Probability riddle (Continued)

Fremen

...nah, just kidding

tireoghain01

So does anybody want to continue?

JSK 252

Fremen wrote: »

...nah, just kidding

Grow up.

im_invisible

tireoghain01 wrote: »

So does anybody want to continue?

first post, huh? welcome to boards,,,

tireoghain01

im_invisible wrote: »

first post, huh? welcome to boards,,,

Thanks.

Pherekydes

JSK 252 wrote: »

Grow up.

Right you:

I've tossed a coin 7 times and I've got 7 heads. What are my chances of getting 8 in a row?

And relax, fremen (the trollspotter) is only having a joke.

ZorbaTehZ

Slow coach wrote: »

I've tossed a coin 7 times and I've got 7 heads. What are my chances of getting 8 in a row?

50% obviously.

M.T. Cranium

If you've got seven heads you're lucky to have any money at all, how do you keep a job? :cool:

Pherekydes

I don't.

Great, zorba.

I'm waiting for thetaxman to say the answer is (1/2)^8 and that all the text books are wrong.

sd123

Jaysus, I'm sorry I ever started that FCUKING thread............. :D

gondorff

Er.. ok.

A bag contains one counter, known to be either white or black. A white counter is put in, the bag shaken, and a counter drawn out, which proves to be white. What is now the chance of drawing a white counter?

Pherekydes

3/4

seamus

Maybe 1/2, but my probability's a little rusty, and that "gameshow" riddle always ****s with me.

gondorff

Try again!

At least you didn't say 1/2.

edit: oops, I was replying to slowcoach.

MathsManiac

A bag contains two white discs and one black disc.
Into the bag are placed all of the people who insist on refusing to believe all of the probabilistic results that are demonstrably true.

The bag is securely tied and repeated bashed with a sledgehammer, wielded by a woman with two children, three of whom are known to be female. It is driven over with a steam-roller, (driven by Monty Hall). Several large weights are then attached to the bag. The bag is then carried out to sea by 367 people, (all of whom were born on different days of the year) and tossed in an extremely biased fashion into the ocean.

What is the probability that peace will finally descend on the rest of us?

gondorff

MathsManiac wrote: »

A bag contains two white discs and one black disc.
Into the bag are placed all of the people who insist on refusing to believe all of the probabilistic results that are demonstrably true.

The bag is securely tied and repeated bashed with a sledgehammer, wielded by a woman with two children, three of whom are known to be female. It is driven over with a steam-roller, (driven by Monty Hall). Several large weights are then attached to the bag. The bag is then carried out to sea by 367 people, (all of whom were born on different days of the year) and tossed in an extremely biased fashion into the ocean.

What is the probability that peace will finally descend on the rest of us?

The same probability as you removing your head from your arse.

Pherekydes

MathsManiac wrote: »

A bag contains two white discs and one black disc.
Into the bag are placed all of the people who insist on refusing to believe all of the probabilistic results that are demonstrably true.

The bag is securely tied and repeated bashed with a sledgehammer, wielded by a woman with two children, three of whom are known to be female. It is driven over with a steam-roller, (driven by Monty Hall). Several large weights are then attached to the bag. The bag is then carried out to sea by 367 people, (all of whom were born on different days of the year) and tossed in an extremely biased fashion into the ocean.

What is the probability that peace will finally descend on the rest of us?

I don't know what your problem is.

gondorff wrote: »

The same probability as you removing your head from your arse.

I don't know what your problem is.

Gimme the answer or I'll ban myself with .01 probability.

MathsManiac

Excuse me for attempting to be mildly amusing.

(I see that it is apparently ok to ban such people for trolling, but not okay to be humourous about how their antics can be somewhat irritating.)

Pherekydes

MathsManiac wrote: »

Excuse me for attempting to be mildly amusing.

I see my attempt at being mildly amusing has fallen flatter than Bertie's accent.

Just give us the answer, someone, before I top meself.

Where's Michael Collins when you need him? Béal na mBláth? Get him back here. He can get assassinated some other time.

