
The question is wrong - peter123
http://paulbuchheit.blogspot.com/2009/01/question-is-wrong.html
======
Eliezer
In the _correct_ version of this story, the mathematician says "I have two
children", and _you_ ask, "Is at least one a boy?", and she answers "Yes".
Then the probability is 1/3 that they are both boys.

See here for why I think this problem confuses people:
<http://www.overcomingbias.com/2008/03/mind-probabilit.html>

~~~
MisterMerkin
If I told someone that in real life and they replied by asking, "Is at least
one a boy?", I would probably back away very, very slowly.

The main point is that given the way Atwood asked the question the correct
answer is 50% which wasn't the answer he intended to discuss and indicates he
himself didn't fully understand the subject he was writing about. It wouldn't
be the first time either.

~~~
lg
Atwood doesn't seem wrong or particularly unclear to me. If you "met someone
who told you they had two children, and one of them is a girl", then
presumably we should imagine this person saying: "I have two children and one
of them is a girl," and not "I have two children, X and Y, and X is a girl."
Obviously in the first case, we don't know if it's X or Y that's the girl, so
the set of possible worlds is [(X-G & Y-G), (X-G & Y-B), (X-B & Y-G)], so we
get the 2/3 answer. But maybe other people don't share my linguistic
intuitions...

~~~
tspiteri
If I met someone who told me "I have two children, and one of them is a girl,"
I would be pretty sure they have one girl and one boy. If they had two girls,
they would have said "I have two children, and both of them are girls." So the
English statement has some implicit information. On the other hand, if I were
given a riddle with that statement, I would know the implicit information may
be intentionally misleading. In this case, as Paul Buchheit says, the
statement does not have enough information.

------
mpk
The author apparently doesn't understand where the confusion in the original
question (or variations) comes from.

One way of phrasing it is asking,

"If you have an unrelated man and woman and they both have two children (one
of which is a boy), where the oldest son of the man is a boy - what are the
odds that they both have two boys?"

(There are any number of variations on this - what are the odds of the man
having two boys? what are the odds of the woman having two boys what are the
odds of the man but not the woman having two boys .. all the same problem, but
stated differently).

This 'loads' the question by implying (superficially incorrect) that there
might be a difference between the chances of a man and a woman having two
boys.

Next up is a bit of probability theory. In the case of the woman, no order is
stated, so the chances of her two children have no connection - the events are
unrelated. The man, however, has as a _first_ child a boy (which eliminates
the possibility of this being a girl).

This is another variation on the Monty Hall problem,
<http://en.wikipedia.org/wiki/Monty_hall_problem>

Read up on some historical background on this here,
<http://www.marilynvossavant.com/articles/gameshow.html>

And as for the overall birthrate of men vs women or the possibilities of
having twins/triplets/etc (and their male/female ratio) ... well, that's
really out of the scope of a fairly trivial question statement such as this.

[edit for non-trivial detail]

~~~
trominos
No, you're wrong. The thing is that there are two kinds of confusion that
arise from Atwood's problem. The first kind of confusion comes from not
understanding how probabilities work, which you discuss. The second kind --
which is what Paul Buchheit is talking about -- comes from noticing that the
statement "I have two children, and one of them is a girl" _can be parsed in
two different ways_. It can be parsed as, "I have two children, and at least
one of them is a girl," or as "I have two children, and the gender of one of
them is #{my_first_child.gender}." Despite what many commenters in this thread
are saying, _these are not the same._

That's the real problem here: the English language is inexact. The words that
Atwood used to describe the scenario actually describe at least two
mathematically distinct scenarios.

The Monty Hall problem suffers from this fact, too, but not as badly --
because both interpretations yield the same conclusion, namely, that you
should switch doors. It's just that under one interpretation, you get a car
1/2 of the time, and under the other you get it 2/3 of the time. Also, under
the interpretation that yields a car 1/2 the time, it's logically implied that
the host is willing to open a door and reveal a car -- which most people use
to rule out that interpretation, if only subconsciously.

~~~
zby
Maybe this is because English is my first language - but I cannot imagine how
"I have two children, and one of them is a girl" could mean: "I have two
children, and the gender of one of them is #{my_first_child.gender}." - if
someone was to confer that meaning I would expect him to say: "I have two
children, and the gender of my first child is girl". But frankly as someone
already pointed out - normally the sentence "I have two children, and one of
them is a girl" would implicitely mean that the other child is a boy.

