Arguing about Probability and Statistics (From: DM's who want you to Roleplay but...)

[Khedrac]

However, actually presenting the question, "at least one child is a girl; are they both girls?" does not provide any more or less information than "the older child is a girl; are they both girls?" does.

Now the statement I have quoted above is where you are going wrong as it is demonstrably false.

I think we all agree that for two children the four possibilities overall are:

1) Girl then Boy
2) Girl then Girl
3) Boy then Girl
4) Boy then Boy

Saying "The older is a girl" covers two of the options (1 and 2) whilst saying "at least one is a girl" covers three possibilities (1, 2 and 3).

If we ignore all the rest of the stats, since they provide for different list of possible outcomes the two statements provide very different information.

Once we have accepted that they provide different information, accepting that they may have different outcomes should be automatic.

Incidentally, once you see just how difficult to understand statistics can be it becomes really easy to see why people who know a little, often do not trust it.

[Pinnacle]

"One of these is heads; what is the other?" is not specifying which one. But "the one on the left is heads; what is the other?" is, somehow.

There are three ways for the former to be true. There are only two ways for the latter to be true.

What makes the specification...specific? Rather than extraneous information no more designating than "one" and "the other?"

It's all about the proportions among the sample. The specification removes another one of the original four possibilities, and now there are only two.

My conclusion in the last one is that, absent information about how you decided it was the older sister you'd tell me about, it is actually no more information than if you just picked one of the two children at random to tell me about. It's no different than if you flipped a coin to decide whether you'd tell me about the older or the younger child.

That's true. As long as you tell me which one you're talking about, the odds are 50/50, because you've always eliminated two possibilities--the two where the thing you told me isn't true.

In fact, the ONLY way to generate the "at least one child is a girl; are they both girls?" 1/3 probability of "yes" is if we explicitly select for situations where that is the case.

I don't see how "all cases where this is true" is more specific than a subset of them. But sure, true 'nuff.

That is, we can safely say that, if we only look at families of 2 children where at least 1 is a girl, the odds that both are girls in any one of those families chosen at random is 1/3.

Correct.

However, actually presenting the question, "at least one child is a girl; are they both girls?" does not provide any more or less information than "the older child is a girl; are they both girls?" does.

Incorrect. The first clue tells us there are three possibilities; the second narrows it to two.

This is illustrated by my earlier blue-coin/green-coin game, wherein I chose the information I gave such that it would always be true (and, based on the information, was not 100% known), but randomized it where it was possible to be truthful with two or more different pieces of information. It broke down to 50% odds, because of the nature of how I generated questions and information.

Yes. You gave me the information that at least one is heads [i]every[i] time there was a match, but only half the time it was a mismatch.
The odds of a match are 50/50 in any given pair of coin tosses (counting tails matches--1/4 otherwise).
The odds of a match are 1/3 in all cases that contain at least one heads.
The set you gave me didn't include all such cases, though; it only included a fraction of the ones where there was no match.

Similarly, "at least one child is a girl" does not tell us whether we would ONLY have been presented this situation if at least one was.

No. If we have only that one piece of information, we can only narrow the possibilities down to the three cases where that information is true.

Hence why the problem is so deceptive. In truth, the "at least one child" version is actually specifying one, because it's asking about one specific sample, not about a collection of samples wherein that condition is true. In any one specific sample of 2 children, if "at least" one is something, we know information about one of them. We can now specify them as "the one we know this about" and "the other, whom we do not."

It's asking about one. There is only one correct answer.
You don't know the right answer, though. We simply do not have the information to state it, since there is only one piece of information complete enough to determine it: The answer itself.
The question wasn't about the correct answer, though, it was about probability. The chances of each answer being correct are determined by the sample.
The sample where "at least one child is a boy" is a larger sample than "the eldest is a girl."
Which single case was the one chosen? No idea, other than it's one of the ones where the single clue is true.

[Lord Torath]

Okay, let's try a slight change. Let's say I have twins Cameron and Kim. Fraternal, but nearly identical pre-pubescent twins. Cameron has a 50% chance of being a girl, and Kim also has a 50% chance of being a girl. There are three possible sex combinations: Boy-Girl (50%), Boy-Boy(25%), and Girl-Girl(25%). I put a sheet over each of them, and put them in my front room. I tell you that one of them, Cameron, is a girl. This eliminates the Boy-Boy possibility. So now the odds are Boy-Girl(66.7%) and Girl Girl(33%). Now you don't know which one is Cameron. But you know that Kim has a 50% of being a girl. So now we've got the paradox. There are two children in front of you (under sheets), and a 33% chance they are both girls. But there's a 50% chance that Kim is a girl, and thus that both are girls. How do you resolve this?

[Pinnacle]

Permutations matter in probability.

It's not:
2 girls 25%
2 boys 25%
one of each 50%

It's:
Both are girls 25%
Cameron is a girl and Kim is a boy 25%
Cameron is a boy and Kim is a girl girl 25%
Both are boys 25%

By telling me that Cameron is a girl, you haven't just eliminated the possibility that both are boys, you have also eliminated the possibility that Cameron is a boy and Kim is a girl. You've eliminated half of the possibility that there's one of each along with the possibility that there's a pair of boys.
There's only two permutations left.

It's basically the same as the "eldest is a girl" premise, only with names instead of ages.


This stuff is really confusing, and I don't understand it too intuitively myself. Write out all possibilities and you'll always find the answer that way, though. Eventually.
Well, assuming you don't make a mistake of course.

Our possibilities here are:
C-boy/K-boy
C-boy/K-girl
C-girl/K-boy
C-girl/K-girl
50% chance of match. 25% chance of match of girls, specifically.

If I know at least one is a girl, we're left with:
C-boy/K-boy
C-boy/K-girl
C-girl/K-boy
C-girl/K-girl
33.3..% chance of match of girls.

If I know Cameron is a girl, we're left with:
C-boy/K-boy
C-boy/K-girl
C-girl/K-boy
C-girl/K-girl
50% chance of match of girls.

[Jay R]

If you want to understand conditional probability problems, take a probability course or read a probability book. Get at least through conditional probability.

If you will not do this, you will not understand conditional probability problems.

Nothing replaces learning the material.

[Segev]

Now the statement I have quoted above is where you are going wrong as it is demonstrably false.

