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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.)
seawasp
Jul. 9th, 2010 07:58 pm (UTC)
Probability that both coins landed heads, AFTER you have told me one of them did, is 1/2.

Probability that both landed heads BEFORE you told me, 1/4.

In the case of the original question, it's analagous to the first case. We've already been told that you have one son. The probability the other child is a son -- and thus that you have two boys -- is then 1/2.
gareth_rees
Jul. 9th, 2010 08:09 pm (UTC)
Probability that both landed heads BEFORE you told me, 1/4.

This is correct: P(A and B) = P(A) × P(B) if A and B are independent events.

Probability that both coins landed heads, AFTER you have told me one of them did, is 1/2.

Have you tried this experiment yourself? It's easy to do, and I hope the actual results will be more convincing than some guy on the Internet. If you're right, then after you've got at least one head 60 times, about 30 of those times you'll have got two heads.
seawasp
Jul. 9th, 2010 10:30 pm (UTC)
Your use of "at least one head" indicates to me that you might be envisioning the experiment differently than I would conduct it -- which would, beyond any shadow of a doubt, produce (approximately) 1/2 odds of 2 heads.

Consider the following sequence of flips:

H T T H H T H T T H T T T T T H H H

If I guess what you're envisioning, it would be better written as

H T
T H
H T
H T
T H
T T
T T
T H
H T

In which all but two of these have at least one head but none have two heads

In my view, you only bother to check probability when you HAVE a head, so the sequence goes

H T T H H T H T T H T T T T T H H H

H: Get head next? T No
Tail, ignore, Head: Get head next? H Yes!
Tail, ignore, Head: Get head next? T No
Tail, ignore, Head: Get head next? T no
Tail, ignore, tail, ignore, tail, ignore, tail, ignore, Head: Get head next? H Yes!
Head: waiting for next flip.

Here, I am not calculating probability or counting until I already HAVE a head (just as in the original), and I have to ONLY calculate the next flip's probability (which, assuming a fair coin, is 1/2) of being a head. After that is done, I start anew, waiting for a head to show up.

If you tell me: I have two kids
And you tell me: One is a son

It doesn't matter to me that there are different orders in which you could obtain the kids. It doesn't matter that there are many other potential sequences in which you could not have had a son. All that matters is that you now have one child whose sex is indeterminate, and the chance that THAT child is male is 1/2 barring any oddities in the gene pool.

gareth_rees
Jul. 9th, 2010 10:51 pm (UTC)
Your use of "at least one head" indicates to me that you might be envisioning the experiment differently than I would conduct it

Yes, I have a particular experiment in mind and I am trying to convey as clearly as I can. If I can successfully convey it to you then we'll be in business! Remember that I described it like this:

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".

So to carry out one trial you need to:

1. Flip two coins
2. If they are both tails, this can't be an instance of the experiment (because I will answer "no" to the question in this case). So go back to step 1 and try again.
3. If both coins are heads, write down "YES"
4. If one coin is a head and the other a tail, write down "NO".

After carrying out sufficiently many trials, you'll be able to answer my original question, What is the probability that both coins landed heads? by looking at the number of times you've written YES and NO.

If you tell me: I have two kids

Let's wait until we've sorted out the coins before moving on to the boys.
seawasp
Jul. 9th, 2010 11:10 pm (UTC)
Er, no, let's not. The question is simple. Given a pair of objects, told what one of them is, the other can be one of two things.

There is literally no logical way for this probability to be other than 1/2. The first object doesn't count. It's fixed, no probabilities can be assigned to it (other than 1).