~~~
randallsquared
It's the difference between the situation where a child is chosen and then the
gender announced, and the situation where a girl is found, and then the gender
announced. If you started out saying "Is one of them a girl?", and _only
continuing if the answer is yes_ , then the probability changes.

------
cduan
Isn't the probability still 2/3 even in the proposed alternative? The fact
that we "arbitrarily announce the gender of one of the children" provides new
information for calculating the probability, which is ignored if you assume
that the original 50/50 distribution is unchanged by the announcement.

In an extreme example, consider this algorithm, analogous to the one in the
article:

1\. Choose a random parent that has exactly two children

2\. Announce the gender of _both_ children

3\. Ask about the odds that the parent has both a boy and a girl

In this situation, it would be ridiculous to say that the odds calculated at
step 3 are still 50/50. Because of step 2, it is either certainly true (100%)
or not true (0%) that the parent had both a boy and a girl. Granted, on
average with repeated trials, half the time the step-3 value will be 100% and
the other half 0%. But still, the step 2 information changes the value at step
3.

~~~
trominos
No.

You've gained no useful information. The person who announces the gender of
one child can say one of two things: "this child is a girl" or "this child is
a boy." However, _these two statements are symmetric_ in their effect on P(one
child is a girl and one is a boy).

There's a huge difference between announcing the gender of one child and
announcing the gender of both children. If the announcer announces both
children's genders, he can say that "both are boys," that "both are girls," or
that "one is a boy and one is a girl." But the last statement is not symmetric
to the first two in how it affects P(one child is a boy and one is a girl).

------
cedsav
Manolis comments on Paul's post summed it up pretty well for me.

What feels counter intuitive is that announcing the gender of one child seems
to increase the chances of the other child being of a different gender, but
it's actually the opposite.

"Family with 2 children" -> chance of having at least one boy: 75% "but I have
at least one girl" -> chance of having at least one boy: 66% "but the first
one is a girl" -> chance of having at least one boy: 50% "but they're both
girls" -> chance of having at least one boy: 0%

------
randallsquared
It's true, as a commenter on this article mentioned, that successive children
are (more-or-less) independent events, but that doesn't mean what that
commenter (or our author) apparently thinks. If you specify that the _older_
child is a girl, then the chances of the younger child being a girl are 1/2.
If you don't specify which child is a girl, only that one is, then it's 1/3.
This and the Monty Hall problem are just examples of the same limited-
information situation.

~~~
paul
Nope. Announcing the gender of one child does not magically alter the gender
of the other child. It's like a print statement in code. If you don't believe
me, try writing some code to simulate this. I will bet you an arbitrary amount
of money that I'm right :)

Here's another way of looking at it: By your logic, if I announce that one of
the children is a girl, then the other child only has a 1/3 chance of also
being a girl. Likewise, if I announce that one of the children is a boy, then
the other child only has a 1/3 chance of being a boy. Therefore, by your
logic, the act of my arbitrarily announcing the gender of one of the children
increases the probability that the other child is of the opposite gender from
1/2 (what it was before I spoke) to 2/3, regardless of whether I said it was a
girl or boy. Hopefully you can see why this is not correct.

~~~
randallsquared
"Announcing the gender of one child does not magically alter the gender of the
other child."

Nor is that what I said. :)

Assuming you meant "chance of being a boy" where you said "chance of being a
girl", I agree that announcing the gender of one of the two children increases
the probability of the other child being the opposite gender from 1/2 to 2/3.

The reason this is so is that some of the original probability was sunk in a
case you've now eliminated: the case where there were two of the gender you
didn't announce.

[Edit: Jeff already wrote some code, and while I haven't bothered to review
it, I assume the lack of outcry about it indicates its correctness. Care to
share where he got that wrong?]

