Just walk it up, with 2 kids its 1/3 as we have discussed with 3 kids its 1/7, with 4 its 1/15 ... Generalized its 1/(2^n - 1)

No, Philcore, it's 1 / (n + 1). The reason is that the only possibilities left, given k children of whom (k - 1) are boys, are: all boys (1), or k - 1 boys and 1 girl. This could be GB...B, BGB...B, all the way up to BB...BG. All in all, there are k of these ones (one for each position of the girl). So altogether, it's 1 / (k + 1).

" I have no instinct for combinatorics! "

Me neither! I think very few people do.

GGG - eliminated based on assumption
GGB - eliminated based on assumption
GBG - eliminated based on assumption
GBB
BGG - eliminated based on assumption
BGB
BBG
BBB

So when n= 3, there are 4 possible not 7. So n+1, rather than 2^n-1, with 1 favorable.

Ok semck, I see it now. See, brute force is your friend (when possible). And just to prove the point with n= 4: should be 5 possible, as @smeck said rather than 15 as I did:

GGGG, GGGB, GGBG, GGBB -

Mind you, yours is the correct formula if we only know that there is *A* boy, not n - 1 boys.

sorry mistaken submit.

GGGG, GGGB, GGBG, GGBB - eliminated
GBGG, GBGB, GBBG - eliminated
GBBB - 1
BGGG, BGGB, BGBG - eliminated
BGBB - 2
BBGG - eliminated
BBGB, BBBG, BBBB - 3, 4, 5

5 possible outcomes, 1 favorable.

So semck is correct again, 1/(n+1)

Well done sir. @draug, see how easy it is to admit you were mistaken?

http://en.wikipedia.org/wiki/Boy_or_Girl_paradox#Second_question

Yes, SD, I've read that. It confirms what we've been saying.

The other situation deals with when you see one of the children, or are randomly told either boy or girl.

there is also a nice probability table for you all.

There is, but the probability table is for a different situation, where we are told that "there is at least one boy" or "there is at least one girl" randomly (as appropriate).

That's not the situation here. It's the situation where, for example, you meet Jones on the street with a child.

" From all families with two children, one child is selected at random, and the sex of that child is specified. This would yield an answer of 1/2.[3][4]

Grinstead and Snell argue that the question is ambiguous in much the same way Gardner did.[11]

For example, if you see the children in the garden, you may see a boy. The other child may be hidden behind a tree. In this case, the statement is equivalent to the second (the child that you can see is a boy)" - here we don't know if the boy is the older child or not either, yet the probability is 50/50? the first situation with 1/3 deals with if the family is picked after the prerequisite of the at least one boy, whereas in the question the family clearly comes first before this note, making it a situation where you have a family with two children, a sex is pickeed at random and specified, and ergo 50/50

"The natural assumption is that Mr. Smith selected the child companion at random. If so, as combination BB has twice the probability of either BG or GB of having resulted in the boy walking companion (and combination GG has zero probability, ruling it out), the union of events BG and GB becomes equiprobable with event BB, and so the chance that the other child is also a boy is 1/2."

No. You're wrong. It's true that in the "garden" example, you don't know if the one you saw was older or not, but you do have a way to distinguish the children that's just as good as older/younger: seen/unseen. The point of that table is that in any such case, the odds go up to 50/50.

"Thus, if it is assumed that both children were considered while looking for a boy, the answer to question 2 is 1/3."

That's our scenario, because you're just told there is "at least one boy." The situation you're describing is where you have complete information on one of the children and no information on the other.

Yes, that is explaining why meeting one of them changes things.

But you didn't meet one of them in our hypothetical. Please don't paste the entire article paragraph by paragraph, by the way. It's quite long.

'The paradox arises because the second assumption is somewhat artificial, and when describing the problem in an actual setting things get a bit sticky. Just how do we know that "at least" one is a girl? One description of the problem states that we look into a window, see only one child and it is a girl. Sounds like the same assumption...but...this one is equivalent to "sampling" the distribution (i.e. removing one child from the urn, ascertaining that it is a girl, then replacing). Let's call the statement "the sample is a girl" proposition "g". Now we have:
P(GG|g) = P(g|GG) * P(GG) / P(g) = 1 * 1/4 / 1/2 = 1/2.'

last copy and paste i promise

Right. The point of that is that, if this actually arose in real life, it would almost always be from a sampling event (like seeing one of the children, etc.), and if that were the case, then it would be 50/50 for the other one.

There's a reason I was careful to specify that ALL we knew was the two things I said.