The answer is 3/4. If it's not, why is it not? Wait, that doesn't sound good.

Anyway, nobody's getting banned on my watch. Leave no poster behind...

MathsManiac

The person who enjoys counting (who was originally in the bag) and who was known at the beginning to be either Caucasian or of darker complexion (presumably with equal likelihood), is, following the first forcible extraction of a person from the bag and thanks to Mr Bayes, now known to have been, with probability 2/3, Caucasian. And, irrespective of whether it is (s)he or his/her latterly inserted colleague that remains in the bag, this is also the probability that the person who emerges will be of that skin-tone.

Accordingly, I will gladly wager up to two hamburgers to one of yours, that the counter is white.

Needless to say, I'm open to being convinced otherwise.

MathsManiac

gondorff wrote: »

The same probability as you removing your head from your arse.

...and I'm not sure how I've managed to provoke this reaction from you. Perhaps I oughn't to have joked about a certain revenue official and his ilk when he's not around to throw one back, but I'm surprised at your leaping to his defence in such a vitriolic fashion.

(And, for what it's worth, I thought the problem you posed is a great one. It really is tempting to say 1/2!)

gondorff

MathsManiac wrote: »

...and I'm not sure how I've managed to provoke this reaction from you. Perhaps I oughn't to have joked about a certain revenue official and his ilk when he's not around to throw one back, but I'm surprised at your leaping to his defence in such a vitriolic fashion.

(And, for what it's worth, I thought the problem you posed is a great one. It really is tempting to say 1/2!)

Lines crossed. Your answer of 2/3 is bang on the money.

After removing a white counter, the bag will contain either the original (white), or the original (black), or the added counter, which is white.

Whilst only one of these possibles is black, the other two are white.

Pherekydes

gondorff wrote: »

Lines crossed. Your answer of 2/3 is bang on the money.

After removing a white counter, the bag will contain either the original (white), or the original (black), or the added counter, which is white.

Whilst only one of these possibles is black, the other two are white.

But won't the bag contain one of two possible counters?

Either the original or the new white one?

Each has probability 1/2.

Further, the original can be either black or white, with assumed equal likelihood. So, for the original counter, white is 1/4 and black is 1/4.

Add up the 'white' probabilities: 1/2 + 1/4 = 3/4

MathsManiac

Slow coach wrote: »

But won't the bag contain one of two possible counters?

Yes

Slow coach wrote: »

Either the original or the new white one?

Yes

Slow coach wrote: »

Each has probability 1/2.

No!

Having picked and observed a white, the chances that you have picked the second counter are greater than that you picked the first one.

Pherekydes

MathsManiac wrote: »

Having picked and observed a white, the chances that you have picked the second counter are greater than that you picked the first one.

Yep, got it now. Thanks.

Mikel

A serious point on the other thread. I realise the answer is 2/3

Quote:
Originally Posted by thetaxman View Post
How does the fact that she has had a girl increase the chance that she will have a boy for second child.
The Law of averages. You do realise that roughly half the population is female and roughly half is male, right?

So you would expect two children to be one boy and one girl, right?

I don't think this reasoning is true, it contradicts what I've learned.
This is like saying if you toss a coin and it lands on heads the probability of tails next time is increased so that they will even out isn't it?
Apologies if I've taken you up wrong

Pherekydes

Mikel wrote: »

I don't think this reasoning is true, it contradicts what I've learned.
This is like saying if you toss a coin and it lands on heads the probability of tails next time is increased so that they will even out isn't it?
Apologies if I've taken you up wrong

Not quite the same thing. If I was to toss two coins, I would expect 1 head and 1 tail at the outset.

If I told you I had already tossed two coins, then you should expect 1 and 1.

If I told you the first toss was a head, then you should expect a tail 2/3 of the time for the second coin.

If I said I tossed a coin and it was heads, and then said I was going to toss another coin, then the expectation of tails is only a half. This is a subtlety of probability that many people can't get their heads around. I made this mistake earlier in this thread.

