Poll: A little math problem

[werepossum]

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.

Your description of #2 sound fishy.

These scenarios are equivalent:
A. I flip a coin and look at it, then I flip another coin and look at it, &c., up to 10 coins.
B. I flip ten coins and then look at them in some specific order (it can be different from the order in which I flipped them).

The probability only changes when you start cherry-picking: searching until you find a male.

-- Alex

Not at all. Remember that the question is not "Is the tenth coin heads?" but rather "Is the tenth coin heads given that the first nine are heads"?

In the first scenario, I'm looking at the possibility of the tenth coin coming up heads KNOWING that the first nine came up heads. At this point, nothing I can do can change the outcome that at least nine of ten are heads. I have selected my set of ten with knowledge that eliminates the vast majority of probability distributions. In fact, only two possible probability distributions remain - nine heads followed by a tails, OR ten heads in a row. In selecting nine consecutive "heads" coin tosses, I have done the lion's share of obtaining ten straight "heads" coin tosses.

In the second scenario, I'm looking at the possibility that I have randomly selected a set of ten fair coin tosses which are all heads. There's a 50% chance the first toss will be tails; therefore half my random sets fail already even if I don't yet know it. I toss the second coin. My chance of being all heads is now 0.50 x 0.50 or 0.25, because both throws have to be heads to meet my criteria of ten heads. I can fail on the first throw, or on the second. It makes no difference whether I look at the values individually or all together at the end; I've selected them as a set. Now I toss the third coin; my chance of remaining all heads is now 0.50 x 0.50 x 0.50 or 0.125 (or 12.5% expressed as a percentage.) Now I toss the fourth coin: my chances of remaining all heads drop to 0.0625. The fifth, sixth, seventh, eighth, and ninth coin tosses take me to 0.3125, 0.15625, 0.0078125, 0.00390625, and 0.001953125 respectively. This number (0.195%) is the chance of a randomly selected set of ten fair coin tosses containing nine heads in any combination. If the first nine tosses were heads, then the tenth has the same base 50% chance to be heads. However, I selected the set randomly. To get the chance that I randomly selected ten coin tosses and they are all heads, I have to multiply 0.195% by 0.50 again - at which point I get 0.00098 or 0.098% - roughly one in a thousand. In this case, I have selected my set with a huge amount of possible permutations and combinations.

I picked ten coin tosses rather than four to emphasize the difference between sequential probability and set probability. If I group items together into a set AFTER I know their value, then the odds of the last item in the set are independent of the other items because I have negated all the possible probability distributions that don't fit that set's values. If I group items into a set BEFORE I know their values, then all possible probability distributions are still valid, and I can only eliminate them as I learn things about the set OR about individual items in the set.

To belabor a point, if I throw nine heads in a row with a coin that I know to be fair, the tenth heads is not remarkable. It's the nine in a row that are remarkable.

EDIT: Forgot to add, the chances are variables in set mathematics. If you alter the percentage chance of a particular permutation occurring, you alter the value distribution of the set. Thus the value of some set items must be changed. Since grouping the items into a set is merely a matter of convenience, it can't change the value of any particular item in the set; set mathematics can only describe probabilities of each particular combination or permutation of values of the items.

 

[werepossum]

SNIP
Ahh good--you're back. Maybe you can explain this.

Those two scenarios? What if Scenario #2 describes some kind of strange casino game--they flip a fair coin ten times, revealing the first nine after the flips are over. But you've been able to cheat the system, and you've been getting the results of the coin tosses as they come.

Why are the odds different for the honest gambler than for the cheater when the same coins are being bet on by both of them? I get why both could be true--it sounds like sequential probability is where you forget about every result but the last result, which you can do because they are independent events. And set theory makes sense, because what it sounds like you're doing is you're looking at one of those probability curves and asking the question 'what are the chances I'm not just four deviations from the middle, but five'.

This is why I kept talking about quantum puppies: the probability changes depending on whether you decide to look at something in sequence or as a set--that the nature of the human observation changes reality.

SNIP

The odds of getting ten heads in a row are exactly the same whether you look at each coin as soon as it is flipped, OR you look at all the coins together.

In the first scenario, you start with almost all possible permutations of ten coin tosses removed from the realm of possibility. I have flipped nine coins and gotten nine heads; what is the chance I'll roll a tenth heads GIVEN I HAVE ALREADY ROLLED NINE? That's the difference. If I've already rolled nine in a row, much of the hard part of rolling ten in a row is already done. By grouping nine known heads into my set of ten, I have eliminated nine possible tails, any one of which keep me from rolling ten heads in a row which is necessary if the tenth coin in the set is to be heads.

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.

This is easily experimentally verified. Take three coins and begin flipping them. When they each turn up heads, toss a fourth coin. Log how many times you get four heads as YES and how many times you get three heads and a tail as NO. If you make thirty tosses (of four coins) this way, you should end up close to fifteen YES. This is sequential probability; the chance of satisfying the question is dependent only (and independently) on the last throw.

Now take four coins and flip them all, recording for each toss either YES (meaning four heads) or NO (meaning less than four heads.) If you make thirty tosses this way, you should end up close to two YES. This is set probability; the chance of satisfying the question is dependent on every throw.

I'll try to go back on the pups problem later tonight.

 

[werepossum]

Say we have 200 pairs of pups. There is a 50% chance each is male. Our possible combinations are thus 25% all male, 50% pair, and 25% all female. We expect two hundred females and two hundred males.

Using 50% chance:
For 50 pairs the response is "No, there is no male." This yields one hundred females and no males.
MATH: 200 * 0.25 = 50 all-female pairs and one hundred females.