I think we all agree that for two children the four possibilities overall are:

1) Girl then Boy
2) Girl then Girl
3) Boy then Girl
4) Boy then Boy

Saying "The older is a girl" covers two of the options (1 and 2) whilst saying "at least one is a girl" covers three possibilities (1, 2 and 3).

If we ignore all the rest of the stats, since they provide for different list of possible outcomes the two statements provide very different information.

Once we have accepted that they provide different information, accepting that they may have different outcomes should be automatic.

Incidentally, once you see just how difficult to understand statistics can be it becomes really easy to see why people who know a little, often do not trust it.

This is actually not correct. We have changed the nature of the problem by reducing the sample size to one, and specifying nothing about the population from which the family was selected.

As a selection criterion, "at least one child must be a girl" does generate a population where only 1/3 of the 2-child families present have two girls.

Similarly, "the older child must be a girl" will generate a population where 1/2 are both girls.

However, as demonstrated by the blue-coin/green-coin game, where the content of the information and the question asked is determined by the nature of the result, a single sample presented to a given individual does not work that way.

You have to be told not, "at least one of them is a girl," but "chosen from a population of 2-child families where at least one was a girl." Because otherwise, it is equally possible that it was chosen from a population of all 2-child families, and the nature of the information given was based on the family chosen.

As demonstrated, that changes things.


Essentially, unless you specify that the information given is about the selection criteria for the sampled population, rather than information chosen based on the sample taken, you have rendered whether or not the older child is the girl irrelevant. By specifying, for a single sample, "at least one is a girl," you have created a situation where you know the answer to one of the two.

That is, you now can identify the children as "the one whose sex is known, and the one whose sex is unknown."

To be as clear as possible, this is because we do not know how you chose whether to ask us about both being girls, nor how you chose to tell us that at least one was a girl. Not knowing that, we do not actually have:

BB at 1/3
BG at 1/3
GB at 1/3

We have

BB "at least one is a boy" at 1/3
BG "at least one is a boy" at 1/6
BG "at least one is a girl" at 1/6
GB "at least one is a boy" at 1/6
GB "at least one is a girl" at 1/6

The stricken-thru ones are additional possibilities you've removed with the information you've given, that at least one is a boy. By telling us that, you tell us that it is not the half of the times BG is the result that you tell us there is a girl, nor is it the half of the times GB is a result that you tell us there is a girl.

That leaves us with normalized probabilities:

BB "at least one is a boy" 1/2
BG "at least one is a boy" 1/4
GB "at least one is a boy" 1/4

BB has twice the chance of the others because it doesn't matter which of them you tell me about; you'll give me the same information (that one is a boy).

Regardless, because it is a singular sample taken from an unspecified population with the criteria for how you chose whether to tell me about the older or younger child (without telling me which), you throw out enough additional possible results that it again becomes 1/2 chance.

[Segev]

If you want to understand conditional probability problems, take a probability course or read a probability book. Get at least through conditional probability.

If you will not do this, you will not understand conditional probability problems.

Nothing replaces learning the material.

I have. I did understand it. This problem has bugged me anyway because of the paradox of the blue-coin/green-coin game as I tried to set it up.

I have come to the conclusion that, because setting up the coin game influences the probability such that the information given is no longer determining of the odds, the problem as presented is poorly set up.

To properly set it up, it would have to be phrased thusly:

"Of all families with 2 children wherein the older child is a girl, what are the odds that a single randomly-chosen family will have both children be girls?"

and

"Of all families with 2 children wherein at least one of the children is a girl, what are the odds that a single randomly-chosen family will have both children be girls?"


Those two questions properly frame it in a way that shapes the probabilities as specified in the "textbook answer."

As phrased earlier in this thread, the question arises about how you chose to give us the information about what "at least one" of the children's sexes was. Without being told that this was a random sample from a population where "at least one is a girl" is always true, we cannot know that the information and the question were not chosen AFTER the sample was taken, thus providing only information that lets us identify the children as "the one whose sex we know" and "the one whose sex we don't know." Or, if you prefer, introducing the now-excluded options of BG "at least one is a boy" and GB "at least one is a boy."

Heck, given how humans usually hand out information, it is reasonable to assume that the information came from observation of the chosen sample, rather than a criterion of the population.

[Talakeal]

No she did not. She did not say that, nor did she mean that.

Hence the word "imply". In my experience you would only use the word "some" to mean "all" if you are deliberately trying to trick or mislead someone, as it is technically true, but not something you would say when there is a much more appropriate word.

[Jay R]

I have come to the conclusion that, because setting up the coin game influences the probability such that the information given is no longer determining of the odds, the problem as presented is poorly set up.

You are absolutely right. It is a poorly designed math question. But when a puzzle has been set up to be hard to solve, it has been well set up, not poorly.

[Segev]

You are absolutely right. It is a poorly designed math question. But when a puzzle has been set up to be hard to solve, it has been well set up, not poorly.

http://xkcd.com/169/

"Difficult to solve" is not the same as "incorrectly specified."

And in this case, as the effort to set up the experiment as a game demonstrated, the incorrect specification actually renders the sought-after answer wrong. Which means that it's as valid as: "What color is a tomato?" "Red!" "No, that's a common misconception. See, I was talking about this picture of a tomato in this coloring book, where my 5-year-old has colored it purple."

Yes, if you change the scenario not to be the one presented, with assumptions not in evidence, the sought-for answer is right. But it's wrong under not just reasonable assumptions, but the only assumptions the person being asked is even capable of making without being a mind-reader (or knowing this specific problem and how it's usually designed).

[SpectralDerp]

{scrubbed}

[Jenerix525]

Hence the word "imply". In my experience you would only use the word "some" to mean "all" if you are deliberately trying to trick or mislead someone, as it is technically true, but not something you would say when there is a much more appropriate word.

In my experience, you I would use the word "some" because there might be dragons I haven't met.
(Or because "all" would make it obvious to every dragon that they have green eyes, and I am too sentimental to guarantee their transformation knowingly. But that falls under deception, so you included that.)

Besides, as an 'infallibly logical' dragon, you are aware that such a word choice is unpopular but possible. Thus, you know it is possible that you have green eyes, and work it out as provided.
You can't choose to ignore the possibility, you're a hyper-intelligent dragon and the conclusion takes less than a milli-second for you. Plus, ignoring the truth is against your draconic nature.