MathsManiac

Slow coach wrote: »

If I told you the first toss was a head, then you should expect a tail 2/3 of the time for the second coin.

Possibly what you meant to say here was:
If I told you that at least one was a head, then there's a 2/3 probability that the other is a tail.

(The phrasing you have above is incorrect: If you're told that the first toss was a head, then it's evens that the second is a head, as the second toss is independent of the first. It's the "at least one but we don't know which" bit that moves you away from the 1/2.)

Mikel

I understand the riddle, no problem with that.

If I told you that at least one was a head, then there's a 2/3 probability that both are heads.

Agreed

My point is what you said about the law of averages, that in a 50 50 situation, the first outcome affects the second so that things 'even out'
I may have misundertood what you said or may have misunerstood the law
or you may have been typing quickly when trying to explain it

MathsManiac

Mikel wrote: »

I understand the riddle, no problem with that.

Agreed

My point is what you said about the law of averages, that in a 50 50 situation, the first outcome affects the second so that things 'even out'
I may have misundertood what you said or may have misunerstood the law
or you may have been typing quickly when trying to explain it

Not sure who you're quoting here when you say "what you said about...". If you're quoting "the taxman" then I'm not surprised you're confused.

I didn't contribute to that thread, but observed it. My assessment of it is that the taxman was asserting throughout the debate something that pretty much any statistician or mathematician would refute, and others were doing a good job af explaining the truth, (to no avail!).

MathsManiac

Yikes!

Sorry, I've just edited my second-last post; (I had the 2/3 for the wrong thing).

Sorry for any confusion or irritation caused.

Mikel

MathsManiac wrote: »

Not sure who you're quoting here when you say "what you said about...". If you're quoting "the taxman" then I'm not surprised you're confused.

I didn't contribute to that thread, but observed it. My assessment of it is that the taxman was asserting throughout the debate something that pretty much any statistician or mathematician would refute, and others were doing a good job af explaining the truth, (to no avail!).

post no 22

MathsManiac

Ok, I see where you're coming from. Perhaps the references to "law of averages" wasn't helpful, since it's a term that, in normal use, refers to a commonly held fallacy, (although I don't think that's how it was intended here). But bear in mind that Slow Coach had given a perfectly satisfactory response in post #4, and was perhaps looking for a different tack to convince the disbeliever, and was (I think) appealing to population considerations.

Perhaps better would have been to say that given that we know the following:
- of all the two-children families that exist, roughly 1/2 have a boy and a girl, and
- of all the two-children families in which the elder is a girl, roughly 1/2 have a boy and a girl, and
- of all the two-children families that have at least one girl, roughly 2/3 have a boy and a girl

...then the fact that you've been told that you're dealing with the last situation (a two-children family with at least one girl) changes your assessment of the probabilities from what it would be if you didn't know that.

Pherekydes

MathsManiac wrote: »

Possibly what you meant to say here was:
If I told you that at least one was a head, then there's a 2/3 probability that the other is a tail.

(The phrasing you have above is incorrect: If you're told that the first toss was a head, then it's evens that the second is a head, as the second toss is independent of the first. It's the "at least one but we don't know which" bit that moves you away from the 1/2.)

Yep, that's what I meant.

MathsManiac wrote: »

Ok, I see where you're coming from. Perhaps the references to "law of averages" wasn't helpful, since it's a term that, in normal use, refers to a commonly held fallacy, (although I don't think that's how it was intended here). But bear in mind that Slow Coach had given a perfectly satisfactory response in post #4, and was perhaps looking for a different tack to convince the disbeliever, and was (I think) appealing to population considerations.

Point to note: thetaxman was trolling, pure and simple. Many people gave perfectly satisfactory explanations, but he wouldn't/couldn't be convinced.

Trying different tacks was, in the end, fruitless.

Mikel

fair enough, thought it might be something like that

pH

MathsManiac wrote: »

Possibly what you meant to say here was:
If I told you that at least one was a head, then there's a 2/3 probability that the other is a tail.

