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34 replies to this topic

#21 Plumbstar Tom

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Posted 15 June 2011 - 05:18 PM

to futher comment on super's resoponce, we are obviouly only interested in the case where a person answers yes to the first two questions. this essentially limits the possibilies to being only yes to the third question or no, with 50/50 chances. but let's say i changed the questions to be... do you have two children? are both childern boys? are both children girls? we know the answer is yes to the first question and no to the second question. we can have one coin toss per question, but count an answer of yes, yes, yes to be a miss-toss. now if you ran the simulation, it would be 1/3. the questions however are very similar.

two questions would be are both boys? is one a boy and one a girl?

if one is a boy and one is a girl then the child is a boy, if both are girls the child is a girl

two outcomes of even probability = 1/2
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#22 wolfgang

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Posted 15 June 2011 - 07:30 PM

According to my theory....
If Teanchi and Beanchi love each other more than before,they will have a(boy).
but if they stay to this level of (love),they will have another(girl).
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#23 Aaryan

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Posted 15 June 2011 - 09:29 PM

Argue all you want but the information given to you about the first child helps in no way. The child is either a boy or a girl so it is 50% 1/2 however you put it.

Absolutely right. The question is, what is the chance of being a girl.
options are:

a. Boy
b. Girl

What is the third option to make it 1/3? It has to be 1/2.
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#24 harpuzzler

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Posted 15 June 2011 - 09:43 PM

Absolutely right. The question is, what is the chance of being a girl.
options are:

a. Boy
b. Girl

What is the third option to make it 1/3? It has to be 1/2.

You are over simplifying the problem....
Think of a punnet square like this

* Sorry the graph didn't turn out well but set up a punnet square with one boy and one girl on each axis and you will see what I mean

_______Boy______________Girl_____
| BB | BG |
Boy| | |
_________________________________
Girl| BG | GG |
| | |

So, yes there are only two options, but the odds of one of those options is greater; as there are three cases with a girl, and one where it is girl-girl, the odds are 1/3

Edited by harpuzzler, 15 June 2011 - 09:43 PM.

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#25 harpuzzler

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Posted 15 June 2011 - 11:04 PM

You are over simplifying the problem....
Think of a punnet square like this

* Sorry the graph didn't turn out well but set up a punnet square with one boy and one girl on each axis and you will see what I mean

_______Boy______________Girl_____
| BB | BG |
Boy| | |
_________________________________
Girl| BG | GG |
| | |

So, yes there are only two options, but the odds of one of those options is greater; as there are three cases with a girl, and one where it is girl-girl, the odds are 1/3

But this is assuming the question is implying the girl does not need to have been born first.
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#26 James22

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Posted 16 June 2011 - 06:32 PM

Where in the puzzle does it say anything about the girl being born first, there is no reason to assume it and this is the main reason as to why people get this wrong. There is a case for the answer being 0.5 but this is based on the method used for selecting the family which is unclear.
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#27 harpuzzler

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Posted 16 June 2011 - 07:34 PM

But this is assuming the question is implying the girl does not need to have been born first.

Yes, I am assuming order doesn't matter because the question states both kids are already born. The fact that one is a girl just means that out of the two, one is a girl.
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#28 superprismatic

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Posted 17 June 2011 - 07:13 PM

My last post(#15) was awkward and ill-worded. Here's a better one:

Suppose many two-child households each filled out a questionnaire truthfully.
The questionnaire contained two questions: "Do you have a daughter?" and
"Do you have a son?". If you take all of the questionnaires with a yes
answer to "Do you have a daughter?", what portion of these would have a
no answer to "Do you have a son?" (this is the same as saying that
both are daughters)?

To simulate this, just use two tosses of a fair fair coin to decide the
genders of the two children. Use this to fill out each questionnaire.
If you then take those with a yes answer to "Do you have a daughter?",
you will find that 1/3 of those have a no answer to "Do you have a son?".
Try it!

I just did it quickly for 12 questionnaires and I got 10 with a yes to
"Do you have a daughter?". Of these 10, 4 (that's 4/10 = .4) had a no
answer to"Do you have a son?". The more questionnaires you do, the closer
this gets to 1/3.
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#29 Twinhelix

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Posted 18 June 2011 - 09:59 PM

My last post(#15) was awkward and ill-worded. Here's a better one:

Suppose many two-child households each filled out a questionnaire truthfully.
The questionnaire contained two questions: "Do you have a daughter?" and
"Do you have a son?". If you take all of the questionnaires with a yes
answer to "Do you have a daughter?", what portion of these would have a
no answer to "Do you have a son?" (this is the same as saying that
both are daughters)?

To simulate this, just use two tosses of a fair fair coin to decide the
genders of the two children. Use this to fill out each questionnaire.
If you then take those with a yes answer to "Do you have a daughter?",
you will find that 1/3 of those have a no answer to "Do you have a son?".
Try it!

I just did it quickly for 12 questionnaires and I got 10 with a yes to
"Do you have a daughter?". Of these 10, 4 (that's 4/10 = .4) had a no
answer to"Do you have a son?". The more questionnaires you do, the closer
this gets to 1/3.

Hmmm I'm gonna write a quick java program and test this quick, but yeah, when you think about it that way, does seem to be a third

BUT I STILL LIKE 1/2 AS AN ANSWER!!! FOOK YOU ALLLLL
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#30 Twinhelix

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Posted 18 June 2011 - 10:14 PM

yaya, you people who say 1/3 are right. I hate you alllllllll!

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