Technically, the problem doesn't state that each dragon knows his peers to be logical. It does state that it is obvious, and leaves you to assume that the dragons are aware of said obvious fact.

The only certain new information is that "every other dragon knows that at least one has green eyes".

Dragon's noses are natural lie detectors, don'cha know.

[BeerMug Paladin]

You are absolutely right. It is a poorly designed math question. But when a puzzle has been set up to be hard to solve, it has been well set up, not poorly.

I think the point of some statistics/logic problems being taught in the first place is to show how easy it is to be misled by intuition. It's a valuable thing to learn, for people dealing with mathematics. Because being that rigorously careful is annoying and difficult. And even if one has a lot of practice with logic, and is quite good at it, it's never a sure thing.

Plus, it's a good demonstration of how easy it is to accidentally make hidden assumptions. That is, if you can convince someone they've actually made that hidden assumption in the first place. (Sometimes it's quite hard to convince people they're absolutely wrong!)

[LooseCannoneer]

People who think the probability is 1/2 are making a mistake. The question isn't worded, "Would you like to keep your door or switch to the UNOPENED door," its, "Would you like to keep your door or switch to ANY door. The not taking of the garbage prize is so obvious that people cut that step.

[Segev]

I think the point of some statistics/logic problems being taught in the first place is to show how easy it is to be misled by intuition. It's a valuable thing to learn, for people dealing with mathematics. Because being that rigorously careful is annoying and difficult. And even if one has a lot of practice with logic, and is quite good at it, it's never a sure thing.

Plus, it's a good demonstration of how easy it is to accidentally make hidden assumptions. That is, if you can convince someone they've actually made that hidden assumption in the first place. (Sometimes it's quite hard to convince people they're absolutely wrong!)

The issue I'm having is that there are hidden assumptions in the asking of the "trick" question.

As evidenced by the attempt to construct a game from it with green and blue coins, you cannot actually reduce it to only the three options:

HH
HT
TH

without also reducing the odds of HT and TH to be half of what that presentation above makes them appear to be. That is:

HH (50%)
HT (25%)
TH (25%)

This is because, unless you specify that you deliberately first constructed the pool of possible families from those wherein at least one is a girl, it is impossible to tell if we don't actually only have a 50% chance of being told "at least one is a girl" vs. "at least one is a boy" in the HT/TH possibilities.

To word them in a way that illustrates, properly, the deceptive nature of intuition, one has to say:

The Smiths are one family in a sample group of 2-child families wherein the oldest child is a girl. What are the odds that both their children are girls?

and

The Joneses are one family in a sample group of 2-child families wherein at least one child is a girl. What are theo dds that both their children are girls?


Phrase it as originally phrased, and the blue-coin/green-coin game problem of having a 50/50 shot at being told about the boy or girl in the mixed cases reduces it again to a 50% chance that both are girls.

[NecroRebel]

Segev, I've looked back at your original two-coins thing, and have noticed a large problem with it that is really, really screwing you up there. Namely, you're considering asking about heads and asking about tails at the same time. You put both "HT asking about heads" and "HT asking about tails" in the same data set, for instance, along with the other 2 possibilities for the two questions, and came up with half being wrong and half being right. The trouble is that those are two different questions, so putting them together is just confusing.

Look back at your post here. You'll note that before you started conflating the two questions, you found that always answering "no" was better for the person who was asked the "at least one" question. If, rather than randomizing the questions asked as you did there, you always asked the same question and simply re-did any flips that didn't fit the question (such as HH on an "at least one tails" scenario), you'd find that answering "no" is correct 2/3 of the time and answering "yes" is correct 1/3 of the time.

And when you get to the part of the post starting "If they always guess 'no...'" you should split the list into two parts, one where you always ask about heads and one where you always ask about tails. As is, in that section, you're adding the 3 possibilities for heads (along with the impossible TT) to the 3 possibilities for tails (along with the impossible HH) and concluding that since 4/8 are correct that it's a 50% chance, when it's really one 2/3 that is correct and another 2/3 that is correct, or total of 4/6 that are correct, for an ~66% chance.

Combining "I flipped these two coins and at least one was heads; what are the chances that both are heads?" and "I flipped these two coins and at least one was tails; what are the chances that both are tails?" is just exactly as valid as "There are twenty mountains in this mountain range; what are the chances one is taller than 10,000 feet?" and "I caught a thousand fish and at least one of them is a tuna; what are the chances all of them are tuna?" You wouldn't add the fish to the list of mountains and say of all of them "this isn't taller than 10,000 feet," because that's just crazy, but from a math standpoint, it's just exactly as crazy to add the HH case on the "at least one was tails" list. They're different questions. Don't conflate them.

[Segev]

Segev, I've looked back at your original two-coins thing, and have noticed a large problem with it that is really, really screwing you up there. Namely, you're considering asking about heads and asking about tails at the same time. You put both "HT asking about heads" and "HT asking about tails" in the same data set, for instance, along with the other 2 possibilities for the two questions, and came up with half being wrong and half being right. The trouble is that those are two different questions, so putting them together is just confusing.

It's not confusing, or at least not "just" so (it may well be confusing). It is, however, going to methodology. Which actually is what deterines the odds, not the information I choose to give you.

Look back at your post here. You'll note that before you started conflating the two questions, you found that always answering "no" was better for the person who was asked the "at least one" question.

True! But it was also always better for the person I asked "the blue one" about.

If, rather than randomizing the questions asked as you did there, you always asked the same question and simply re-did any flips that didn't fit the question (such as HH on an "at least one tails" scenario), you'd find that answering "no" is correct 2/3 of the time and answering "yes" is correct 1/3 of the time.

Also true. But, again, if I do that, the person I tell "the blue coin is heads" to has the same odds of getting it right as the person I tell "at least one is heads."

The information being given has to be about the population from which the sample is taken, not about the sample itself. Because if the information given is about the sample itself, it truly does bring us back to the problem where the question is semi-randomized.

It's the difference between rejecting samples that do not fit the question, and shaping the question to fit the sample.

