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When intuition and math probably look wrong

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Re: When intuition and math probably look wrong

#11

Post by RiderOfGiraffes »

bad_user wrote: post #8
RiderOfGiraffes wrote: post #7

Earlier quoted context omitted.

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.

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 :)

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.

Re: When intuition and math probably look wrong

#12

Post by bad_user »

RiderOfGiraffes wrote: post #11
bad_user wrote: post #8

Earlier quoted context omitted.

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 :)

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 yo…

> 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.

Re: When intuition and math probably look wrong

#13

Post by NeilCJames »

bad_user wrote: post #8
RiderOfGiraffes wrote: post #7

Earlier quoted context omitted.

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.

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 :)

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)].

Re: When intuition and math probably look wrong

#14

Post by bad_user »

NeilCJames wrote: post #13
bad_user wrote: post #8

Earlier quoted context omitted.

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 :)

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…

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.

Re: When intuition and math probably look wrong

#15

Post by RiderOfGiraffes »

bad_user wrote: post #12
RiderOfGiraffes wrote: post #11

Earlier quoted context omitted.

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 yo…

> 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…

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.

Re: When intuition and math probably look wrong

#17

Post by NeilCJames »

bad_user wrote: post #14
NeilCJames wrote: post #13

Earlier quoted context omitted.

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…

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.

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.

Re: When intuition and math probably look wrong

#18

Post by RiderOfGiraffes »

bad_user wrote: post #6
bena wrote: post #2

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…

> 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 y…

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?

Re: When intuition and math probably look wrong

#20

Post by shasta »

Daniel_Newby wrote: post #5

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…

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?

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