well known proba problem

[Bill Spight]

It seems to me that the question/answer is confused by the fact that you/people seem to care which of the two stones in the BB box has been picked, as if they were different.

Actually, the fact that the stones are distinct matters. Probability with fermions, which are distinct, and bosons, which are not, is different. Coins and go stones are distinct.

[EdLee]

Actually, the fact that the stones are distinct matters. Probability with fermions, which are distinct, and bosons, which are not, is different. Coins and go stones are distinct.

Bill, for this question, clearly we must assume the stones and coins OF THE SAME COLOR are like bosons, completely indistinguishable from one another.

[Bill Spight]

But we have not used all the information that we got from drawing a Black stone from a box.

Bill, exactly -- this is also my question (confusion): P = X/Y. In the original wording of " (3) What is the probability of...? ",
I want to know what is X and what is Y for that wording of (3) ?

( Because, from the perspective of the little green alien, X ~= half a million, and Y = 1 million exactly; so for her P ~= 50%. )

This gets into what is called pragmatics. Humans are good communicators, and one reason for that is that we do not state everything in each sentence. The pragmatic meaning of "probability" in (3) is the "probability, given (0), (1), and (2)".

That is why the little green alien misinterprets (3). It was not meant for her. She is unaware of the context.

[Bill Spight]

Actually, the fact that the stones are distinct matters. Probability with fermions, which are distinct, and bosons, which are not, is different. Coins and go stones are distinct.

Bill, for this question, clearly we must assume the stones and coins OF THE SAME COLOR are like bosons, completely indistinguishable from one another.

I see what you mean. Suppose that one of the Black stones is chipped, and we do not know in which box it is. Then if the two boxes contain different number of stones, and we pick the chipped stone, the probability of each box is proportional to the number of Black stones in the box.

[EdLee]

The pragmatic meaning of "probability" in (3) is the "probability, given (0), (1), and (2)".

Yes, I see; this is key:

If we start counting the 1 million from after (2) -- her perspective -- X will be ~= half a million.
If we start counting the 1 million at (0) -- our perspective -- X will not be ~= half a million. (TJ's simulation).

I now rephrase the question, completely removing "pragmatics" (for my own sanity and clarity):

(0) 3 bowls: ( ), ( ), and ( ).
(0a) We start our counter X = 0.
(1) Suppose you randomly pick a bowl, then take one of the 2 stones out from the bowl.
(2) This stone turns out to be .
(3a) IF second stone in the bowl is also , we increment X.
(4) Go back to (1) -- NOT (0) -- repeat this 1 million times. What is the approx. value of X after 1 million iterations ?

Now I see it clearly -- (2) SUBSTANTIALLY lowers X from ~= half a million.

(I still think the original wording is not as good as in the Monty Hall problem --
there, they ask "should the contestant switch" instead of "what is the probability of..." ?
I think this way of asking "should the contestant switch" is less ambiguous than the "pragmatics" of "what is the probability of...". )

( Sanity check for myself: if WE do the experiment 1 million times, the little green alien
will not be invited approx. 333,333 times. This is the part SHE does not know.
Conversely, for HER to be invited 1 million times, WE would have to repeat 1.5 million times. )

[Bill Spight]

OK. Here is an similar problem. Suppose, first, that the number of boys born is equal to the number of girls born. Suppose also that you know that I have two children, but you do not know their genders. One afternoon you (randomly) run into my wife, along with one of my children, who is a girl. What is the probability that my other child is also a girl?

Edit: To be clear, my wife is equally likely to be with one child as the other.

[HermanHiddema]

OK. Here is an similar problem. Suppose, first, that the number of boys born is equal to the number of girls born. Suppose also that you know that I have two children, but you do not know their genders. One afternoon you (randomly) run into my wife, along with one of my children, who is a girl. What is the probability that my other child is also a girl?

1/3

There are four options:

Oldest = Girl, Youngest = Girl
Oldest = Girl, Youngest = Boy
Oldest = Boy, Youngest = Girl
Oldest = Boy, Youngest = Boy

Since I have observed you with one girl, I can eliminate the fourth option. Of the three remaining options, the other child is a boy in two cases, a girl in one case, hence there is a 1/3 chance the other child is a girl.

[perceval]

OK. Here is an similar problem. Suppose, first, that the number of boys born is equal to the number of girls born. Suppose also that you know that I have two children, but you do not know their genders. One afternoon you (randomly) run into my wife, along with one of my children, who is a girl. What is the probability that my other child is also a girl?

Edit: To be clear, my wife is equally likely to be with one child as the other.

Ehee i was going to ask that one next, that one drives me crazy . Its the same, except with children and family we instinctively rebel against the idea of retrying until we get the one we want... to be more clear : to erase the ambiguity in the bowl + stones case, , we explain with an algorithm to pick the stones, and state that we only accept to count a subset of the experiments: the one where indeed we drew a Black stone.
With your 2 children we refuse to make this experience of thought as you have 2 children and you are stuck with them

i am still not clear but that one gives me headaches

[TheBigH]

OK. Here is an similar problem. Suppose, first, that the number of boys born is equal to the number of girls born. Suppose also that you know that I have two children, but you do not know their genders. One afternoon you (randomly) run into my wife, along with one of my children, who is a girl. What is the probability that my other child is also a girl?

