When intuition and math probably look wrong

When intuition and math probably look wrong(sciencenews.org)

25 points by 3hoss 16 years ago | 30 comments

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.

dkarl 16 years ago |

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.

nazgulnarsil 16 years ago |

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.

bena 16 years ago |

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 16 years ago | |

> 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 16 years ago | | |

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?

RiderOfGiraffes 16 years ago | | |

You can't mean that. There are four possibilities:

   Older child boy, younger child boy
   Older child boy, younger child girl
   Older child girl, younger child boy
   Older child girl, younger child girl

Of those only one is precluded by saying (at least) one child is a boy.

dododo 16 years ago |

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 16 years ago | |

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 16 years ago | | |

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 16 years ago |

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

Daniel_Newby 16 years ago |

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 16 years ago | |

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?

#!/usr/bin/python import random def toss(): if random.random() <= 0.5: return 'Head' return 'tail' trials = 0 two_heads = 0 for trial in xrange(10000): coin0 = toss() coin1 = toss() if coin0 != 'Head' and coin1 != 'Head': continue trials += 1 if coin0 == 'Head' and coin1 == 'Head': two_heads += 1 print trials,float(two_heads)/trials