I think in this statement is all the confusion about this topic, why even clever people can debate this topic for ages and why, as phrased, Taxman had a valid point on the original thread (as the question was posted).

Now let's be clear here, I'm not disputing that 2/3 is the correct answer when the question has been properly phrased:

http://www.faqs.org/faqs/puzzles/faq/ wrote:

2.3. ==> oldest.girl <== [probability]
If a person has two children, and truthfully answers yes to the question
"Is at least one of your children a girl?", what is the probability that
both children are girls?

The answer is 1/3, assuming that it is equally likely that a child will be
a boy or a girl. Assume that the children are named Pat and Chris: the
three cases are that Pat is a girl and Chris is a boy, Chris is a girl and
Pat is a boy, or both are girls. Since one of those three equally likely
possibilities have two girls, the probability is 1/3.

As phrased above, the answer is definitely 1/3 (or 2/3 that the other child is a boy), it's counter intuitive and most people will get it eventually.

However as phrased in the Original thread :

Random woman meets you and tells you she has two children, at least one of which is female, What is the liklihood of the other child being male?

It's unclear whether she volunteered that she had "at least one female child" or if she was asked. Without this enquiring step (made absolutely clear in the rec.puzzles faq) then either 1/2 or 2/3 are correct answers depending on assumptions you make.

Example A (enquiry)

I throw 2 coins and look at them, you ask me is one a Tail - I say "yes" then the probability that the other one is a head is indeed 2/3.

Example B (discovery)

I throw 2 coins - push a random one towards you and say "Look it's a Tail" (presuming that if it had been a head then I'd have said "Look it's a tail") have I given you any information about the second coin? Presume it's still covered and neither you nor me has seen it, I've just pushed the first coin towards you and shown you it's a tail. I think we all can agree that in this scenario the odds on the 2nd hidden coin are completely independent - they're still 50/50.

That's why this problem can get so confusing, I've seen it phrased as follows also:

"My neighbour has 2 kids, she sends one to my door for some sugar and I see it's a girl, what is the probability that the other child is a boy?"

Would anyone defend an answer of 2/3 for the question as phrased above?

MathsManiac

I understand your point entirely, and I've also seen such riddles poorly phrased (notably Monty Hall) but I don't think that the phrasing in this case was especially poor.

I think it's quite proper for the reader to assume that this:
Random woman meets you and tells you she has two children, at least one of which is female.
can be deemed synonymous with this:
Random woman walks up to you and says: "I have two children, at least one of which is female."

Whether or not she gave that information as a result of being asked is hardly relevant, unless you're going to start making probabilistic judgments about the likelihood of women in the diffferent categories offering such information to complete strangers out of the blue.

To argue that a person can assume that some further information might have been given that we're not being told about is pushing it a bit. Are we to demand that every such riddle explicitly tie down all the things that did not happen? "There weren't two girls shouting 'Mom' from across the street" "She hadn't just bought an Action Man in the shop.", "She wasn't lying"...

Surely one may assume that the information given in the question is a complete statement of the information available?

pH

MathsManiac wrote: »

Whether or not she gave that information as a result of being asked is hardly relevant, unless you're going to start making probabilistic judgments about the likelihood of women in the diffferent categories offering such information to complete strangers out of the blue

I disagree, whether she was asked is entirely relevant, if she wasn't asked then the answer is 1/2, there is no way that this puzzle results in a 2/3 1/3 probability without the enquiry being made explicit.

Again, let's return to my coin example, this time let's image they have a girl on one side and a boy on the other.

I throw two of these 'coins', look at one and tell you it says 'girl', what is the probability that the other coin is 'boy', what is the probability that the other coin is 'girl' ?

First let me ask you do you agree with the answer 50/50 for the above phrasing?

MathsManiac

Of course that's 50:50, but that is not by any means an analogy to the situation described. You have identified a particular coin and told me that it is a head/girl/whatever. There was nothing in the riddle posed to make it analagous to this.

If you looked at both coins, and told me that at least one is a head, then we are in a 2/3 to 1/3 situation. On the other hand, if you identify a particular coin and inform me that it is a head, then it's 50:50 that the other is a head.