~~~
paul
So if I ask all of my friends who have two children to "tell me the gender of
one of their children", then you think that after they answer the question,
there is a 2/3 chance they have both a boy and a girl? (but before answering
the question the probability was 1/2) Doesn't that seem a little absurd to
you?

If Jeff wrote code that yielded 2/3, then he was implementing my "algorithm 1"
(which has selection), not the second algorithm (which does not do any
selection).

I just updated my post with an explanation that may clear things up for you.

~~~
nostrademons
I thought it might be instructive to implement both algorithms side-by-side in
the same loop:

    
    
        import random
    
        jeff_stats = { 'boy_and_girl': 0, 'two_girls': 0 }
        paul_stats = { 'boy_and_girl': 0, 'other': 0 }
    
        for i in xrange(100000):
            children = [random.randint(0, 1), random.randint(0, 1)]
    
            # Jeff's algorithm
            if children[0] or children[1]:
                print 'I have a girl'
                if children[0] and children[1]:
                    jeff_stats['two_girls'] += 1
                else:
                    jeff_stats['boy_and_girl'] += 1
            else:
                print "Oops, I'm a liar - I have no girls, so don't count me"
    
            # Paul's algorithm
            print 'I have a ' + ('girl' if children[0] else 'boy')
            if children[0] != children[1]:
                paul_stats['boy_and_girl'] += 1
            else:
                paul_stats['other'] += 1
    
        print "Jeff stats: " + str(jeff_stats)
        print "Paul stats: " + str(paul_stats)
    

Output (ignoring the actual printouts for each person):

    
    
        Jeff stats: {'boy_and_girl': 50027, 'two_girls': 24949}
        Paul stats: {'boy_and_girl': 50027, 'other': 49973}
    

I liked tromino's explanation best: it's really an ambiguity in the English
language. Under your 2nd algorithm, 1/4 of parents can't truthfully announce
"I have a girl" - but in ordinary conversation, this doesn't matter, because
they'd just truthfully say "I have a boy" and we don't make a distinction.
It's only when we explicitly filter out parents who don't have a boy that we
change the probabilities.

------
peakok
You are right about the question ("Announcing the gender of one child does not
magically alter the gender of the other child"), but wrong about the
statistics, even in your second formulation the probability is 2/3, not 1/2.
This problem, the way Jeff has formulated it, has nothing to do with the Monty
Hall problem. But still, the end probability is the same : 2/3 (that's the
only common point). Tell me if there is a mistake :

The question is : What are the odds that the person has a boy and a girl, if
we already know that one of the child is a girl ? (If we agree on the
question, then we must agree on the following probabilities).

Possibilities are :

1/ boy / boy

2/ boy / girl

3/ girl / boy

4/ girl / girl

Since one of them is a girl, we must remove possibility number 1. That leaves
us with 3 possiblities, and 2 of them have a boy. Probability : 2/3

(edited for correction, we search the probability of having a boy, not a girl
:p)

~~~
timr
Ordering is irrelevant here. Options 2 and 3 are identical.

~~~
randallsquared
Options 2 and 3 each still have the same probability weight as each of options
1 and 4.

~~~
timr
No, they don't.

Again, ordering is irrelevant in this problem. We want to know _only_ the
probability that there will be a boy/girl pair, not the probability that the
boy/girl pair was born in a particular order.

But the same result can be obtained when taking ordering into account -- the
key observation is that if you subdivide options 2 and 3 to account for
sibling ordering, then you _also_ must subdivide options 1 and 4 to account
for sibling order, resulting in 6 total possibilities, of which 2 are M/F
sibling pairs, 2 are M/M pairs, and 2 are F/F pairs:

1) M/m

2) m/M

3) M/F

4) F/M

5) F/f

6) f/F

Given the knowledge that one sibling is female, you then exclude 2 of the 6
possibilities (the m/M and M/m pairs), to obtain 2/4 = 50% probability that
the pair of siblings is of mixed sex.

