Hacker News Reader

Comment by comex | Hacker News Reader

Wow. That's really nonintuitive to me. Like the Monty Hall problem: you have something that's obviously an irrelevant detail that can't affect the probability… except… somehow it does.

View on HN

There are problems that I find even more counterintuitive than Monty Hall, such as the sister's paradox.

    Your neighbours has two children, one of which is a girl.
    What is the probability of the other child to be a girl?

A. One may say 1/2, but actually, using basic Bayes theorem: P(2 girls | at least 1 girl) = P(2 girls AND 1 at least girl) / P(at least 1 girl) = (1/4) / (3/4) = 1/3.

I think this problem is actually equivalent to the Monty Hall problem. But it can get much weirder.

    Your neighbour has 2 children, one is a girl.
    You know the other child is born on a Sunday,
    what is the probability that this other child is a girl?

A. Similarly, counting all possible cases, one does not get 1/2 or 1/3, but rather 13/27.

  Your neighbours has two children, one of which is a girl.
  What is the probability of the other child to be a girl?

This is oft-mentioned but with this wording it's ambiguous. Here are three possible interpretations:

1. Your neighbor has two children. We observed both and found that there was exactly one girl. What is the probability that both are girls? (Answer: 0)

2. Your neighbor has two children. We observed one and found that she was a girl. What is the probability that the other (unobserved) child is a girl? (Answer: 1/2)

3. Your neighbor has two children. We observed both and found that there was at least one girl. What is the probability that both are girls? (Answer: 1/3)

Interpretation #1 is kind of silly but it serves to illustrate the ambiguity of the plain English wording. Interpretation #2 is how most people interpret the problem, and it's bolstered by the use of the word "other" in your prompt. For there to be an "other" child, there must have been a specific girl observed originally, rather than a general observation that there was at least one girl. Interpretation #3 is how you intended the problem.

yup, you're right.

You are horribly butchering Bayes' Theorem. Both your examples are incorrect because your two events ('2 girls' and 'at least 1 girl') are not oberved independently of one another, which is a necessary condition to using Bayes' Theorem with a valid result. Great tool but wrong job.

What? Bayes' theorem does not require anything about the events, even more so it is useful when the events are not independent. If events are independent, Bayes' theorem has no usefulness. https://en.wikipedia.org/wiki/Bayes%27_theorem#Statement_of_...

Edit: however now that I look into it, I should rather have used "conditional probabilities" name rather than Bayes'. I always mix both of them, my bad.

Looks like this is the Boy or Girl paradox [0], which has ambiguity problems.

[0] https://en.wikipedia.org/wiki/Boy_or_Girl_paradox

I love how Wikipedia helpfully informs us that 'Biologically, according to studies of the human sex ratio, it is more likely for any one person to be born into the male sex than the female sex.' as though it is somehow relevant or has anything to do with the actual paradox being discussed...

NOTE I've since edited the Wiki page, and deleted the sentence - since it isn't relevant and there was no citation for the claim!

> Your neighbours has two children, one of which is a girl. What is the probability of the other child to be a girl?

The way you have phrased the question, the probability is 1/2.

Actually, it's ever so slightly higher than 1/2 because there is a non-zero chance of them being identical twins. If we assume an identical twin incidence rate of 0.4%, then the actual probability would be 63/125 that the second child is also a girl.

In the first case, I guess what's happening is that when we pick two people randomly with uniform distribution, we expect, with equal probability, any combination of boys and girls: MM MF FF FM. When we observe a girl, we can exclude the case of two boys, so the remaining cases to consider are MF FF FM, where obviously only 1/3 of the cases is two girls.

So I guess what makes this counterintuitive is that we think of the example as independent variables actually affecting each other, but what happens is really that we're drawing from a collection without replacement, and obviously the mix of the collection is going to change as we do so.

This should apply also on a humankind, world-wide scale: if you have met three women in a row, the fourth person you meet is more likely to be a man. Gambler's fallacy! Or just drawing without replacement. A roulette table is drawing with replacement. :)

But MF and FM in this case are the same?

No because children are distinguishable.

If you flip two coins, you get HH, HT, TH, and TT with equal probability. Try it.

But the sequence is irrelevant.

HT is equal to TH, because in this situation we only care whether the two coins are equal to each other or different from each other.

It's the percent chance that matters, so it doesn't matter if you can distinguish them. The cases still exist with equal chance.

To elaborate, the situation as stated in the question is currently Hx, so HH and HT are the only possible results.

You can calculate 50/50 but it's not by ignoring the difference between HT and TH.

The first child is not an influencing factor on the second child. It's either a boy or a girl, 50/50.

My favourite quote from the wonderful David Spiegelhalter, Winton Professor of the Public Understanding of Statistics:

> Why do people find probability unintuitive and difficult? Because it is unintuitive and difficult.

Another counter intuitive probabilistic property that may be relevant here:

- If you make a distribution of the first digit of prices in a webshop you will notice that 1 appears more frequently than other digits.

- If you convert the prices to another currency and redo the distribution, the most frequent first digit will still be 1!

So it's not a given that you can assume the properties of a random bit stream will hold for the calculation in the article.

(As an aside: this weird property explains why you need slightly fewer coins in a currency that has quarters than those with 20 cent coins)

The logic behind this is more or less the same idea behind the Knuth-Morris-Pratt algorithm (https://en.wikipedia.org/wiki/Knuth%E2%80%93Morris%E2%80%93P...), which might help with your intuition.

Reply on news.ycombinator.com