home. If you want, run the experiments with coins. Or look up a guide to the SAT or ACT, assuming those questions are still asked.

 

[cookyt]

How about this attempt at an explaination?

If we count this as a coin problem, and we are assured that one of the coins will almost magically turn up heads on our next flip, then one assumes they can put down one coin heads up on the table and flip the coin. 50% right? Wrong.

Although we know one of the coins will be heads, we don't know which one, thus we cannot simply pick out a coin, and label it heads. Here we view it as a problem involving exclusively the set, but not either coin individually.

the permutations are:
heads, tails : male, female
tails, heads : female, male
heads, heads : male, male

if one must be heads, then the only way the other is heads is by them both being heads. Tails, tails is impossible, so the answer is 1/3.

 

[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. This is the group which is not allowed in this problem. We have prevented them from getting any female female pairs by removing the female female pairs. It's as simple as that. Now, all the rest of the pairs contain at least one male, these are the ones the washer woman is looking at when she says "yes, there is at least one male." If you take the problem at its word, and I'm assuming you are, they are a random pair of puppies. Therefore, this must be one of the random pairs that has at least one male.

You've beaten Werepossum out without even exposing your major fallacy, your lack of belief in the validity of a reflip.

 

[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. Then, we only used the pairs with at least one male in the problem. So we are applying the probabilities only to pairs which satisfy the premise of the problem. So if we apply the 50 percent probability to this, we get a certain number of females and males in the 150 pairs with at least one male. Adding that to the dogs in the female female pairs should end up with 200 males and 200 females.

But no. Because 50 percent is an incorrect ratio. If you use 1 third, however, you come up with the correct proportions, a 1 to 1 ratio of males to females.

 

[mnimmny]

uh... wtf.

probability is information about the state of the world NOT the actual state of the world.
i.e. chance of heads if i flip it and cover it is 50%, chance of it being heads after i look at it ends up being 100% or 0%.

the list of all possible sets of 2 beagles is:

1 set male beagles (mm).
1 set female beagles (ff).
1 set male female (mf).

and you know that one dog is male
therefore the only way the other dog can be male is if both are.

the chance of randomly picking the set of male beagles from all the possible sets is 33%
because (1 set of male beagles)/(3 possible sets).

Note:
I have female male and male female compressed into one set because she's randomly selected a male from her set of two, these aren't ordered pairs .


if you write male male as male male it doesn't change anything, b/c the ratios stay the same.
you get:

1 mm
1 mm
1 ff
1 ff
1 fm
1 mf

and if the unknown dog is male then you must have a male male set. Your chance of selecting that is 2(there are two sets of males) out of 6(all possible sets) == 2/6 ==1/3.

 

[Samirat]

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%.

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

?

Because if so, rethink. Then 1/3 of the pairs would be male male. Now I know you know that's not true.

Say you have a coin tossing experiment. Flip 4 coins at once, at least two of them are heads. What are the chances that the other two are heads.

Is HHHH the same as HHHH, and HHHH as well? What about HHHH, forgot that one. There are actually 6 of these. By this logic, it would actually be equally likely to two heads two tails, even though two heads, two tails is an exponentially more likely solution when you flip 4 coins. It goes like this:

6/11 chance 2 tails two heads
4/11 chance 1 tail 3 heads
1/11 chance 4 heads

 

[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? He just has random puppies, therefore there will 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.

 

[mnimmny]

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.