onenickelmiracle
onenickelmiracle
Joined: Jan 26, 2012
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January 14th, 2014 at 3:28:40 AM permalink
Quote: Jeepster

You're opening a can of worms with the Monty Hall problem.
I'm not going near that one, most people will not accept the correct answer.

I've always explained this one as switching from odds 1/3 to 1/2.
I am a robot.
Canyonero
Canyonero
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January 14th, 2014 at 3:56:51 AM permalink
Quote: gameterror

ok, then i don't understand how you come to a different solution. 4 variants. each variant of birth has the same probability. now the fact that we see at least one boy eliminates the GG variant and only this one. so we are left with the 3 others BB, BG, GB. leaving the second kid we don't see as a B, G, G ...so 1/3.



the fact that we would only see a boy in a GB or BG combination 50% of the time is no longer relevant...as we have already SEEN him.



It is very relevant.

You must not just eliminate the GG combination (= adjust to 0%) and ignore the adjustments to the other combinations.

If you meet a boy, it is twice as likely he came from the BB combination - it might be either of the two boys. You mustn't assume that all combinations BB GB BG have an equal likelyhood, since after already meeting a boy the odds are now clearly skewed towards the BB combination.


Can somebody with better explanatory skills back me up on this please. The Monty Hall reference totally applies btw...

Btw. the independent event approach also totally works and I have not seen any flaw in that logic. Imagine this: You meet a dad with only one child - a boy. That night he goes home and impregantes his wife. The next day you meet him again. What is the chance of the other child becoming a girl? - Exactly!
gameterror
gameterror
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January 14th, 2014 at 4:00:12 AM permalink
Quote: AZDuffman

You mostly beat me to this. Here is another way of putting it:

There is a "no zero" roulette game, the last time the wheel was spun it landed on red. WThat are the chances it will land on red the next spin?

The chances of the other child being a boy should be 50%. The sex of the child with the parent at the time should not matter.



Yes, but this is a different question. If the question was : i already have one kid what is the chance the second is going to be a boy too ? the answer is 50% because the sex of the first kid has no influence on the second kid.

but the possible distribution of having 2 kids is BB, BG, GB, GG. all with the same probability of 25%. See tree above. Now I tell you that one of the kids is a boy. that eleminitates the GG variant. and leaves you with 3 possible distributions BB, BG, GB.... so 1/3 that it's two boys.
Things have never been so swell I have never failed to fail
gameterror
gameterror
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January 14th, 2014 at 4:16:49 AM permalink
Quote: gameterror


edit:
I guess the problem that Canyonero has (and i also had for a short moment just now) is that he "creates" another solution where the order of the BB solution matters. Creating two variants B1B2 and B2B1 which doesn't exist. See table above. There is only one BB possibility and it doesn't matter that we don't know if we see the old or young boy of the boy-boy combo.



I think you are right.... after rereading your early statement that's exactly what i referenced in this edit.

The combination BB has two different boys which creates 4 scenarios for what we can see.

B1B2 -> can be either B1 seen, B2 hidden or B1 hidden and B2 seen.
and then the two BG GB variants which in fact have no hidden boys...as we see the boys of those variants.

now - and i can't wrap my head around it - we have to figure out if it really matters which boy we see of the BB combination. your pregnancy example strongly indicates it doesn't :-)

i'm now also in the 50/50 camp :-)
Things have never been so swell I have never failed to fail
s2dbaker
s2dbaker
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January 14th, 2014 at 4:17:55 AM permalink
Quote: Jeepster

24Bingo, I never said the boy with me was born first.
The only other info I gave you was I had two children.
Therefore you cannot eliminate GB
BG, GB, and BB are all valid and equal possibilities

I disagree. Here's a table of possibilities:
Child With YouChild At Home
BoyBoy
BoyGirl
GirlBoy
GirlGirl

Here's the question:
Quote: Puzzle

You see me out with one of my two children, who happens to be a boy.
What's the chances of the other child at home also being a boy?

The only info you have and are allowed to know is-
I have exactly two children, the boy with me and another child at home.
Assume a world wide boy/girl birth ratio of 50/50.

Keep in mind that the child that you brought with you is completely irrelevant according to this wording. You may as well have said, You see me out with my Subaru which happens to be Red, what are the chances that my child at home is a boy? So..
Because the child that you brought with you is a boy, the last two rows are impossible, you must choose from what remains and that makes the child at home a 50/50 chance of being a boy.
Someday, joor goin' to see the name of Googie Gomez in lights and joor goin' to say to joorself, "Was that her?" and then joor goin' to answer to joorself, "That was her!" But you know somethin' mister? I was always her yuss nobody knows it! - Googie Gomez
JB
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JB
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January 14th, 2014 at 4:21:05 AM permalink
Before knowing anything about either child, here were the four possibilities:

Child #1 Child #2 Probability
Boy Boy 0.25
Boy Girl 0.25
Girl Boy 0.25
Girl Girl 0.25
Total 1

Since we know that at least one of them is a boy, we can eliminate Girl-Girl as a possibility, leaving the following:

Child #1 Child #2 Probability
Boy Boy 0.25
Boy Girl 0.25
Girl Boy 0.25
Total 0.75

However, the probabilities of the remaining possibilities only sum to 0.75, so we need to scale each probability by dividing them by 0.75. This is the same thing done in craps to calculate the probability of making or missing a point; we don't care about rolls of 2, 3, 11, 12, etc., and therefore we remove those possibilities from the calculation and scale the probability of the only outcomes that matter such that they sum to 1. So after doing this, we are left with:

Child #1 Child #2 Probability
Boy Boy 0.333333...
Boy Girl 0.333333...
Girl Boy 0.333333...
Total 1

And since the only remaining possibility that allows for the other child to be a boy has a probability of 1/3, that is the answer.
s2dbaker
s2dbaker
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January 14th, 2014 at 4:25:46 AM permalink
Quote: JB

And since the only remaining possibility that allows for the other child to be a boy has a probability of 1/3, that is the answer.

Read the puzzle carefully. The boy that is with the parent is completely irrelevant.
Someday, joor goin' to see the name of Googie Gomez in lights and joor goin' to say to joorself, "Was that her?" and then joor goin' to answer to joorself, "That was her!" But you know somethin' mister? I was always her yuss nobody knows it! - Googie Gomez
gameterror
gameterror
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January 14th, 2014 at 4:36:59 AM permalink
Child #1 Child #2 Child with meChild at home
Boy1 Boy2Boy1Boy2
Boy1 Boy2Boy2Boy1
Boy Girl BoyGirl
Girl Boy BoyGirl
Girl1 Girl2 N/A
Girl1 Girl2 N/A


50/50
Things have never been so swell I have never failed to fail
JB
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JB
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January 14th, 2014 at 4:43:34 AM permalink
I concede, it actually is 1/2 when picked apart word by word. I was incorrectly reading the "spirit of the puzzle" as opposed to the "letter of the puzzle."
tringlomane
tringlomane
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January 14th, 2014 at 4:45:25 AM permalink
Quote: JB

Yes, I see now that the semantics of the original wording are ambiguous. (First, there is no indication that the child with him is a boy, next there is.) So the answer is either 1/2 or 1/3 depending on how the question is interpreted.



Agreed. And as currently written, it's 1/2. Hope we didn't kill Monty Hall with this discussion.

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