[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

[RiderOfGiraffes]

Your first error is here:

  P(BH or RH) = 1

This is supposed to be the probability without conditionals, in which case P(BH or RH) = 3/4. Using the Bayesian formula is what is taking into account the fact that at least one head has shown.

Your second error is here:

    P( blue_head ) * P( red_head ) (independent events)
    = 1/4 * 1/4

That's probably a typo, because you probably know that P(BH)=1/2.

Your third error is here:

    1/4 * 1/4 = 1/2

1/4 * 1/4 = 1/16

Even if you correct this to the 1/2 * 1/2 that you intended, you now get that P(BH and RH)=1/4. Surely you realise that this must be wrong. I've already told you that there's at least one head, so you can't get the same answer as when I've given you no information.

Here's the correct Bayesian sum:

    P(RH and BH | RH or BH)
  = P(RH and BH and (RH or BH)) / P(RH or BH)
  = P(RH and BH) / (3/4)
  = P(RH) * P(BH) / (3/4)
  = 1/2 * 1/2 / (3/4)
  = 1/4 / (3/4)
  = 1/3

The fact that I'm using the conditional probability is what uses the information I've given you, so you can't use that again and claim P(RH or BH)=1.

Show me the error.

Here, do an experiment. Toss two coins of different denominations. Consider those cases where there's at least one head. How often are they both heads? Are you a programmer? Run the simulation - show me the code.

Here, let me help:

  #!/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

[RiderOfGiraffes]

Now replying to the modified version. My response to your first version is here: http://news.ycombinator.com/item?id=1473088

Anyway ...

Don't take this the wrong way ... probabilities are easy to get wrong ... I'm only having the conversation with you because you're getting it completely wrong, and yet you appear to want to learn.

You say:

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

This is an incorrect application of Bayes' Theorem. The whole point of taking P(X|Y) isn't that P(Y) can be then taken as being 1. P(Y) remains the probability of Y. The point is that when we then only consider those events where Y occurs, then we have conditioned our probabilities, and the formula gives us what we want. Specifically, relative to Y, the probability of X changes.

Using "BH" for "Blue coin shows Head" and "RH" for "Red coin shows Head" we have:

    P(BH and RH | BH or RH)
  = P(BH and RH and (BH or RH)) / P(BH or RH)
  = P(BH and RH) / P(BH or RH)
  = 0.25 / 0.75
  = 1/3

You then go on to say:

  P(BH OR RH) = P(BH) + P(RH) - P(BH) * P(RH) = 1

Taking just the second part of this:

  P(BH) + P(RH) - P(BH) * P(RH) = 1

and adding P(BH) * P(RH) to both sides we get:

  P(BH) + P(RH) = 1 + P(BH) * P(RH)

Since by symmetry P(RH)=P(BH), and letting x=P(BH)=P(RH) this simplifies to

  x + x = 1 + x^2

The only solution to that is x=1, so P(BH)=P(RH)=1

Your math is clearly screwed at this point.

So I've given you the correct interpretation, I've shown you where your calculations are wrong, and I've explained your incorrect use of Bayes' Formula.

Twice.

[tlammens]

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

[tlammens]

did you just prove that 1 == 1/2?

[bena]

You're making the mistake he mentions in the article.

You're only selecting two child families.

Flip one million coins. Select two at random. If one of them is heads, what is the probability that the second coin is heads?

[RiderOfGiraffes]

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.

[bad_user]

No there aren't ... notions of order like older/younger don't enter the equation (as the problem was stated).

Oh well, I guess this is what the article is talking about :)

[NeilCJames]

Older/younger doesn't exactly enter into it, but some ordering is necessary to properly describe the set of probabilities. This can be arbitrary, but birth order makes the most intuitive sense. The set of probabilities of sex distribution in two-child families is [(B, B), (B, G), (G, B), (G, G)], for any arbitrary ordering, this cannot be reduced to [(B,B), (B,G), (G,G)] as you have done without adding a variable to double the probability of (B,G).

Assuming a coin-flip probability for boy/girl distribution, you get the 1/3 answer if we select for two-child families where at least one child is a boy: [(B,B), (B,G), (G,B)]. If we don't pre-select for having at least one boy, (i.e. if we select the family because we just met the father socially), the probability rises to 1/2, because we have two cases to consider, each with a 1/2 probability: [(B,B), (B,G)], and [(B,B), (G,B)].

[bad_user]

Yes, but this is wrong ... the sex of the second child is in no way connected to the sex of the first child.

He didn't ask ... what's the sex of the second child? No ... he asked ...

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

Choosing a distribution that involves both children in this case is wrong, hence my answer that ordering doesn't matter. 33% is just wrong.

[NeilCJames]

What I wrote is not wrong unless, when you wrote: "So given 2 children, there are only 3 possibilities ... boy, boy girl, girl boy, girl" you are asserting that in the absence of any information about the sex of either child, 2 boys is a 1/3 probability, instead of (1/2)*(1/2)= 1/4.

[RiderOfGiraffes]

Let's toss two coins until at least one shows a head. By your reasoning the odds of them both being heads is 1/2. It's not. Try it.

Suppose I roll two dice until at least one of them shows a 6. What's the odds of both being 6's? I've said nothing about the red die versus the blue die, but the underlying truth requires that the situations are kept separate. It's only - as far as we know - in quantum mechanics where you deliberately lose the distinction.

I've done these as real world experiments as I explore them with kids, and I have a lot of direct experience. If you disagree then I'd be delighted to gamble with you.

[bad_user]

> Let's toss two coins until at least one shows a head. By your reasoning the odds of them both being heads is 1/2. It's not. Try it.

You haven't read the article then ... the problem as stated in the article is that you know one coin is going to be a head, so what's the probability of the other one also being a head?

Of course ... the events aren't connected ... the second coin toss doesn't depend in any way on the first coin.

That's why I think there's something wrong about the article ... saying that the probability is 33% fails both intuition and elementary probabilistic.

[RiderOfGiraffes]

Here's the problem as Gary originally stated it, and as the article quotes it:

    > I have two children, one of whom is a son born
    > on a Tuesday.  What is the probability that I
    > have two boys?

You say:

the problem as stated in the article is that you know one coin is going to be a head,

No. The point of the article is that you don't know how or why you are given this information.

Suppose I toss two coins until I get one that's a head, then I tell you that I have two coins, and one is a head. I've complied with the problem as described. The probability that they are both heads is 1/3.

I was there when this problem was posed. I was in the room when the questions were asked, and Gary clarified. I had lunch with Gary afterwards, and he said it was deliberate that it was ambiguous.

It seems to me that you're missing the point. Perhaps you should explain clearly exactly how you think the situation arises where the information given is as described.