The problems as presented initially are given in a form such that the one being asked does not know whether this is a truth about the sample population (and thus he would be asked the same question no matter what), or it is a question with information given after the results of THIS sample were known to the question-asker.

Not knowing, he has to assign probabilities to the question-asker's likelihood of asking a particular question and giving particular information. This brings us, as far as the information provided to the one being asked the question goes, back to the "50% chance he'll ask about this and tell me about that" consideration that alters probabilities to 50%.

The question has to be phrased to specify that the information is about the population from which the sample is taken in order for it to have the "textbook" answer as correct.

Combining "I flipped these two coins and at least one was heads; what are the chances that both are heads?" and "I flipped these two coins and at least one was tails; what are the chances that both are tails?" is just exactly as valid as "There are twenty mountains in this mountain range; what are the chances one is taller than 10,000 feet?" and "I caught a thousand fish and at least one of them is a tuna; what are the chances all of them are tuna?" You wouldn't add the fish to the list of mountains and say of all of them "this isn't taller than 10,000 feet," because that's just crazy, but from a math standpoint, it's just exactly as crazy to add the HH case on the "at least one was tails" list. They're different questions. Don't conflate them.

Except that it again goes to methodology of setting up the problem.

And your analogy is highly flawed, because "There are 20 mountains in this range; what are the chances all of them are tuna?" is not something that even makes sense in the context of possible results. Fish are not mountains.

Coins, on the other hand, can be heads or tails.

The two questions you contrast regarding chances of heads or tails are both questions you can arrive at if your methodology is, "I flipped two coins, and provide information about one of them, and then ask a question about something that has a non-zero chance."

You can see this clearly if you try the experiment your way:

Let's say that you reject all results where both coins are tails. You then take one of your results that has at least one heads and ask me the question, "I flipped two coins, one of which came up 'heads;' did both come up 'heads?'" I will always answer "no."

You then ask Pinnacle the question phrased thusly: "I flipped two coins, and the green one came up 'heads;' did both come up 'heads?'" She will also always answer "no," just to give us a solid basis of comparison.

By the postulate of the two questions originally posed about the Smiths and the Joneses, I should have a 2/3 chance of being right, and Pinnacle should have a 1/2 chance of being right. However, since both of us are choosing the same answer, we will be right (and wrong) exactly the same percentage of the time (which, statistically, should be 2/3).

This is because the information you are giving Pinnacle is extraneous and changeable based on the result. You will tell her about the blue coin if the blue coin came up "heads," but the green one didn't. And presuably will pick one arbitrarily (50% chance) if both are Heads.

The result is determined by your methodology, which you did not tell either of us in the question you asked.

Similarly, if you eliminate the possibility of changing any information given to Pinnacle, you will reject, say, all results where the green coin came up "tails."

Then, you will still phrase your question to me thusly: "I have flipped two coins, at least one of which came up 'heads;' did they both come up 'heads?'" and your question to Pinnacle thusly: "I have flipped two coins, and the green one came up 'heads;' did they both come up 'heads?'"

If we again both answer "no," we again will have statistically the same results on repeated experiments. This time, it will be 1/2 chance of being right.

Again, it's because of your methodology, of which you told neither of us anything.

Therefore, the two phrasings, "at least one of which" vs. "the green one," are functionally identical until they are used as selection criteria to form your population from which valid samples may be drawn.

In the questions as initially phrased about the Smiths and Joneses, there is no indication that the sample populations were restricted such that the questions could be asked as they were. To assume that they were is an unstated assumption which is no more correct than to assume that they were not, but rather than the information to be given about the Smiths and Joneses was chosen after they were selected, and thus is information specifically derived from their conditions.

In fact, because of how the questions are phrased, it is more reasonable to assume that the information to be given and the question asked was chosen based on the sample taken. No indication was given that the Smiths and Joneses were selected nor screened for anything. The information given therefore does not actually contain selection criteria about the population.

So far as the person being asked knows, you flipped two coins, looked at the results, then decided what to tell him and what to ask. That is, in fact, what the scenario sounds like, since specific selection criteria being applied is generally stated when problems are formulated.

That's why the problems are badly worded; they have their own unstated assumptions (which are less obvious, colloquial, or common than the unstated assumptions for which those who give the "wrong" answer are chided), without which the "textbook correct" answer isn't.

[NecroRebel]

it was also always better for the person I asked "the blue one" about.[/i]

What? No it wasn't; it was a 50/50 chance for the person you asked "the blue one" about.

Also true. But, again, if I do that, the person I tell "the blue coin is heads" to has the same odds of getting it right as the person I tell "at least one is heads."

That's not what your writeup in the linked post says.

The information being given has to be about the population from which the sample is taken, not about the sample itself.

The only relevant thing is the population from which the sample was taken. The sample is the whole population. You're including things that aren't part of the sample - on a list of coin flips wherein at least one was heads, TT never appears, because there is not at least one heads in it.

It's the difference between rejecting samples that do not fit the question, and shaping the question to fit the sample.

Yes, and meaningful probabilities only come when the samples all fit the question. Mountains are not fish and asking questions about what sort of fish a mountain is is nonsensical. A TT flip is not a flip with at least one heads, and asking questions about what sort of flip with at least one heads a TT flip is is nonsensical.

The question has to be phrased to specify that the information is about the population from which the sample is taken in order for it to have the "textbook" answer as correct.

And it is. At least one of the coins was heads - that tells you what population the sample was taken from: flips wherein at least one of the coins was heads.

And your analogy is highly flawed, because "There are 20 mountains in this range; what are the chances all of them are tuna?" is not something that even makes sense in the context of possible results. Fish are not mountains.

Coins, on the other hand, can be heads or tails.

Yes, that's exactly my point - fish are not mountains, mountains are not fish. Flips wherein at least one flip is heads are not flips wherein at least one flip is tails.

We're not discussing coins. We're discussing pairs of flipped coins with certain properties. There's a difference. The fact that HT and TH flips are both pairs of coins wherein at least one is heads and pairs of coins wherein at least one is tails is irrelevant; we're discussing only one or the other.

If I said that I had a thousand fish, at least one of which was a tuna, and asked about what chance there was 200 of them were tuna, it would not be relevant to say that at least one of those same thousand fish was a shark.