Your approach assumes a sequence of pairs, mine assumes a single instance in which I am presented with a single set of facts: there are two things. Thing one is in state 1. Thing two may be (with equal probability) in state 1 or state 2. What is the probability that both are in State 1? Well, we KNOW the first Thing is state 1, so we can ignore it. Thus we have only to consider Thing two, which is stated to have a 50/50 chance of being state 1 or state 2. Thus the probability of both being in State 1 is (1) [the certainty that the first is in State 1] * 0.50 [the probability that Thing 2 is in State 1], for a total of 0.5 or 1/2 that both are State 1.

gareth_rees
Jul. 9th, 2010 11:19 pm (UTC)
Well, the reason for my not wanting to move on is that we have several unresolved difficulties, and it's better to solve them than to ignore them.

The most important of these difficulties is that I am not convinced yet that you understand what I mean when I ask my 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?)

If we can't agree on what this means, or how a question like this can be turned into series of trials, then we've got no business attempting anything more complicated!

So, do you agree that the experimental procedure I outlined above will answer my question?
seawasp
Jul. 9th, 2010 11:30 pm (UTC)
No, not really, because the question as I view it is not a probabilistic series (which you outline) but a very simple situation. I have two things which can be A or B. I uncover one of them, find that one is A. I then have to bet on whether the second one is A.

The first one no longer matters. All that matters is that I have one remaining unknown which can be A or B.

Any other interpretation implies that prior events influence future ones -- the Gambler's Fallacy in which "The next one's GOTTA come up heads! The last three have been tails!".

Also, actual experiments in this area are far too messy (and long-drawn-out if you want the statistics to be reasonably close to the expected). I've seen wayyyyyy too many completely improbable sequences of dice rolls to trust flipping coins (and that's assuming I am able to flip coins reliably AND randomly).
nwhyte
Jul. 10th, 2010 04:56 am (UTC)
I have two things which can be A or B. I uncover one of them, find that one is A. I then have to bet on whether the second one is A.

That isn't quite right. What Gareth and I have in mind is more like this:

I have two things which can be A or B (with equal probaility). I uncover them both, find that at least one is A, and tell you. You then have to bet on whether the second one is A.

Do you see the difference?
seawasp
Jul. 10th, 2010 06:12 am (UTC)
Um... no, I'm afraid I don't. Before you uncover one, the probability of both being A is going to be 1/4. You uncover both and tell me one of them is A. This eliminates that one from contention in probability; I now have one remaining unknown (from my point of view; you have of course seen both) and that unknown may be either A or B, thus 1/2 probability.
nwhyte
Jul. 10th, 2010 06:18 am (UTC)
But what I actually do when I tell you that one of them is A is that I remove the BB combination from probability, leaving AA, AB and BA. That's where the 1/3 comes from.
gareth_rees
Jul. 10th, 2010 07:59 am (UTC)
If you don't agree that the procedure I outlined will answer my question, then I think the two coins problem is too hard for us at the moment.

Let's take step one back, and consider a simpler problem: I roll one six-sided die, you ask me "is it showing an even number of spots?" I answer "yes." What is the probability that it landed showing six?
raycun
Jul. 9th, 2010 12:30 pm (UTC)
"You first give one ball at random ... You then pass out to these five hundred one more ball each"

You're introducing a sequence.

If I have two kids, and the eldest is a boy, then the other kid is 50:50 boy:girl
If I have two kids, and one of them is a boy, I could have boy/boy, boy/girl or girl/boy. Each alternative is equally likely, and only one alternative has two boys.


seawasp
Jul. 9th, 2010 08:01 pm (UTC)
No, YOU are introducing a sequence. I'm pointing out that the important point is that I know you have one boy. You thus have only one child that is unknown in sex, and there are only two choices then: either that child is a boy, or the child is a girl. (barring the miniscule hermaphrodite probability, etc.). Thus, the chance you have two boys is 1/2, given that you have TOLD me one of your children is a boy. At that point, THAT child is no longer part of the equation, so to speak.
tortoise
Jul. 9th, 2010 11:18 pm (UTC)
(reposted because I somehow forgot a rather important "not")
The thing is, "at least one of my children is a boy" is not a fact about either of my children individually; it's a fact about both of them in conjunction. So it's not reasonable to say I have one child that is unknown in sex, because (if I have two boys) I might not have been talking about a specific one of my children in the first place!