It is the precise nature of the information given that is relevant, not whether it was volunteered or sought.

Are you seriously suggesting that in the following phrasing, the answer would be 1/2 ???
Random woman walks up to you and, without being asked anything about her children, says: "I have two children, at least one of whom is female." etc.

pH

MathsManiac wrote: »

Of course that's 50:50, but that is not by any means an analogy to the situation described. You have identified a particular coin and told me that it is a head/girl/whatever. There was nothing in the riddle posed to make it analagous to this.

If you looked at both coins, and told me that at least one is a head, then we are in a 2/3 to 1/3 situation. On the other hand, if you identify a particular coin and inform me that it is a head, then it's 50:50 that the other is a head.

It is the precise nature of the information given that is relevant, not whether it was volunteered or sought.

Are you seriously suggesting that in the following phrasing, the answer would be 1/2 ???
Random woman walks up to you and, without being asked anything about her children, says: "I have two children, at least one of whom is female." etc.

Yes it's exactly the same, she in effect has told you the sex of one of her children, nothing more. This is a *completely* different situation from the one where she has been asked if she has at least one girl.

Without an explicit question the answer is 1/2, I'm amazed that people can't see this.

Let's walk through a random person walking up to you in the street.

The set of 2 child families (presuming equal boy/girl chance) is
GG, GB, GB, BB

So let's say a random person walks up to you (from a 2 child family) - no selection has gone on here at all.

Obviously 1/2 of them will say "I have at least one girl" and the other half will say "I have at least one boy" Remember they're making a random statement - we haven't asked them to tell us about boys or girls, they've just volunteered this information to us.

There is no reason to believe they'll prefer the girl phrasing that is used in your question - this is the crux of the issue.

If they've been asked the "at least one girl" question then the sets GG, GB and GB are in play - hence the 2/3 question.

However presuming they're volunteering the information:

GG - Will always say "I have at least one girl" - they have no other option
BB - Will always say "I have at least one boy" - they have no other option
GB,BG - should split randomly 50/50 - there's no reason to think these people would prefer the girl phrase to the boy.

Hence - in the *volunteered* scenario - if you hear "at least one girl" you're dealing with the set GG + 50% of (GB, BG) (if you get my meaning)

Therefore in a volunteered information scenario - the probability is 50/50

MathsManiac

But you have now introduced a new element into the question that was neither stated nor implied before: an assertion that the woman will select a statement to make to you at random from a set of comparable statements that she might make. In essence, you are asserting that some people in the sample space are more likely to volunteer to you the information that you have received than others.

You are making an assumption about the motivation of the woman in making the statement.

(ref my earlier caveat: "unless you're going to start making probabilistic judgments about the likelihood of women in the diffferent categories offering such information to complete strangers out of the blue.")

pH

You are making an assumption about the motivation of the woman in making the statement.

Sorry, it's not just me, but also you making an assumption about the motivation of the woman making the statement. Let me explain:

Random woman meets you and tells you she has two children, at least one of which is female, What is the liklihood of the other child being male?

Imagine a room with 100 women with 2 children per woman, and let's say they cover the set space perfectly. 25 have GG, 25 have BB and 50 have GB

Let's quickly cover the question version of this problem:

You ask those with "at least one girl" to raise their hands, you should see 75 hands - and of those 75 women, 25 have 2 girls and 50 have a boy and girl, giving us the 1/3, 2/3 probability.

But a random woman from that room coming up to you and making the statement from the OP (in bold above)?

The only way you can achieve the 2/3 1/3 probability split is if to assume that all the women with a girl and boy will also make that statement - you can't just accuse me of making assumptions, because you also make that assumption.

The only difference is my assumption is reasonable (a woman with a boy and girl would have an equal chance of saying "I have at least one boy" as "I have at least one girl").

So if all those women were to make a random statement, you'd expect 50 of them to say "I have at least one girl" and 50 to say "I have at least one boy", and in the 50 who said "one girl" you'd find 25 with a boy and 25 with a girl.