By the postulate of the two questions originally posed about the Smiths and the Joneses, I should have a 2/3 chance of being right, and Pinnacle should have a 1/2 chance of being right. However, since both of us are choosing the same answer, we will be right (and wrong) exactly the same percentage of the time (which, statistically, should be 2/3).

Yes, because both answers are the same. You're running into the trap of the Monty Hall problem - the probabilities of a given answer for a given event are fixed when that event happens. In the classic Monty Hall problem, when you choose the first door, it's probability of being correct is set to 1/3, so when another door is opened and shown to have a 0/3 chance of being correct, your door is at 1/3, the now-open door is at 0/3, and the third door is at 2/3.

You're adding information to the scenario when you guess the first time. Thereafter, the two possibilities are no longer equally likely. This is to be expected - how would it make sense if you, answering "no," had a 2/3 chance of being correct, while someone else giving the same answer had a 1/2 chance of being correct?

If, on the other hand, you did the test and one person guessed with some information and then you did the test again and another person guessed with different information, the probabilities would be as expected - 2/3 and 1/2 of "one heads" being correct. Otherwise, you're no longer considering mathematical probabilities, but rather real-world scenarios which are much messier.

[SpectralDerp]

Let's see how badly worded these problems actually are.

Problem 1: Mr. Jones has two children. The older child is a girl. What is the probability that both children are girls?
Problem 2: Mr. Smith has two children. At least one of them is a boy. What is the probability that both children are boys?

Let's make the following assumptions:
- Any child is either a boy or a girl.
- For any child, the probability of that child being a boy is 1/2.
- The sexes of any two children are independent.

Let's make one more:
- The question "What is the probability that X?" is to be interpreted as "What is the probability that X, given the information you were just given?".

Given these assumptions, the answers to these problems are 1/2 and 1/3 (and their calculations have no need for notions like population and samples).

All of them are unstated, yet everyone figured them out. That's because they are typical conventions people use when discussing probability.

Your examples of coin games involve scenarios where "What is the probability that X?" means "What's the probability that X, given a probability space I am misleading you about by straying away from these basic conventions?". That's not evidence that the original problems are badly worded, the ones in your scenarios are.

[Pinnacle]

Segev, I'm not really sure what you're trying to ask/argue/prove.

If you change the scenario, sample, assumptions, and question, you get different odds.
Well, yeah.

[Segev]

What? No it wasn't; it was a 50/50 chance for the person you asked "the blue one" about.



That's not what your writeup in the linked post says.

Er... the linked post was shown to have an error in it later, which I acknowledged. The end result, however, is this:

You can construct the scenario such that it has either a 1/2 or 1/3 chance of somebody always answering "no" to the question asked being correct. Once you have established the scenario, however, you do not get different odds of correctness based on the information you give the people answering the question.

If I decide what to tell and what to ask after seeing the results of the coin flips, the chances are 1/2.

If I decide on my question but not which coin I will give information about, and reject results which would make the question always "no" given the information, then choose the coin about which I will give information, the chances are 1/3.

If I decide on my question and which coin I will give information about, then reject any results which would always be "no" given the information and coin chosen, the chances are 1/2.

In all three cases, I can tell Alice "at least one coin is [result]; are they both [result]?" and Bob, "the [color] coin is [result]; are they both [result]?"

It will not make Alice be right a different fraction of the time than Bob.

The only relevant thing is the population from which the sample was taken.

Correct.

The sample is the whole population.

That is the most obvious assumption, and it actually comes out to the 1/2 result. This is because, if the sample is the entire population, then it must be the case that you were given this sample before you decided on what information to give and what question to ask.

It, in fact, makes the question part of the sample, which is the point I think I was missing way back when. That is, the possible scenarios, given that you have only this one sample and had it before you decided what information to give and what question to ask, are:

BB "the older is a boy, are both boys?"
BB "the younger is a boy, are both boys?"
BB "at least one is a boy, are both boys?" (You are telling me about the older one)
BB "at least one is a boy, are both boys?" (You are telling me about the younger one)
BG "the older is a boy, are both boys?"
BG "the younger is a girl, are both girls?"
BG "at least one is a boy, are both boys?" (You are telling me about the older one)
BG "at least one is a girl, are both girls?" (You are telling me about the younger one)
GB "the older is a girl, are both girls?"
GB "the younger is a boy, are both boys?"
GB "at least one is a boy, are both boys?" (You are telling me about the older one)
GB "at least one is a girl, are both girls?" (You are telling me about the younger one)
GG "the older is a girl, are both girls?"
GG "the younger is a girl, are both girls?"
GG "at least one is a girl, are both girls?" (You are telling me about the older one)
GG "at least one is a girl, are both girls?" (You are telling me about the younger one)

Now, I do have one more piece of information than the list of possible scenarios: I know which question and which information you gave me.

Let us now examine the two forms the examples took:

"The older child is a girl; are both girls?"

This gives us:

GB "the older is a girl, are both girls?"
GG "the older is a girl, are both girls?"

as the only two possibilities left. They had even odds of being the case before I knew which question and info you would give me. It is, as the textbook answer expects, a 50% chance that I will guess correctly.

"At least one child is a girl; are both girls?"

This gives us:

BG "at least one is a girl, are both girls?" (you are telling me about the younger one)
GB "at least one is a girl, are both girls?" (you are telling me about the older one)
GG "at least one is a girl, are both girls?" (you are telling me about the older one)
GG "at least one is a girl, are both girls?" (you are telling me about the younger one)

Again, each of those possibilities has an even chance. This is because it was only a 50% chance, if it was BG or GB, that you would have even told me about the girl and asked the associated question.

This again gives me a 50% chance of guessing correctly.


To reitterate: this is true because your "one sample" situation means you had zero controls on how you generated your sample. Thus, the true population is about the way you presented me the information and question, since this was generated based on the single sample.

To test this, consider the situation where you flip a blue coin and a green coin. Decide on your question and the information you will give. Give half the people you ask the information about WHICH coin you're telling them about, and don't give it to the other half. You will find 50% of them get it right.

You could also do this by flipping new pairs of coins each time, and generating your question as described in the post you linked; considering again that I'd made a mistake and not considered all cases then (I have, now, in this post), you'll find that you still get 50% right answers.