1/3

There are four options:

Oldest = Girl, Youngest = Girl
Oldest = Girl, Youngest = Boy
Oldest = Boy, Youngest = Girl
Oldest = Boy, Youngest = Boy

Since I have observed you with one girl, I can eliminate the fourth option. Of the three remaining options, the other child is a boy in two cases, a girl in one case, hence there is a 1/3 chance the other child is a girl.

There are indeed three remaining options, but the first of the three is twice as likely as the other two. The probability is therefore 50/50.

edit: ...actually, looking at it again I think a case can be made for 60%. But I still think it's 50%.

edit 2: The 60% argument goes as follows. Think about it this way. Let's say there's four families: one with two girls, another with two boys, and two families with one of each. You've met one of the girls, so the all-boy family is eliminated. So there's six kids left altogether, four girls and two boys. You're talking to one of them, a girl. So the "missing" five must be three girls and two boys.

[perceval]

there is a whole page on it on wikipedia (spoiler, obviously):

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

and after rereading it i would tend to say .... 0.5 given your phrasing:

following your phrasing, and the wiki page, i think that we are in the case when we observe 1 Girl when sampling the distribution.
so P(GG|g) = P(g|GG) * P(GG) / P(g) = 1 * 1/4 / 1/2 = 1/2.
here P(g)= probability of seeing a girl when shown the child of soemone in hte street, 1/2 by symetry.

it is different of the proposition h:"at least one is a girl" which have proba 3/4 (true for GG,GB,BG, false for BB)and would give:
P(GG|h)= P(h|GG) * P(GG) / P(h) = 1 * 1/4 / 3/4 = 1/3.
(that is just copied from wikipedia but i **think** i agree)

[mitsun]

OK. Here is an similar problem. Suppose, first, that the number of boys born is equal to the number of girls born. Suppose also that you know that I have two children, but you do not know their genders. One afternoon you (randomly) run into my wife, along with one of my children, who is a girl. What is the probability that my other child is also a girl?

1/3

There are four options:

Oldest = Girl, Youngest = Girl
Oldest = Girl, Youngest = Boy
Oldest = Boy, Youngest = Girl
Oldest = Boy, Youngest = Boy

Since I have observed you with one girl, I can eliminate the fourth option. Of the three remaining options, the other child is a boy in two cases, a girl in one case, hence there is a 1/3 chance the other child is a girl.

If we are going to differentiate the children by age, then we have to include that in the enumeration of possible observations. The following four options fit the observation and are equally likely:

Oldest = Girl, Youngest = Girl, Observed = Oldest
Oldest = Girl, Youngest = Girl, Observed = Youngest
Oldest = Girl, Youngest = Boy, Observed = Oldest
Oldest = Boy, Youngest = Girl, Observed = Youngest

[shapenaji]

How about a variation on the original?

We have 3 boxes,
the first contains an unknown number of black stones
the second contains an unknown number of black and white stones,
the third contains an unknown number of white stones

We pick a box, pull a stone and it's black,

What is the probability that if we pull another stone from the box, that it will also be black?

[Unusedname]

Box 1:
Box 2:
Box 3:

If the stone you picked is ____ then the other stone will be _____:

If then
If then
If then
if then
if then
if then

You know you have in your hand a white stone, you do not know whether the white stone is 1, 2, or 3, therefore one of the following is true.

If then
If then
If then


2 in 3 chance to have two white stones.

this is how i would explain it :]

[perceval]

How about a variation on the original?

We have 3 boxes,
the first contains an unknown number of black stones
the second contains an unknown number of black and white stones,
the third contains an unknown number of white stones

We pick a box, pull a stone and it's black,

What is the probability that if we pull another stone from the box, that it will also be black?

well if we do not know anything its hard to compute something.

With N B stones in the all black box, M Black + K white in the B and W box, and L W sotnes in the W bowl -then we can compute something:

proba to pick a B stone in the all B box is 1/3N in the B+W bo is 1/3(K+M).

Then when we know we indeed had a B stone: proba to be in the all B box is
N*(1/3N)/(N/3N+K/3(K+M))=1/[1+K/K+M) = (K+M)/(2K+M)
P (2nd box)= K/(2K+M)

So P (2nd is B is:

p=(K+M)/(2K+M)+ K.(K-1)/[(2K+M).(K-1+M)]

(first: proba of box1 * 1, second term: proba of box 2 * K-1/ (K-1+K) ie ratio of remaining B stones in seconf bowl)
.... maybe


sanity check: first problem as N=2,K=1=M
p=2/3+0 =>OK

fun fact: its independant of N AND L (if N>1 of course)

might be wrong too...

[EdLee]

Bill,

Suppose, first, that the number of boys born is equal to the number of girls born.

Why Beautiful People Have More Daughters