For the other 150 pairs, there is at least one male and a 50% of a second male. This yields seventy-five females and two hundred twenty-five males.
MATH: Females: 150 * 0.50 = 75 females
MATH: Males: 150 + (150*0.50) = 225 males

The total is one hundred seventy-five females and two hundred twenty-five males.

Using 33% chance:
For 50 pairs the response is "No, there is no male." This yields one hundred females and no males.
MATH: 200 * 0.25 = 50 all-female pairs and one hundred females.

For the other 150 pairs, there is at least one male and a 33% of a second male. This yields one hundred females and two hundred males.
MATH: Females: 150 * 0.67 = 100 females
MATH: Males: 150 + (150*0.33) = 200 males

The total is two hundred females and two hundred males.

QED?

EDIT: Do I hear crickets?

 

[Samirat]

Hehe, Werepossum, that's awesome. I never thought of doing something like that.

 

[Samirat]

Stuff

What if the first one's not male? I'll name the dogs Sparky and Othello.

There are three situations here, where at least one dog is male:

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

So, you are neglecting the third choice. If the dog washer checks Sparky first, he is the first dog. You don't know which dog is male, therefore the order is independent of that. You can't define the order based on information you do not, in fact possess.

In a random pair of two dogs, you have this distribution:

Male Male: 25 percent
1 Male 1 Female: 50 percent (Because this includes both options Male Female and Female Male)
Female Female: 25 percent

Since the only situation not including at least one male here is female female, it's knocked out of the problem. So the remaining probabilities are redistributed to equal:

Male Male: 33 percent
1 Male 1 Female: 67 percent

 

[Samirat]

*chirp chirp*

Say we have 200 pairs of pups. We expect two hundred females and two hundred males.

Using 50% chance:
For 50 pairs the response is "No, there is no male." This yields one hundred females and no males.
MATH: 200 * 0.25 = 50 all-female pairs and one hundred females.

For the other 150 pairs, there is at least one male and a 50% of a second male. This yields seventy-five females and two hundred twenty-five males.
MATH: Females: 150 * 0.50 = 75 females
MATH: Males: 150 + (150*0.50) = 225 males

And no two females can wind up with each other: like you said, "For the other 150 pairs, there is at least one male" So you subtract the predicted seventy-five females from the predicted two hundred twenty-five males because:

1) the females have to wind up in a pair,

and

2) no two females can be in the same pair,

and you wind up with one hundred and fifty males

225-75=150

in seventy five pairs, and seventy-five females and seventy-five males in seventy-five pairs, which means the chances of getting a double male pair are...50%?


I think what you did there is you applied the "No, there is no male" condition to the probable pairs of puppies from the original 400, then you forgot to apply that same rule to the matching up of the left over 300 (which is the puppies left over when we take out the expected all-female pairs) so you don't see that your prediction gets all-female pairs in your group where all-female pairs are not allowed.

No, you don't understand.

He took 200 random pairs of puppies. 50 of them included two females, because 25 percent of pairs of random puppies will be two females. So he disregarded them, because they aren't the same as the pair in the problem. For the pairs that did include at least one male, he displayed what would happen if you used each of our probabilities, 33 percent and 50 percent. And 33 percent gives you the correct answer.

Surprised?

What he's doing there is displaying what would happen if we used the probability of 50 percent. And it comes out with the wrong total number of males to females. We get, out of 400 random pups, 225 males to 175 females. Whereas using 33 percent as the answer, we get 200 males to 200 females. Which is correct.

 

[werepossum]

Say we have 200 pairs of pups. We expect two hundred females and two hundred males.

Using 50% chance:
For 50 pairs the response is "No, there is no male." This yields one hundred females and no males.
MATH: 200 * 0.25 = 50 all-female pairs and one hundred females.

Actually, what I don't understand is how we wind up with 50 F/F puppy pairs when the guy told us at least one puppy in each pair is a male.

If the guy in the problem is talking about all the puppy pairs in the problem when he says 'there is at least one male' then why is he only talking about 150 out of 200 pairs when he says 'there is at least one male' about all of werepossums pairs?

We know from the problem that the sex of the pups was unknown when the set was made. Otherwise the pet shop woman would not have had to make the call to tell the customer if there was a male pup in the pair she had. Therefore we know that this set had a possibility of having two females. Since we know the distribution is random, the number of items (two pups) in the set, and the chance of each possible value, we know the probable distribution.

I set up two hundred pairs from which this woman could have selected this particular pair. You could run the equations with any number of pup pairs from one up; I chose two hundred because it yields whole dogs and made the math easy to follow. Also, two hundred pairs means we could run this test enough times to get a statistically significant answer at almost any common confidence level you'd care to use.

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. In other words, if I took two hundred pup pairs and queried two hundred groomers "Is there at least one male?" I would expect fifty "NO" and one hundred fifty "Yes" answers, since one or two pups both trigger a "Yes" answer. You can easily experimentally verify that distribution with two coins and pen and paper as it is pure combinational; permutations are not used.

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. And although the call let us know there was a male, it was not capable of altering sex. Therefore I know that if this experiment is repeated there will be pairs with no males. And I know that if this experiment is repeated enough times, the number of those all-female pairs will settle around 25% of the total number of pairs.

That leaves me with 150 pairs which could all potentially be the pair in the problem, as we are left without knowing if there are in fact two males. I then ran the equations for both percentages and compared results. Instead of percentages, we get whole dogs which makes it easier to follow. We're simple counting how many times each formula returns a second male. And I used pure combinational theory because it's simpler to follow and explain.

Sleep on it and look it over tomorrow.

EDIT: 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.

 

[Bulletinmybrain]

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..

 

[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.

 

[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.