By not rejecting results - which you cannot do when you have literally one sample that is your population - you have made it a 50/50 proposition

If I said that I had a thousand fish, at least one of which was a tuna, and asked about what chance there was 200 of them were tuna, it would not be relevant to say that at least one of those same thousand fish was a shark.

No, but it would be relevant to know what the odds of a given fish out of all possible fish were tuna, and whether you had looked at your sample of 1000 fish before deciding to a) tell me that at least one is tuna and b) to ask me if all of them were tuna.

If you did, in fact, wait to see what was in there, then my chances are really the chances of any collection of 999 fish having 199 tuna in it.

Just as, if you did, in fact, wait to see what the Jones's two children were before deciding to a) tell me that at least one was a girl, and b) to ask me if both were girls, my chances are really the chances of any collection of 1 children having 1 girl in it.

If, on the other hand, you tell me that you sought out a collection of 1000 fish that had at least 1 tuna in it, you have the case you're looking for.

That is, you'd have to tell me that the Joneses were selected because they have at least one daughter.

This tells me that you would have rejected any samples that had two boys. You already knew what you were looking for, and THAT criterion gives a 2/3 chance of one boy and one girl, with only a 1/3 chance of two girls (and no chance of two boys).

Whereas, without knowing you actively sought "at least one girl" in the pair, and therefore rejected all BB samples, I only know this was a random sample taken from all possible samples (BB, BG, GB, GG), and that you generated the question based on its results, leading to my lengthy discussion earlier in this post.

If, on the other hand, you did the test and one person guessed with some information and then you did the test again and another person guessed with different information, the probabilities would be as expected - 2/3 and 1/2 of "one heads" being correct. Otherwise, you're no longer considering mathematical probabilities, but rather real-world scenarios which are much messier.

That again depends not on the information you give, but on the criteria you have for choosing what information and question to give and for accepting or rejecting your sample.

If you choose your information and question before accepting the sample, and reject samples that don't fit, you will get the results expected for the acceptence criteria you chose. If your acceptence criterion was "at least one heads," then it's a 2/3 chance soembody guessing "no" will be right, even if you tell them WHICH is heads (because there's a 50/50 shot of which coin you're telling him about - green or blue). If your criterion was "the blue coin is heads," then it's a 1/2 chance they'll get it right either way. Even if all you do is tell them "at least one was heads," they still have a 50/50 shot at a right answer because the only coin that could possibly not be heads is the green one.

IF you instead accept any sample, as my lengthy discussion shows, and then choose your information and question, they again have a 50/50 shot. That's the situation of "only one sample," as the question we're discussing asks it about the Joneses' two children. Absent information that the Joneses were specifically accepted because at least one was a girl (and thus would have been rejected and the Johnsons or the Jacksons or the first family that had at least one of their two children being a girl was accepted), it is more likely that you've flipped two coins/found the first two-child family you could, and asked a question and gave information based on what you found the sample to be.

And this post has gotten enormous, so I'll answer the otehrs in at least one separate post.

[Segev]

Let's see how badly worded these problems actually are.

Let's make the following assumptions:
- Any child is either a boy or a girl.
- For any child, the probability of that child being a boy is 1/2.
- The sexes of any two children are independent.

Let's make one more:
- The question "What is the probability that X?" is to be interpreted as "What is the probability that X, given the information you were just given?".

Given these assumptions, the answers to these problems are 1/2 and 1/3 (and their calculations have no need for notions like population and samples).

No, they're not.

Because again, the information I've been given does not include how the Smiths and Joneses were selected as samples. I do not know if you chose your question and information first, and rejected samples that would render the information false, or if you chose the samples and then picked whati nformation about the samples you would give me, and a question that was associated with that information.

(i.e., if it's BG or GB, 50/50 shot that you'll tell me about teh boy or the girl, and the question will ask if they're both the sex of the child you told me about.)

In fact, the base assumption that should be made, given how the questions were asked and the information presented, is that you did, in fact, just pick a 2 child family and formulate the information and associated question based on the facts of that sample.

To get the 1/3 result, you have to tell me not that at least one of the Joneses' two children is a girl, but that you selected the Joneses because at least one of their two children is a girl. i.e. that you have a population of families that consists solely of 2-child families with at least one daughter.

All of them are unstated, yet everyone figured them out. That's because they are typical conventions people use when discussing probability.

Er...yes and no. This is a classic "trick question to show how normal people don't understand probability," so the formulation is such that anybody who knows the textbook answer understands the unstated assumption. The trouble is, it is probable that many don't actually realize the unstated assumption is being made. They've been given the textbook explanation, which leaves the assumption unstated even as it relies on it.

Thus, they will not appreciate the difference between "look at sample; formulate question" and "formulate question; choose sample about which it is a valid thing to ask."

This is exacerbated by the fact that the actual way people generally expect information to be given about a sample is as informative, rather than restrictive, details.

Your examples of coin games involve scenarios where "What is the probability that X?" means "What's the probability that X, given a probability space I am misleading you about by straying away from these basic conventions?". That's not evidence that the original problems are badly worded, the ones in your scenarios are.

Except that the original problems are straying away from the conventiosn of standard information-giving. That is precisely why they are badly worded.

The coin-game uses exactly the base assumptions that are expected by standards of giving information.

To formulate the Joneses' question in a way that is not misleading by straying away from the assumptions, it would have to actually be, "The Joneses are participating in a program for 2-child families with at least one daughter. What is the probability that both of their children are girls?"

Or, alternatively, "I have sought out the Joneses because at least one of their two children is a girl. What is the probability that both of their children are girls?"

Phrased as it was, "The Joneses have 2 children. At least one is a girl. What is the probability that both are girls?" implies that you have first selected the Joneses and THEN started to give information about them. Not that there was a selection criterion in place that required them to match this information.

Segev, I'm not really sure what you're trying to ask/argue/prove.

If you change the scenario, sample, assumptions, and question, you get different odds.
Well, yeah.

What I'm trying to show is that the original question is badly formulated. It implies one set of assumptions based on how people usually report information, and then uses - without stating - a different set of assumptions.

Implied: "The Joneses were selected at random. There were no rejection criteria. We are giving you the following additional pieces of information: They have two children; at least one is a girl. You know, based on the fact that there were no rejection criteria, that if one is a girl and one is a boy, there is only a 50% chance that we would be asking you this question, and a 50% chance that we could be asking you instead if both were boys (and telling you at least one was a boy). What is the probability that both are girls?"