The mistake you're making is that you're including ordering on the mixed-sex
pairs, but not including ordering on the same-sex pairs.

~~~
peakok
If you include m and f in the universe of possibilities, then you must add the
following combinations as well :

7) m/f

8) f/m

You are not allowed to skip arbitrarily some of the combinations of your
universe (wich is now [M, F, m, f]). Probability : 2/3 =]

~~~
timr
Don't be silly -- do you think I invented a couple of new sexes by adding
lower-case letters? You're just getting thrown by the notation. I could have
written the options as:

1) M/M

2) M/M

3) M/F

4) F/M

5) F/F

6) F/F

but I thought that was confusing, so I introduced a symbol to more clearly
illustrate the differences between the ordering of the same-sex options.

~~~
peakok
"Somebody is wrong on the internet". I'm wasting my time and this is my last
answer. If you haven't noticed, I only use strict mathematical arguments and I
invite you to do the same if you intend to answer. Pure and clear maths
please, no litteral arguments about the sexes or god knows, this is the only
field where we can verify it.

Lacking elementary probabilities knowledge isn't as dramatic as refusing to
learn it, please teach yourself now since nobody will look at this thread
again.

Here is my last point, and you can ask any teacher of formal logic,
probabilities or maths to verify it :

For universe [M,F], the table of possibilities is :

MM

MF

FM

FF

And that's it. I'm sorry but the table you just made up doesn't exist at all,
please go ask one of your teacher about it. If you still believe you are
right, and you can prove it, you just discovered a new field in mathematics
and probabilities, congratulations.

~~~
timr
Rather than insulting me, take a moment to think about the consequences of
what you're saying: you're arguing that by announcing the sex of one child in
a pair, the probability of the other child's sex being a particular value
changes to 2/3. Does that make sense to you? Really?

Again, this has nothing to do with symbols or notation. There are two sexes,
two symbols: M, F. The undergrad probability 101 mistake you're making is that
options:

MF, FM

take order into account, while options:

MM, FF

do not. This is incorrect. If you take order into account for the mixed-sex
case, you _must_ take order into account for the same-sex cases. MM and FF
encompass _four_ options with ordering, not two.

~~~
peakok
I'm sorry if I sounded offensive, and I indeed was, I was tired when I wrote
my last comment and my words didn't reflect my tought. I'd like to elucidate
this problem once and for all, I really do believe we can both agree on a
conclusion.

I'd like you to notice that your new table of probabilities imply that I have
a chance of 1/6 to guess the gender setup of a family of 2 children. I don't
think that makes sense either to you.

Let's make the experiment a bit clearer :

\- We gather a number of families who have 2 childs.

\- For each family, we announce the gender of one of the child, but we don't
know wich one.

\- We are then asked to guess the sex of the other child.

At this point, you believe that the probability to guess right is 1/2, and
that the 2/3 probability doesn't make any sense. My claim is that you fall in
the Monty Hall problem trap, wich is very counter-intuitive and doesn't seem
to make sense at first.

But here is some clarification of the problem :

\- What we are really asked is to guess the _gender_ setup of the family. So
we need to establish the universe of possible family setups before answering.
What are they ?

Even if we don't care about the order, we must acknowledge that there are 2
childs in the family, so there must be a first child, and a second child.

Setup 1 : both childs are boys : M/M

Setup 2 : both childs are girls : F/F

Setup 3 : the first child is a boy, and second child is a girl. M/F

Setup 4 : the first child is a girl, and the second child is a boy. F/M.

Why order matters in setup 3 and 4 ? Because M does not equal F, while M=M and
F=F. We investigate not the individual itself, but the property of the
individual (in this case, the gender). Therefore, M/F is not equal to F/M, and
in the real world there must be a first and a second child.

If you are asked to write down all the possible setups of a family of two in
the real world, you would write the same table. You'd say that :

Some families have 2 boys = 1 setup

Some families have 2 girls = 1 setup

Some families have one girl and one boy = 2 possible setups (1st one is a girl
OR a boy).

Your argument of F/M = M/F implies that all families have _either_ one of the
2 setups, every first child is a boy, or a girl. But it doesn't work like this
in real life. That is why order matters.

Conclusion : if we agree that there are 4 possible setups in a family of 2
childs, then we have a probability of 1/4 to guess the correct setup of the
family given NO information. But if we are informed of the gender of one of
the child, then one solution of the setup is removed (F/F or M/M), and we have
a chance of 2/3 to guess right IF we chose the opposite gender (see Monty Hall
problem).

And if we are informed of the gender of one specific child (1st one or 2nd
one), then it leaves us with only 2 solutions ! And here, the probability
becomes 1/2.

------
Andi
It's all a question about the correct precodition definition. I think people
who heard about conditional probability know that the correct answer is "2/3".
Most people will just imagine the first precondition "they have two children"
and forget the second one.

What about going in the opposite direction and simply ask "What is the
probability that a mother has a boy and a girl?" I think many people would
answer "1/2", too, simply adding the precodition "she has 2 children" ;)

The origin for this is - I think - an implicit context people are thinking of
without real awareness of it. You think quicker with implicit context.

------
BvS
If Jeff Atwood argument would be right, couldn't it than be applied to
roulette as well?

If you play two games for example, there are 4 different orders of black/red
possible (neglecting 0/00):

b,b

b,r

r,b

r,r

Each has the same possibility of 1/4. If you place 1$ in the first round on
black, your chance to win is 1/2. If it is black (or red), only 3 of the
original 4 option are left and two will lead to red...