Anyway -- thanks for discussing it, SD. I'm off for tonight. Catch you later.

http://www.sciencenews.org/view/generic/id/60598/description/When_intuition_and_math_probably_look_wrong

later, semck.

This post is quite funny as in a way it has nothing to do with math(s) and everything to do with interpretation of a question.

Let's look at the coin toss example. Somewhere abge had volunteered to conduct the experiment. Suppose he picked a copper coin and a silver coin to conduct his experiment. We all hopefully know he can only get four outcomes

CH SH
CH ST
CT SH
CT ST

When we discount CT ST from the equation we only have three possible outcomes, all of which contain at least one head. Most people on here seem to interprete this as being 1 in 3 probability of the other coin being a head. They are correct in their interpretation.

Some people however see the following possible outcomes.

CH SH
SH CH
CH ST
CT SH

Following this interpretation they argue that there is a 2 in 4 (50%) probability that the 'other' head is also a head. They are correct in their interpretation.

I know that some people will jump up and say ST CH and SH CT should be included in the second list resulting in a 2 in 6 probability. They are of course right if, and only if, they see those outcomes as being separate outcomes.

I think you can all take a bow for getting the right answer. You could take another bow if you can now see that people who got a different answer are also right.

So it comes down to "how did you learn he has at least one son" the problem is that it could have come out in a simple question like "Where you off to?" "Gonna see my son pitch hia first major league game with the Reds!". It establishes nothing beyond him havimg one soon at a minimum but it does imply position: the order in which you discover the sex of the kids. Anytime you put it into real world concrete examples and not groups, you add in order of discovery which is an.identifying characteristic in and of itself.

Ok so I think I found a way in a concrete example for us to not have any identifying info on kid A or B. Perhaps it came out in conversation that Mr. S is just glag his house isn't overrun with teenage girls. Then we aren't dealing with the other child but with both in toto giving us the 1/3 odds. My argument has been all along that the existence of "the other child" and making everything concrete or discrete results in us only determining the sex of one discrete child even if we don't know exactly which one.

http://www.youtube.com/watch?v=qXIdamBEUJE

Draugnar, you are correct in both counts, but that is also what others have been saying all along in this thread.

To put it another way, suppose you have two children, whose names are A and B. One framing of the question is:
"A is a boy. What is the probability that B is also a boy?" This is of course 1/2, and the interpretation of the question Draugnar has been taking. It is not a wrong interpretation, as it is ambiguous whether the statement "one is a boy" is being specific about one of the children or whether at least one (but unspecified) of the children is a boy.

Another framing is:
"At least one of A or B (or both) is a boy, but we do not know which is a boy. What is the probability that both A and B are boys?" As put numerous times, the probability is 1/3 on account of there being 3 options, only one of which has both as a boy. This framing interprets "one of the children" and "the other child" in a general sense rather than being specified to individual children.

Draug, all of this is why we moved pretty quickly to coin examples, where the ambiguities largely don't exist.

So back to post 6, does not thevuse of "other child" make this concrete and distinct by implying order to the children and therefore making the odds 50/50? We know one childbis a boy and the "other child" would then be 50/50. That has been my argument. It may seem to be semantics, but the use of other child provides the order leaving us with BB and BG as the only options.

No, Draug, it does not, because of the explicit fact that ALL we know is that "at least one" is a boy. "The other" is colloquial.

And even if it did, as I said yesterday -- you can wiggle all you want, but I was far too explicit about coins far too many times for you to pretend now that you were living in this ambiguity. I'll you quote for you:

" 'If you know only that (a) two coins were tossed and (b) at least one was heads, there is a 50% chance that both coins were heads.' "

Yes! Yes! Yes! "

And I stand by that statement because the coins are concrete. Now, if you run a series of coin toss pairs, remove all those with TT, then randomly draw a pair, the odds are 1/3 that thr coins will be HH. A single tossed set, once examined even by another observer, changes the odds through the observation.

Are you still arguing this variation of the Monty Hall problem?

"A single tossed set, once examined even by another observer, changes the odds through the observation."

If at least one coin is heads and if you know which coin is heads, you know more about the coin toss than if you simply know that one of the coins was heads but don't know which one. Hence the former has has a 1/2 probability of being double heads, while the latter has only a 1/3. The important fact is that former has more information.

But Draug, taken literally, the statement of yours quoted by Semck83 is of the latter category (1/3). The statement (b) "at least one was heads", says nothing about which coin was heads.

... "The statement (b) "at least one was heads", says nothing about which coin was heads." That is, does not imply the *exact* results of the coin toss was observed. Merely that it is known that at least one coin is heads.