With Actually-Used Assumptions Stated: "The Joneses were selected at random, but would have been rejected if we could not truthfully give you the following information about them: They have two children; at least one is a girl. You know, because we would have rejected any which did not make that statement true, that this information restricted the possible results, rather than being determined by the results. What is the probability that both are girls?"

The answer to the implied question is the same as with the two-coin game wherein I flipped the coins THEN decided what information (and associated question) to give. The answer to the question with the actually-used assumptions stated is the textbook "1/3."

[SpectralDerp]

No, they're not.

Yes they are and every step in the calculation is trivially correct.

Spoiler

Show

Given the fourth assumption, the problem wants us to calculate the value of

P(Both children are boys | At least one child is a boy)

This is a conditional probability. Conditional probabilities are calculated via

P(A|B) = P(A and B) / P(B)

This mean the solution to the problem can be calculated via

P(Both children are boys | At least one child is a boy)
= P(Both children are boys and at least one child is a boy) / P(At least one child is a boy)

First, we calculate P(At least one child is a boy). From the first assumption, we can deduce that

P(At least one child is a boy)
= 1 - P(the first child is a girl and the second child is a girl)

because the two events here are complementary. Given the third assumption, it follows that

P(the first child is a girl and the second child is a girl)
= P(the first child is a girl) * P(the second child is a girl)

and given the second assumption, it follows that

P(the first child is a girl)
= 1/2

as well as

P(the second child is a girl)
= 1/2

Putting all of this together, we get

P(At least one child is a boy)
= 1 - (1/2) * (1/2)
= 3/4

Next, we will calculate P(Both children are boys and at least one child is a boy). First, we will simply rewrite it as

P(Both children are boys and at least one child is a boy)
= P(Both children are boys)
= P(the first child is a boy and the second child is a boy)

From the third assumption, it follows that

P(the first child is a boy and the second child is a boy)
= P(the elder child is a boy) * P(the younger child is a boy)

From the second assumption, it follows that

P(the first child is a boy)
= 1/2

and

P(the second child is a boy)
= 1/2

Putting this together, it follows that

P(the first child is a boy and the second child is a boy)
= (1/2) * (1/2)
= 1/4

Putting in these values gives us

P(Both children are boys | At least one child is a boy)
= (1/4) / (3/4)
= 1/3

Q.E.D.

The first problem is solved in an analogous way.

You need to drop the notions of populations, samples and selection. Probability theory does not need them. These concepts primarily play a role in statistics and are used to make educated guesses about probabilities.

[Segev]

Yes they are and every step in the calculation is trivially correct.

Spoiler

Show

Given the fourth assumption, the problem wants us to calculate the value of

P(Both children are boys | At least one child is a boy)

This is a conditional probability. Conditional probabilities are calculated via

P(A|B) = P(A and B) / P(B)

This mean the solution to the problem can be calculated via

P(Both children are boys | At least one child is a boy)
= P(Both children are boys and at least one child is a boy) / P(At least one child is a boy)

First, we calculate P(At least one child is a boy). From the first assumption, we can deduce that

P(At least one child is a boy)
= 1 - P(the first child is a girl and the second child is a girl)

because the two events here are complementary. Given the third assumption, it follows that

P(the first child is a girl and the second child is a girl)
= P(the first child is a girl) * P(the second child is a girl)

and given the second assumption, it follows that

P(the first child is a girl)
= 1/2

as well as

P(the second child is a girl)
= 1/2

Putting all of this together, we get

P(At least one child is a boy)
= 1 - (1/2) * (1/2)
= 3/4

Next, we will calculate P(Both children are boys and at least one child is a boy). First, we will simply rewrite it as

P(Both children are boys and at least one child is a boy)
= P(Both children are boys)
= P(the first child is a boy and the second child is a boy)

From the third assumption, it follows that

P(the first child is a boy and the second child is a boy)
= P(the elder child is a boy) * P(the younger child is a boy)

From the second assumption, it follows that

P(the first child is a boy)
= 1/2

and

P(the second child is a boy)
= 1/2

Putting this together, it follows that

P(the first child is a boy and the second child is a boy)
= (1/2) * (1/2)
= 1/4

Putting in these values gives us

P(Both children are boys | At least one child is a boy)
= (1/4) / (3/4)
= 1/3

Q.E.D.

The first problem is solved in an analogous way.

You need to drop the notions of populations, samples and selection. Probability theory does not need them. These concepts primarily play a role in statistics and are used to make educated guesses about probabilities.

Alright. Perform the experiment. Since you said we have no populations to worry about, we'll just do it with one pair of coins, one green and one blue. I will flip them and hand you the results.

You decide whether you're going to ask about heads or tails, and pick one of the coins which matches that result to give information about. The question needs to be in yes/no form, not "what is the probability" form.

Now, ask 200 people the question you've chosen to ask. Give every other person you ask the information, "at least one is [side]," and the rest the information, "the [color] one is [side]." Again, you choose which coin to tell them about, just be consistent across all whom you tell.

Count how many of the 100 you gave one kind of information to get it right. Count how many of the other 100 people get it right.

If your hypothesis is correct, you should see the "at least one" group get a 1/3:2/3 split on right/wrong answers (which is right and which is wrong depends, theoretically, on whether the coins match or not), and the "the [color] one" group get it right 1/2 the time (regardless of whether the coins match or not).


If you see a problem in the methdology with which the experiment is set up, please explain. The experiment is designed, here, to simulate you walking up to random people after having gotten the single sample, "the Joneses," and asking them questions about whether their children are the same sex based on the information you give about the sex of at least one/the older child.

[huttj509]

Alright. Perform the experiment. Since you said we have no populations to worry about, we'll just do it with one pair of coins, one green and one blue. I will flip them and hand you the results.

You decide whether you're going to ask about heads or tails, and pick one of the coins which matches that result to give information about. The question needs to be in yes/no form, not "what is the probability" form.

Now, ask 200 people the question you've chosen to ask. Give every other person you ask the information, "at least one is [side]," and the rest the information, "the [color] one is [side]." Again, you choose which coin to tell them about, just be consistent across all whom you tell.

