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( 38 comments — Leave a comment )
seawasp
Jul. 9th, 2010 12:38 am (UTC)
WTF? The probability is 1/2. We already know that you have one boy, thus that does not matter in terms of calculating probability. There are only two possibilities then, that the other child is a boy, or the other child is a girl. Assuming having a girl is equally likely as having a boy, the chance you have two boys is one in two.
nwhyte
Jul. 9th, 2010 05:23 am (UTC)
You would be right if the statement was "the older is a boy". But the lack of clarity about which child is known to be a boy makes a big difference.

Do read through the debate between Pete Birks (who has convinced me) and Pete Doubleday (who takes your side) at the link. Pete Birks puts the first half of the case well:
I have 1000 black snooker balls and 1000 green snooker balls

1000 people come to a cocktail party. I give each of them two snooker balls at random.

This should result in roughly 250 people getting two green balls, 250 people getting two black balls, and 500 people getting one green ball and one black ball. If you disagree with that analysis, there's not much hope I fear.

Half way through the party I ask all of those with at least one black ball in their pocket to raise their hands.

About 750 people will raise their hands.

I now ask all of those with two black balls in their pocket to raise their hands.

About 250 people will raise their hands.

Therefore if someone has "at least" one black ball in their pocket, the chance of them having two black balls in their pocket is 250/750, or 1/3.

The situation with two children (= two snooker balls, equally likely to be black or green) and the black ball ("B" = Boy) and Green Ball ("G" = Girl") is the precise logical equivalent to the Parent with two children.
Then the days of the week argument takes the same technique to a slightly different place.
communicator
Jul. 9th, 2010 05:50 am (UTC)
His methods are wrong, because he is adding a hidden variable (action sequence). This is a well known error.
seawasp
Jul. 9th, 2010 12:19 pm (UTC)
No. This is not at all similar.

His use of "about" shows that he is not doing the same thing.

In the original case, we **KNOW** that one of the children is a boy. Whether the boy is older or younger is utterly irrelevant, because we know two facts:

1) You *already have* two children (so no variation of probability applies to the number of children)
2) We *already know* that one of the children is a boy (thus no variation of probability applies to whether *at least* one of the children is a boy.

Thus, the ONLY variable is the sex of the other child. And, barring very odd genetic variations, there are only two options -- boy and girl -- and these are generally considered to be equally likely (50/50).

Therefore, there is a 1/2 probability that you have two children.

In the example above, both black and green balls are randomized. To be equivalent to the ORIGINAL scenario, you must proceed as follows;

You first give one ball at random to all your 1,000 guests. Then everyone who has a black ball is asked to raise their hands.

At that point, anyone who has a green ball is asked to leave. This leaves you with ("about") 500 people in the room *known* to have one black ball. You then pass out to these five hundred one more ball each, at random. The number who will then raise their hands to say they have now got TWO black balls will be ("about") 250, or, as I said, 1/2.

THIS is the equivalent of the original case.
gareth_rees
Jul. 9th, 2010 12:28 pm (UTC)
Let's start with the two coins version of the puzzle and see if we can come to agreement on that. There's no point in thinking about the "two boys" variant (let alone the "Tuesday" variant!) until we've got the simple version sorted.

So, I flip two coins randomly and hide them from you. You ask me, "did at least one of the coins land heads?" I answer "yes". What is the probability that both coins landed heads?

(For a mathematician, this is a straightforward application of Bayes' theorem, but if you're not intimate with this theorem, you might need to think about it a bit.)
(no subject) - seawasp - Jul. 9th, 2010 07:58 pm (UTC) - Expand
(no subject) - gareth_rees - Jul. 9th, 2010 08:09 pm (UTC) - Expand
(no subject) - seawasp - Jul. 9th, 2010 10:30 pm (UTC) - Expand
(no subject) - gareth_rees - Jul. 9th, 2010 10:51 pm (UTC) - Expand
(no subject) - seawasp - Jul. 9th, 2010 11:10 pm (UTC) - Expand
(no subject) - gareth_rees - Jul. 9th, 2010 11:19 pm (UTC) - Expand
(no subject) - seawasp - Jul. 9th, 2010 11:30 pm (UTC) - Expand
(no subject) - nwhyte - Jul. 10th, 2010 04:56 am (UTC) - Expand
(no subject) - seawasp - Jul. 10th, 2010 06:12 am (UTC) - Expand
(no subject) - nwhyte - Jul. 10th, 2010 06:18 am (UTC) - Expand
(no subject) - gareth_rees - Jul. 10th, 2010 07:59 am (UTC) - Expand
(no subject) - raycun - Jul. 9th, 2010 12:30 pm (UTC) - Expand
(no subject) - seawasp - Jul. 9th, 2010 08:01 pm (UTC) - Expand
(no subject) - raycun - Jul. 10th, 2010 07:05 am (UTC) - Expand
(no subject) - raycun - Jul. 10th, 2010 07:11 am (UTC) - Expand
communicator
Jul. 9th, 2010 05:36 am (UTC)
I agree with you. Assuming you have two children, and you know one is a boy, the chances of having two boys is one in two.

