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A math question

(No ending time set)
Either; it doesn't make a difference  21%  [ 6 ]
Change it! the other cup has better odds!  71%  [ 20 ]
Don't change it, the one he chose at first is likely the right one!  7%  [ 2 ]

Author Message
Brightguy

Posted: Thu Jun 05, 2008 3:21 am   Post subject: Re: A math question

Reality Check: The answer to your problem depends on how exactly you obtained the information "one of the kids is a boy". When you looked out the window, were you able to obtain the genders of both of the kids, or just one?

Reality Check

Posted: Thu Jun 05, 2008 7:15 am   Post subject: Re: A math question

Yea Tony but this man isn't doesn't just do Physics. He's pure genius that is amazing at Math, English, History, Sciences, and Programming. He is a member of Mensa in fact so while I understand what you are saying, I find it very hard to doubt him. But hey, I suppose he could have been wrong.

You know what, I'll ask him again on Monday and present to him what you told me and I'll see what he has to say.
Insectoid

Posted: Thu Jun 05, 2008 7:58 am   Post subject: RE:A math question

Here's another one:

10 computer nerds argue over a question. they all vote 1 of 2 ways. if 3 vote one way, what are the odds that the others will argue?

I suppose I should mention that my mom knew it was the 'wrong' cup that was removed, and that it was intentionally removed as an incorrect cup.

This thread's a fighter, that's for sure...(as in, it won't go away...not that it has to or anything...)
zylum

Posted: Thu Jun 05, 2008 8:45 am   Post subject: RE:A math question

The sample space (in both cases) is:

{BB, BG, GB, GG} where the first letter is the first born and second is second born.

In conditional probability, you have P(B|A) = P(A n B)/P(A) -> 'n' is the intersection symbol
ie. the probability of B such that A is equal to the probability of A intersect B divided by the probability of A.

Situation 1:

event A = one kid is a boy
event B = one kid is a girl

P(A n B) = 2/4 = 1/2
P(A) = 3/4

Thus P(B|A) = (1/2)/(3/4) = 2/3

Situation 2:

event A = the eldest kid is a boy
event B = the younger kid is a girl

P(A n B) = 1/4
P(A) = 2/4 = 1/2

Thus P(B|A) = (1/4)/(1/2) = 1/2

This is what you do in the first week of intro stats
Insectoid

Posted: Thu Jun 05, 2008 9:14 am   Post subject: RE:A math question

Confusing...

Well, I guess we could put biology & trends into this to adjust the results (a girl is more likely to be born than a guy, I believe..)

But that would be immature (or overly-mature?) and though usually I don't mind acting a bit like a dufus, I feel this thread has had enough twisting and shall refrain.
Brightguy

Posted: Thu Jun 19, 2008 4:49 pm   Post subject: Re: A math question

Reality Check @ Thu Jun 05, 2008 7:15 am wrote:
You know what, I'll ask him again on Monday and present to him what you told me and I'll see what he has to say.

So? Did he concede that Tony is correct?
Reality Check

Posted: Thu Jun 19, 2008 5:26 pm   Post subject: Re: A math question

I Smell Death

Posted: Thu Jun 19, 2008 6:35 pm   Post subject: Re: A math question

you have to look at it for things that are not specified.

i'll look at it a bit at a time
After one has been selected by Jeremy, (Good for Jeremy he chose a cup, it doesn't say that he took the cup or looked under it, just that he chose one above the rest)

one cup is removed and identified as a cup without the token. (it doesn't say that the cup that was removed was the one that Jeremy chose, infact it doesn't specify which cup it was at all, all that we know is that it is not the right one)

Now there are 2 cups on the table, one the right one, one a wrong one. (this would seem obvious that there are two cups on the table with a wrong one having been removed)

Jeremy still doesn't know if his is the right one. (here we can determine that Jeremy's cup was infact not the one removed from the table and determined to be the wrong one)

Jeremy is given a choice: Change his selection to the other cup, or keep his current selection. (Knowing that a wrong one was removed from the table, leaving the one that he chose and an other one, with one of them being the right one and the other not. Jeremy could choose either of the two remaining cups as there is a 50-50 chance that the cup he chose is the right one)

but that's just my view

Reality Check

Posted: Thu Jun 19, 2008 7:59 pm   Post subject: Re: A math question

I don't know Tony every single site I've searched says the answer is 2/3 and not 1/2 like you suggested. They all use the sample space of:
BB BG GB GG.
Your solution does make sense but I'm sure there must be a reason they all say 2/3.
Brightguy

Posted: Thu Jun 19, 2008 9:42 pm   Post subject: Re: A math question

Oh the answer is 1/2 alright. Either you misunderstand the problem or you aren't using the "most plausible" interpretation (this was what my previous question was about).
Tony

Posted: Thu Jun 19, 2008 10:42 pm   Post subject: Re: A math question

Reality Check @ Thu Jun 19, 2008 7:59 pm wrote:
Your solution does make sense but I'm sure there must be a reason they all say 2/3.