Count how many of the 100 you gave one kind of information to get it right. Count how many of the other 100 people get it right.

If your hypothesis is correct, you should see the "at least one" group get a 1/3:2/3 split on right/wrong answers (which is right and which is wrong depends, theoretically, on whether the coins match or not), and the "the [color] one" group get it right 1/2 the time (regardless of whether the coins match or not).


If you see a problem in the methdology with which the experiment is set up, please explain. The experiment is designed, here, to simulate you walking up to random people after having gotten the single sample, "the Joneses," and asking them questions about whether their children are the same sex based on the information you give about the sex of at least one/the older child.

The difference is in how often you can ask which question.


"At least one is heads, are they both heads" can be asked after 3/4 of the initial flips (HH, HT, TH)
"The Green one is heads, are they both heads" can be asked after half the initial flips (HH, HT)
"The Blue one is heads, are they both heads" can be asked after half the initial flips (HH, TH)

And similar for tails.

So how many questions can you ask for each flip?

HH: (at least H, green H, Blue H) = 3
HT: (at least H, Green H, Blue T, at least T) = 4
TH: (at least H, Blue H, Green T, at least T) = 4
TT: (at least T, Green T, Blue T) = 3

How often do you ask each question, if it's chosen randomly after the initial flip?

At least 1 Heads: (1/3)(1/4) + (1/4)(1/4) + (1/4)(1/4) = 1/12 + 1/16 + 1/16 = 10/48
Green Heads: (1/3)(1/4) + (1/4)(1/4) = 7/48
Blue Heads: 7/48
Green Tails: 7/48
Blue Tails: 7/48
At least 1 Tails: 10/48

Total: 48/48, good, math checks for that part.

So the odds of each question being applicable to be asked are not the same.

[Pinnacle]

There being exactly one case, there is exactly one correct answer.
The amount of times any one answer will be correct is either 0 or all.

The question has nothing to do with how likely people are to give an answer to itself, it's about how likely an answer is to be correct

[obryn]

You don't need any of that for an experiment, Segev.

Get a randomly generated sample of ten thousand pairs of coin flips.

ask, "At least one of these coins came up Heads. What are the odds that both are heads?"

Eliminate all tt results accordingly.

Count HH vs HT vs TH. You will find (obviously) that 1/3 of the remainder are HH.

[Segev]

There being exactly one case, there is exactly one correct answer.
The amount of times any one answer will be correct is either 0 or all.

The question has nothing to do with how likely people are to give an answer to itself, it's about how likely an answer is to be correct

You cannot determine probabilty emperically without repeated experiments. The only way to verify that the math is being applied correctly is through empirical tests. If you find that your empirical tests do not match the hypothesis established by your math, then you have either set up the experiment wrong, or you have the math wrong.

I am arguing that the math is wrong for the case presented as the Joneses' family, because the math presented is for a different problem. One set up differently, under different assumptions than those provided in the problem itself.

You don't need any of that for an experiment, Segev.

Get a randomly generated sample of ten thousand pairs of coin flips.

ask, "At least one of these coins came up Heads. What are the odds that both are heads?"

Eliminate all tt results accordingly.

Count HH vs HT vs TH. You will find (obviously) that 1/3 of the remainder are HH.

Except that that isn't what the "Jones question" presents.

It does not say, "We selected the Jones because we can tell you truthfully that at least one of their 2 children is a girl."

It says (or at least heavily implies by its wording), "Having selected the Joneses, we can tell you that at least one of their two children is a girl."

If you declare that the implication is not present, then you cannot at the same time declare that the implication is present that the first is the case.

Which means that, if you deny the implication, the question literally lacks sufficient information to be answered accurately.

Note: you are running your experiment by pre-determining the question and rejecting runs which generate combinations which are invalid with the question.

The claim I am making is that this order of decision-making - question first, reject families which do not match this question - is not in any way implied by the "Jones question" as presented. Because of this, it is a bad question, and does not illustrate the point it is trying to get across. Particularly if the point involves revealing "common misconceptions about probability."

[Segev]

The difference is in how often you can ask which question.


"At least one is heads, are they both heads" can be asked after 3/4 of the initial flips (HH, HT, TH)
"The Green one is heads, are they both heads" can be asked after half the initial flips (HH, HT)
"The Blue one is heads, are they both heads" can be asked after half the initial flips (HH, TH)

And similar for tails.

So how many questions can you ask for each flip?

HH: (at least H, green H, Blue H) = 3
HT: (at least H, Green H, Blue T, at least T) = 4
TH: (at least H, Blue H, Green T, at least T) = 4
TT: (at least T, Green T, Blue T) = 3

How often do you ask each question, if it's chosen randomly after the initial flip?

At least 1 Heads: (1/3)(1/4) + (1/4)(1/4) + (1/4)(1/4) = 1/12 + 1/16 + 1/16 = 10/48
Green Heads: (1/3)(1/4) + (1/4)(1/4) = 7/48
Blue Heads: 7/48
Green Tails: 7/48
Blue Tails: 7/48
At least 1 Tails: 10/48

Total: 48/48, good, math checks for that part.

So the odds of each question being applicable to be asked are not the same.

All of this is accurate, and part of my point.

But what you're missing, I think, is that I am asserting that the way the "Jones question" is presented, the implication is that the above experiment is the one you've performed, and you've provided me with the information based on the Joneses family composition. That you would have presented the Joneses to me even if you had to tell me "at least one of them is a boy," and that if the children are of different sexes, there was a chance you'd have told me "at least one of them is a boy" rather than what you did tell me.

This is because the question as presented implies that you are telling me something about the Joneses, specifically, rather than telling me something about how you selected the Joneses to ask me about.

Again, if the question were worded such that it made clear that the probability you're asking about is regarding a single sample from a population pre-determined to have the given information be true, it would be fine. As-is, it does not suggest this to be the case. At best, you can say there is not enough information, because you have not been given the information about HOW it was determined what information to give nor what question (both boys/both girls) to ask.

[SpectralDerp]

I am arguing that the math is wrong for the case presented as the Joneses' family, because the math presented is for a different problem.

So, in terms of conditional probability, if the problem doesn't ask us to calculate

P(Both children are boys | At least one child is a boy)

then what are we supposed to calculate instead?