You can tell peterbirks is starting with some extremely flawed assumptions because he says 'the obvious answer is one in three' WTF??? He doesn't understand probability. No actually, he is making additional assumptions which are over-complicating.

Put it like this - you have two coins. One is hidden. The one you can see is heads-up. What is the probability both are heads-up? Not 13/27, not 1/3. It's one in two, because it depends entirely on a 50/50 chance.
nwhyte
Jul. 9th, 2010 05:42 am (UTC)
Put it like this - you have two coins. One is hidden. The one you can see is heads-up. What is the probability both are heads-up? Not 13/27, not 1/3. It's one in two, because it depends entirely on a 50/50 chance.

But if I have two coins which you can't see, and tell you that one of them is heads-up, the chance that the other is heads-up is 1/3 - because you cannot know which coin I am talking about. See the black/green snooker ball analogy.
(no subject) - communicator - Jul. 9th, 2010 05:53 am (UTC) - Expand
(no subject) - artw - Jul. 9th, 2010 06:19 pm (UTC) - Expand
(no subject) - communicator - Jul. 9th, 2010 06:29 am (UTC) - Expand
(no subject) - seawasp - Jul. 9th, 2010 10:34 pm (UTC) - Expand
(Anonymous)
Jul. 9th, 2010 06:04 am (UTC)
I am slightly too thick to join in any mathematical discussion, but thought I'd pop in and say I just loved the Daleks in your home. As Eccleston would say, fan tastic.
quarsan
gareth_rees
Jul. 9th, 2010 09:21 am (UTC)
It all depends, of course, on how this particular problem came to be stated in the first place. It's if it was in response to a query like, "tell me how many children you have, and then pick one child at random and tell me their sex and the day of the week they were born" that we get the 13/27 answer.
communicator
Jul. 10th, 2010 09:06 am (UTC)
Yes. I think there is a hidden ambiguity in the original framing and that is why people are arriving by legitimate means at different conclusions.

'A person has two children; you know one is a boy' can mean

- you only have information about one child, and he is a boy
- you have information about two children, but you are only revealing that one is a boy

From the initial interpretation of the sentence in one of these two ways, all the rest flows. Nick is using the second interpretation, which is why the bowling ball analogy works. Using the first interpretation the maths is wrong.

This explains why intelligent people are coming to such wildly different conclusions. It's about English not maths basically.

gareth_rees
Jul. 10th, 2010 10:12 am (UTC)
It's partly a difference in conventions about stating problems. Mathematicians have a set of informal conventions when they state problems: objects are point masses, ropes are weightless and inextensible, people make only true statements (or negations of true statements), coin tosses and die rolls are fair, you can't use the principle of relevance when analysing a statement (only its literal content), and so on. Without these conventions, mathematical problems would largely consist of tedious boilerplate.

The trouble comes, of course, when you take a problem for mathematicians and present it to non-mathematicians, people who aren't familiar with the conventions. 90% of any ensuing conversation involves getting the non-mathematician up to the point where they understand the problem (in its intended form). My exchange above with seawasp show how difficult this can be, even for problems much simpler than the "Tuesday" problem.
(no subject) - communicator - Jul. 10th, 2010 10:29 am (UTC) - Expand
artw
Jul. 9th, 2010 06:08 pm (UTC)
The maths is so basic I'm pretty sure I even did it at O-level, but what confuses people is that it's such an artificial and meaningless scenario that one automatically thinks of the more natural, mathematically distinct question "I already have a son born on a Tuesday, what's the chance the next child will be a boy?".
seawasp
Jul. 9th, 2010 08:11 pm (UTC)
I'm not sure that it's mathematically distinct, as that's the same as I interpreted the original question.
artw
Jul. 9th, 2010 09:11 pm (UTC)
Me too. The original question is not clearly stated. I assume this was deliberate as I understand it was posed at a conference to stimulate debate.
(no subject) - seawasp - Jul. 9th, 2010 10:35 pm (UTC) - Expand
nickbarnes
Jul. 10th, 2010 07:11 am (UTC)
Oh, dear God. That link has somebody getting the Monte Hall problem wrong, over and over again. Isn't there a circle of hell like that?

I would beg anyone who is disagreeing with the numbers presented by Peter Birks, to please go away and *do the experiment* described by Gareth Rees in the thread here, to answer this question:

I flip two coins randomly and hide them from you. You ask me, "did at least one of the coins land heads?" I answer "yes". What is the probability that both coins landed heads?

The answer to this is 1/3. It really, really is. If you are convinced that it is not 1/3, then you are incorrect, *but* it will not be possible for us to convince you of this. The only way to convince yourself is to do an experiment. It will take about the same amount of time as replying to this message.

You can do this yourself, with paper, pen, and two actual coins:

1. repeat sixty times:
- flip both coins.
- write down HH, HT, TH, or TT on a fresh line.

2. now go down the column, crossing out any TT lines.

3. now go down the column, ticking HH lines and crossing HT and TH lines.

4. how many ticks and crosses do you have?

Many years ago I didn't understand probability very well. Despite a good maths degree, I used to get into arguments about Monte Hall on Usenet. Finally I did a Monte Hall experiment, and then I understood.
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