What happens with a lot of sites, is that they often repost the same story, if it sounds interesting enough (2/3 sounds counter-intuitive). It'd be interesting if you were to still follow up with your source who could offer his own explanation, not just refer to "all the sites".

@zylum -- that is indeed what one does in intro stats. Though your sample space for the first situation is wrong. It's {BB, BG, GG}, where the order of age doesn't matter. Or if you insist on age differentiation, then (by same reasoning as in my previous posts) it should be {B1B2, B2B1, BG, GB, G1G2, G2G1}.

Either way, P(B|A) = (1/3)/(2/3) = 1/2 or (2/6)/(4/6) = 1/2.

To state my reasoning again, if elder boy in (boy & girl) and younger boy in (boy & girl) count for 2 states, then elder boy in (boy & boy) and younger boy in (boy & boy) should also count for 2 states.
Tony's programming blog. DWITE - a programming contest.
zylum

Posted: Fri Jun 20, 2008 1:02 am   Post subject: RE:A math question

Tony, I would have to disagree. If your solution space is {BB, BG, GG} then you are saying that the probability that the neighbor has two boys is the same as having a boy and a girl which is a false statement. If you want to use that sample space, then you have to state that the distributions are {0.25, 0.5, 0.25} which is the same as {BB, BG, GB, GG} = {0.25, 0.25, 0.25, 0.25} .
richcash

Posted: Fri Jun 20, 2008 2:14 am   Post subject: Re: A math question

Reality Check @ Wed Jun 04, 2008 8:42 pm wrote:
You have a neighbour with two kids.

Situation 1
You see outside the window and notice that one of the kids is a boy. What are the odds that the other is a girl?

Brightguy wrote:
.. The answer to your problem depends on how exactly you obtained the information "one of the kids is a boy". When you looked out the window, were you able to obtain the genders of both of the kids, or just one?

Yeah, depending upon interpretation zylum and Tony are both right.

If someone looks outside, sees both kids and tells you that "one is a boy" then the probability that the other is a girl is:
zylum wrote:
Situation 1

event A = one kid is a boy
event B = one kid is a girl

P(A n B) = 2/4 = 1/2
P(A) = 3/4

Thus P(B|A) = (1/2)/(3/4) = 2/3

However, if you look out and see 1 of the kids, and notice it is a boy (which would be the more common interpretation of the problem statement) then age has nothing to do with this problem. The sample space is {BB, BG, GB, GG} where the first variable is the "kid seen" and the second is the "kid not seen".
A = kid seen is a boy
B = kid not seen is a girl
And then it is the exact same problem as "Situation 2", yielding P(B|A) = 1/2
zylum

Posted: Fri Jun 20, 2008 7:24 am   Post subject: RE:A math question

I don't quite understand your reasoning.. Let me redo situation 1 with a sample space where age is not a factor:

Sample space = {BB, BG, GG} with probability distribution of {0.25, 0.5, 0.25}

Event A = one kid is a boy
Event B = one kid is a girl

Thus were are looking for the probability that one of the kids is a girl given that you know one of the kids is a boy. Whether you look and see one kid and its a boy or someone else looking outside and seeing both and telling you that one is a boy makes no difference. In both cases all you know is that one kid is a boy. Hence we look for P(B|A) (probability of one kid being a girl such that one kid is a boy).

P(A n B) = 0.5 -> P(BG)
P(A) = 0.75 ->P(BB) + P(BG)

P(B|A) = 2/3
richcash

Posted: Fri Jun 20, 2008 11:53 am   Post subject: Re: RE:A math question

You are right for one interpretation of the problem.
zylum @ Fri Jun 20, 2008 7:24 am wrote:
Whether you look and see one kid and its a boy or someone else looking outside and seeing both and telling you that one is a boy makes no difference. In both cases all you know is that one kid is a boy.

We have to arrange the kids by the order they were seen instead of age. If I look out and see a boy, then the First Kid Seen is a boy, and the Second Kid Seen is an individual event (exactly same problem as Situation 2). If I am just told that one of them is a boy, then I don't know if the First Kid Seen is a boy or girl, I only know that there's at least one boy.

If 2 coins are flipped and I see the first one is heads, then the second one has a 1/2 probability of tails. If two coins are flipped and I am told "one of the coins landed heads" then the other one being tails is 2/3 being the same calculation as you showed above.

zylum wrote:
Sample space = {BB, BG, GG} with probability distribution of {0.25, 0.5, 0.25}

Yes, that is the sample space if we know one of the kids is a boy (order doesn't matter), but order does matter if we look out and the first kid we see is a boy.

I'm pretty sure it's just an interpretation difference. I thought the more common interpretation would be that the person looks out and is only able to identify one of the child's genders (which would be First Kid Seen). But I guess you could easily interpret it the other way.
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