Would you change your mind if the question was phrased as "my children are not both girls"? Note that this is logically equivalent to "at least one of my children is a boy"...
seawasp
Jul. 9th, 2010 11:33 pm (UTC)
Re: (reposted because I somehow forgot a rather important "not")
If the question was: My children are not both girls, what's the chance that I have two boys , you mean?

I don't see it makes any difference. You thus have one boy or you have two boys. One is established therefore to be a boy, leaving one that could be either, with a 50/50 chance of it being a boy.
raycun
Jul. 10th, 2010 07:05 am (UTC)
In the cocktail party example that Birks gave, the attendees get two balls together. The balls are black and green, so they could have
Ball 1 Black / Ball 2 Green (at least one black ball)
Ball 1 Black / Ball 2 Black (at least one black ball)
Ball 1 Green / Ball 2 Green
Ball 1 Green / Ball 2 Black (at least one black ball)

That's four possible combinations, three of which include at least one black ball. Only one of those three combinations is 'two black balls'

In your example, they get one ball at a time, and only the people with a black ball first get a second ball. So there are only two possible combinations of people with two balls.
Ball 1 Black / Ball 2 Green
Ball 1 Black / Ball 2 Black

There are two different questions being asked.
You are asking "If my first child is a boy, what are the chances that my second child is a boy?"
But the question above is, "If I have two children, and at least one of them is a boy, what are the chances that both of them are boys?"
raycun
Jul. 10th, 2010 07:11 am (UTC)
Or to go with the coin-flipping example above...

You are being asked to flip two coins together, and see if either one comes up heads.
You're saying, "No, I'm going to flip this one coin. If it comes up heads I'll flip it again, but if it comes up tails I'm stopping there."
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.
communicator
Jul. 9th, 2010 05:53 am (UTC)
Sorry, our comments crossed. It doesn't matter which one you are talking about - it's an irrelevance. The thought experiment is conflating set-sequence with linear sequence this isn't a helpful description, I'll think of better words in a minute. Next paragraph still stands.

If you are undecided whether I am onto something consider this - how likely do you really think it is that having one child born on Tuesday affects the chances that one of your children is a girl? In reality you know it does not. Therefore if someone thinks they have invented a mathematical proof that it does, I would be very sceptical of their maths.

Edited at 2010-07-09 06:06 am (UTC)
artw
Jul. 9th, 2010 06:19 pm (UTC)
The 13/27 answer is not about the chance that something will happen, it's about a (totally weird and unlikely in this case) gambling scenario where you are given incomplete information about something that has already happened and are guessing at the missing information.
communicator
Jul. 9th, 2010 06:29 am (UTC)
To re-express the point I crossed out - there is abstract sequence (AA, AB, BA, BB), and there is a linear sequence to the act of revealing (in the thought experiment, one person observes two items simultaneously but reveals the results sequentially to a second observer). This maths confuses the two in my opinion. And as I say, the Tuesday thing gives us a big clue that something is wrong.
seawasp
Jul. 9th, 2010 10:34 pm (UTC)
The Tuesday thing is, to me, what was always called a Red Herring. Word problems or puzzle books would often throw in details which turn out to be UTTERLY IRRELEVANT to solving the problem, but are there to make you waste time and effort looking for the connection, the key that you're missing -- when in fact it's just there to make you over-think the problem.
(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.
communicator
Jul. 10th, 2010 10:29 am (UTC)
True, but I actually think 'you know one is a boy' is an ambiguous sentence regardless of conventions and next time I am confronted with one of these on-line impasses, that's where I will look first for the cause of the controversy.
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.
seawasp
Jul. 9th, 2010 10:35 pm (UTC)
It seemed clearly stated enough, just with an extra detail (Tuesday) that's absolutely irrelevant to the problem.
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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