Poll: A little math problem

[Aurora219]

This is a very simple logic puzzle. You have to realise these are separate issues, the fact that the first one is male doesn't change the probability of the other one being male or female. So say 63% of the population of all beagles was female, then this'd have a 37% chance of being male.

Who says the first one is male? What if the second one is male? You, like most others here, are making the mistake of assuming that the male dog must be the first one, so out of this set:

MM
MF
FM
FF

You are ruling out both FF and FM, when the question only prohibits FF. Since either dog could be male, there are two situations where the other dog will be female and only one where it will be male.

Does it matter which is which?

The first one's male. It's the first one simply because it's mentioned first. If you wanna call the second one the first one, you do that. The second one's male then. Still doesn't affect the other one.

 

[Lukeje]

All right, let's name the dogs.

There are three situations.

Sparky could be male, Othello could be male.
Sparky could be female, Othello could be male.
Sparky could be male, Othello could be female.

The fact that they refer to "the other dog," doesn't mean that they've defined which dog that is. For instance, in situation 1 and 3, Sparky is the first male dog she found, and hence, the other dog is Othello. In situations 2, the second dog she checked, Othello, is male, and so the other dog is Sparky.

My point was that the fact there is another dog that isn't male suggests that the probability is zero. If, as you assume, there is 'at least one male dog', then the other must be female, because if they are both male, which is the 'Other'?
This is from a literal reading of the question (which I think is poorly transcribed from somewhere else from the ambiguities). [I do not actually think that there is a right answer, I'm just trying to prove the absurdity of this thread by showing how one could concieve of the answer being 1/2, 1/3, and 0.]

 

[Alex_P]

Does it matter which is which?

The first one's male. It's the first one simply because it's mentioned first. If you wanna call the second one the first one, you do that. The second one's male then. Still doesn't affect the other one.

If it doesn't matter which is which, then you get this distribution instead:

1/4 MM
1/2 MF
1/4 FF

Either it always matters which is which or it never matters. You can't mix the two in the same probability matrix and get an intelligible result out.

-- Alex

 

[Aurora219]

Does it matter which is which?

The first one's male. It's the first one simply because it's mentioned first. If you wanna call the second one the first one, you do that. The second one's male then. Still doesn't affect the other one.

If it doesn't matter which is which, then you get this distribution instead:

1/4 MM
1/2 MF
1/4 FF

Either it always matters which is which or it never matters. You can't mix the two in the same probability matrix and get an intelligible result out.

-- Alex

I don't get it. Why does one being male affect the other one?
Read what you wrote and come to terms that one of your options requires both to be female. Maybe you need to put that 25% onto the top option?

 

[thahat]

no im not reading a gazilion posts:

50. damn. percents. %%%%%

seriusly people, its like tossing a coin, the coin dous not have a memory, so its 50%

 

[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..