The only way to achieve 2/3 & 1/3 in this "random" statement scenarios is to *assume* that all those with a boy and girl will say "one girl" hence you'd have 75 who said "at least one girl" and 25 who said "at least one boy", leading us to the unlikely conclusion that : if a random woman meets you and tells you she has two children, at least one of which is male, the liklihood of the other child being male is 100%!

There's no way you can achieve any probability or likelihood without looking at the probability she made the original statement.

I'm not trolling, I've been through this before, many times, the rec.puzzles faq has the questioning step because it's absolutely essential, without it the puzzle is either meaningless or has an answer of 50/50.

gondorff

pH wrote: »

"My neighbour has 2 kids, she sends one to my door for some sugar and I see it's a girl, what is the probability that the other child is a boy?"

Would anyone defend an answer of 2/3 for the question as phrased above?

If you know she has two kids, then you are already pre-informed.

A girl turns up at your door for sugar.
There are only three combinations of kids that your neighbour could have:

GB
BG
GG

In two out of three of these, the girl's sibling is a boy.

If you don't know if the girl has a sibling or not, then you can speculate as to whether she is an only child or part of a large brood and would have to calculate probabilities accordingly before you even begin to wonder about how many are boys/girls.

Good luck with that!

pH

gondorff wrote: »

If you know she has two kids, then you are already pre-informed.

A girl turns up at your door for sugar.
There are only three combinations of kids that your neighbour could have:

GB
BG
GG

In two out of three of these, the girl's sibling is a boy.

If you don't know if the girl has a sibling or not, then you can speculate as to whether she is an only child or part of a large brood and would have to calculate probabilities accordingly before you even begin to wonder about how many are boys/girls.

Good luck with that!

I'm afraid this is incorrect.

In this case you know the sex of the child at your door, the sex of the other child is completely independent.

For example I throw 2 coins and show you one, you see it's a head, what's the probability the other is a head?

If you think that seeing one coin affects the probability of the other being a head or a tail then you're mistaken.

If you think that seeing the sex of one child affects probability of the sex of another child then once again you are mistaken.

pH

In case anyone thinks I'm trolling - here are some links:

Query sensitivity

Consider this example:

My neighbor has two children. Assuming that the gender of a child is like a coin flip, it is most likely, a priori, that my neighbor has one boy and one girl, with probability 1/2. The other possibilities---two boys or two girls---have probabilities 1/4 and 1/4.

Suppose I ask him whether he has any boys, and he says yes. What is the probability that one child is a girl? By the above reasoning, it is twice as likely for him to have one boy and one girl than two boys, so the odds are 2:1 which means the probability is 2/3. Bayes' rule will give the same result.

Suppose instead that I happen to see one of his children run by, and it is a boy. What is the probability that the other child is a girl? Observing the outcome of one coin has no affect on the other, so the answer should be 1/2. In fact that is what Bayes' rule says in this case. If you don't believe this, draw a tree describing the possible states of the world and the possible observations, along with the probabilities of each leaf. Condition on the event observed by setting all contradictory leaf probabilities to zero and renormalizing the nonzero leaves. The two cases have two different trees and thus two different answers.

This seems like a paradox because it seems that in both cases we could condition on the fact that "at least one child is a boy." But that is not correct; you must condition on the event actually observed, not its logical implications. In the first case, the event was "He said yes to my question." In the second case, the event was "One child appeared in front of me." The generating distribution is different for the two events. Probabilities reflect the number of possible ways an event can happen, like the number of roads to a town. Logical implications are further down the road and may be reached in more ways, through different towns. The different number of ways changes the probability.