~~~
randallsquared
No, you've reintroduced ordering.

~~~
mpk
randallsquared is correct. You are misunderstanding sequences of _related_
events.

Running the roulette wheel twice is a sequence of unrelated events. The
(ideal) roulette wheel is totally random. The results of run #1 are unrelated
to the results of run #2.

However, in this case of the man and woman both having two children, one of
which is a boy, but the man having as an oldest child a boy we are talking
about two different scenarios. The events are suddenly related. If you want it
spelled out, the matrices are,

The childbirth options are this,

BB BG GB GG

However, for the man it's been stated that his oldest child is a boy, so it
becomes,

BB BG XX XX

For the woman (who has as a condition 'at least one boy, but not necessarily
the eldest') it is,

BB BG GB XX

(Where 'B' is Boy, 'G' is Girl and 'X' is a non-option).

------
mixmax
In Jeff's post the question was posed to show how bad humans are at
probabilities, and it got the point across very well. I know that I fell right
into the trap...

This post takes what was meant to be an enlightening and refreshing take on
how bad we as humans are at probabilities and turns it into a numbers-duel.

No the original questions doesn't clearly state the definition of who is
asked, it doesn't take into account that slightly more girls than boys are
born, and it doesn't answer the question about whether this all happened in a
cult where all boys are killed at birth. It was a simple metaphor. And it
worked...

Here is a TED talk about the subject of how we are fooled by probabilities:
[http://www.ted.com/index.php/talks/peter_donnelly_shows_how_...](http://www.ted.com/index.php/talks/peter_donnelly_shows_how_stats_fool_juries.html)

~~~
jerf
Really, a (not _the_ , but _a_ ) problem is that almost everybody is right. We
are certainly fooled by probability. But also, if someone volunteers that one
out of their two children is a girl, then it's effectively-100% that the other
is a boy, _in real life_.

Slightly amending a quote in a similar context from "Principles of Economics,
Explained" (<http://www.youtube.com/watch?v=VVp8UGjECt4>), nobody says "I have
a girl. I have another girl. I have another girl."

One way of interpreting this perennial debate is as proof of how just how
fuzzy English can be. People aren't just arguing about the answer, they argue
about the _question_ too.

~~~
mixmax
_Really, a problem is that almost everybody is right._

Yes - this is why I pointed out that in Jeff's post this riddle was used as a
metaphor. Otherwise we could spend all our time bickering about the correct
spelling of "colour" or other trivialities instead of addressing the
interesting questions.

The interesting question in Jeff's post was the poor human understanding of
probabilities.

