

When intuition and math probably look wrong - 3hoss
http://sciencenews.org/view/generic/id/60598/title/When_intuition_and_math_probably_look_wrong

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RiderOfGiraffes
I know Gary, the person who presented this, and I was there when it happened.
He fully intended this to ignite the argument it has.

Firstly, as presented it is clearly ambiguous. It is intended to be ambiguous,
but in such a way that people who are familiar with the original version will
get suckered into believing that it's well formed.

Secondly, if presented precisely, the answer usually given is either 13/27 or
1/2, depending on which version.

Finally, this is like the Monty Hall problem all over again. There are people
arguing vehemently and without listening at all, demonstrating clearly that
they are excellent at missing the point.

In case you're wondering, here's one statement and answer.

Suppose on knock on people's doors and ask - Do you have exactly two children?
If they answer no, I move on. If they answer yes I then ask - Is at least one
of them a boy born on a Tuesday? If they say no, I move on.

If they look surprised and say "Yes," what is the probability that they have
two boys?

Answer: 13/27.

Yes, it really is.

If you replace the second question with "Is at least one of them a boy with
red hair, left-handed, plays piano, was born on Tuesday, and has a cracked
left upper incisor" then if the answer is "Yes" then the probability of both
children being boys is almost exactly 50%.

If, instead, you replace the second question with "Is at least one a boy" then
the probability of two boys is 1/3.

Finally, suppose you see a parent that you know has two children in the park
with a boy. Now the probability of two boys is 50%, because, assuming uniform
probabilities, having two boys makes it more likely you see them with a boy.

tl;dr: It's hard, and depends precisely on the assumptions you make.

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dkarl
Another riddle where the math is easy, but translating the riddle into math is
hard. Which is to say, it's not really a math riddle at all. In fact, the
riddle is complicated by the fact that it's posed as a riddle -- if you had to
answer the question "in real life," you would probably know how you came about
the information, so the problem would be straightforward.

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nazgulnarsil
I had a math teacher that got mad when I used the word "intuitively". Probably
because most people don't understand that when your intuition doesn't match
the mathematical result you have to hit your intuition with a hammer until it
does.

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bena
Ok, I was about to rage about yet another article going on about the Two
Children problem and getting it wrong by leaving out a whole host of children
(i.e. children in families of more or less than two children). Then it
surprised me by not only acknowledging it, but acknowledging that the the 50%
answer is correct when we are selecting from an arbitrary family (as the
original problem is usually presented).

It then goes on to acknowledge that the 33% answer would be correct if we
specifically choose a two child family that fits the parameters of the problem
beforehand.

It's all about why the information was selected.

~~~
bad_user
> _But since the boy could be either the younger or the older child, the
> analysis is more subtle. Devlin started by listing the children’s sexes in
> the order of their birth_

Personally, when I read that I could spot the error. For lazy people (tl;dr
types) ... order doesn't matter as you're not given any info about that order.

So given 2 children, there are only 3 possibilities ...

    
    
      boy, boy
      girl, girl
      boy, girl
    
     So if you're told one of them is a boy ...
    
      boy, girl
      boy, boy
    

And that's it ... a 50% probability that both children are boys.

~~~
RiderOfGiraffes
Actually, I'm going to stop now: <http://xkcd.com/386/>

It's pretty clear I won't convince you.

Let me leave you with these questions:

If I toss two coins until at least one shows a Head, what's the probability
that both are Heads?

If I roll two dice until at least one shows a 6, what's the probability that
both are 6's?

If I spin two roulette wheels until at least one shows a Red-23, what's the
probability that both are red?

Are you sure?

~~~
bad_user
It's been fun :-)

> _If I toss two coins until at least one shows a Head, what's the probability
> that both are Heads?_

I answered in my last reply ...

    
    
      P(head_a AND head_b | head_b) = 1/3
    

The problem is that head_a and head_b are independent events, and that
P(head_a) = 1 (you made sure of that).

    
    
       P(head_a AND head_b | head_a) = 
       P(head_a AND head_b) / P(head_a) =
       P(head_a AND head_b) =
       P(head_a) = 1/2
    

QED :)

~~~
RiderOfGiraffes
I have no idea what problem you think you're solving, but it's not the one I
posed. I never said a specific coin was heads, I just said at least one of
them was heads.

Given that I make money on these sorts of questions (doing it for real in a
semi-military context where it's important to be right, and you get tested
against reality) I feel pretty confident that I know what I'm talking about.
You're clearly talking about something completely different, and I really
don't understand what you're saying.

And that's the last I'll say.

~~~
bad_user
Look, don't take this the wrong way ... probabilities are easy to get wrong
... I'm only having this conversation with you because my math skills are
rusty.

> _I never said a specific coin was heads_

OK, so let's make it mathematically correct (let's say we're painting them)
... making an effort here :)

    
    
      P( blue_head AND red_head | blue_head OR red_head)
      P(blue_head OR red_head) = 1  
          you said that you're retrying until this happens
          normally this would be 3/4
    

Making the problem ...

    
    
      P( blue AND red | blue OR red) 
      = P( blue OR red | blue AND red ) * P( blue AND red ) / P(blue OR red)
          (applied bayes)
      = 1 * P(blue) * P(red) / 1
     

But ...

    
    
       P(blue OR red) = P(blue) + P(red) - P(blue) * P(red) = 1 =>
       P(blue) * P(red) = 1 - P(blue) - P(red) = 1 - 1/4 - 1/4 = 1/2
    

Show me the error.

[EDITED] ... modified the stuff as I've totally fucked up the previous version
:) ... as I said, I'm rusty

~~~
tlammens
multiplication and addition are not the same thing 1/4 * 1/4 = 1/16

~~~
bad_user
Oh shit ... I'm tired :)

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dododo
isn't this the same kind of "intuitive" reasoning that fails at simpson's
paradox? <http://en.wikipedia.org/wiki/Simpsons_paradox>

conditioning upon more events can lead to a higher probability. p(boy = 2 |
boy >= 1) < p(boy = 2 | boy >= 1, tuesday) (or more precisely, conditioning
upon more events can yield a distribution with less entropy)

~~~
bad_user
You know, the problem with that conditional probability is that the sex of the
second child is in no way conditioned by the sex of the first child, so ...

    
    
      p(boy = 2 | boy >= 1) = p(any child = boy)
    

And this was the original problem that led them to the 33% probability.

~~~
dododo
in the sample space i was intending, that would not be the case. i was
imagining boy as a random variable that counts instances in an order tuple of
genders (the underlying sample space).

you're right, without this explicit construction, it's problematic.

------
seanMeverett
This is why I'm glad I was a mathematics major :)

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Daniel_Newby
The answer given is not even wrong.

The statement "I have two children, one of whom is a son born on a Tuesday" is
semantically ambiguous. It can mean (1) "I have two children, and the quantity
of them who are males born on a Tuesday is exactly one", (2) "I have two
children, at least one of whom is a male born on a Tuesday", or even (3) "I
have two children, and the maximum quantity of males born on the same Tuesday
is one".

~~~
shasta
I agree with you that it's not even wrong, but I disagree with your reasoning.
I think (2) is the natural intended meaning of the question.

The problem I have is attempting to assign a probability to something that
isn't reasonably known to be based on randomness. Did the questioner arrive at
this statement by picking an arbitrary child and then declaring his gender and
day of birth? Or did he pick his favorite child and declare his gender and day
of birth? Would he have used this same question if they had both been born on
Tuesday, or would have have picked a different distinguishing feature?