This property of probability theory, which is different from logic, is discussed at length by Pearl (p58). In logic, it does not matter how a proposition was arrived at. But in probability, the query cannot be ignored.

http://research.microsoft.com/~minka/papers/nuances.html


Also explained here:
http://www.mathpages.com/home/kmath036.htm

And here's a quote from Scientific American (1950s) which talks about the ambiguity in the statement "at least one boy"

Another example of ambiguity arising from a failure to specify the randomizing procedure appeared in this department last May. Readers were told that Mr. Smith had two children, at least one of whom was a boy, and were asked to calculate the probability that both were boys. Many readers correctly pointed out that the answer depends on the procedure by which the information "at least one is a boy" is obtained. If from all families with two children, at least one of whom is a boy, a family is chosen at random, then the answer is 1/3. But there is another procedure that leads to exactly the same statement of the problem. From families with two children, one family is selected at random. If both children are boys, the informant says "at least one is a boy." If both are girls, he says "at least one is a girl." And if both sexes are represented, he picks a child at random and says "at least one is a ..." naming the child picked. When this procedure is followed, the probability that both children are of the same sex is clearly 1/2. (This is easy to see because the informant makes a statement in each of the four cases -- BB, BG, GB, GG -- and in half of these case both children are of the same sex.) That the best of mathematicians can overlook such ambiguities is indicated by the fact that this problem, in unanswerable form, appeared in one of the best of recent college textbooks on modern mathematics.

http://www.wiskit.com/marilyn/boys.html

I just wanted to point out that it's not just me, which is why this 'puzzle' has 3 nested layers and causes so much debate!

There is an obvious 'wrong' "50/50" which most people get to first

Using logic and a simple set approach (GG,GB, BB, BG) most people can then be convinced the answer is indeed the counter intuitive 2/3 & 1/3

However unless the the question is extremely tightly specified (involving an explicit selection step) then the correct probabilistic answer is either "undefined" or 50/50, it is not possible to defend 2/3 1/3 using probability theory unless the puzzle is presented with an explicit selection step.

MathsManiac

I would hope that nobody was about to accuse anyone on this thread of trolling, and I for one never took you as a mad lone ranger!

I am certainly prepared to accept your assertion that the problem was, strictly speaking, underspecified, and that you can therefore argue for the "undefined" answer that you mention in your most recent post. You have certainly demonstrated another randomising procedure that is consistent with the problem statement and which leads to a different answer.

I would still stand by my assertion that the original phrasing by the OP was "not especially poor".

I base that statement on a belief that it is not unusual for people posing such riddles to operate on the assumption that, (in the language of the Minka article,) the maximum entropy principle applies.

To assert that the answer is unambiguously 1/2, or that 1/2 is a more reasonable answer, does require one to make assumptions about which kinds of statement that a random person approaching you in the street might make. For example, in arguing for 1/2 in the way you described, one is asserting that a person with one boy and one girl is equally likely to approach a stranger in the street and say: "...at least one of whom is a boy" as they are to say "...at least one of whom is a girl", and yet one is not considering whether a person with two girls might have said, for example, "I have two children, and both of them are girls" or "Excuse me, do you know where the nearest tube station is?".

I would contend that such considerations lead to a quagmire of possibilities of varing degrees of unfothomability, and, in the absence of any information about the randomising procedure in these theoretically defined situations, any assumption other than the maximum entropy one is ultimately futile.

If the problem specification gives no information about the randomising procedure, then one can only move away from maximum entropy by starting to think that this describes an actually plausible real-life scenario and then attempting to analyse it as such. I believe that this is ludicrous, (since, for example, the probability that a randomly selected person in the relevant sample space would actually do this is so small as to be dwarfed by the probability that it is actually a statistics student up to no good, who probably doesn't have any children at all.)

Accordingly, the problem is clearly intended to be theoretical, and the person posing it is, I believe, not unreasonable in expecting you to assume that "random person" means a person randomly selected (with uniform distribution) from the set of people satisfying the criterion.

It is in this sense that I argued that specifying that a question was asked does not add information.

However, in light of the points you raise, I will agree that the version with the question can be deemed "better", since you have demonstrated the existence of well-informed and thoughtful people who would not accept the maximum entropy principle should be assumed in this case, and the version that specifies the randomising process brings them into line with the questioner's (in my view undoubted) intentions.

gondorff

Take a look at this: http://www.bbc.co.uk/dna/h2g2/A19142246