Help with very counterintuitive problem

Footer · science forum beginner Joined: 24 Sep 2005 Posts: 2

Mike · science forum addict Joined: 17 Sep 2005 Posts: 74

I think you're exactly right--the (b,b) sample point should be counted twice
if you want to distinguish birth order because the King could be the younger
brother or the older brother. If you want to ignore birth order then you
should count (b,g) and (g,b) as a single sample point too. Either way, the
answer is the intuitive answer 1/2. What book are you using?

"Footer" <footer105@yahoo.com> wrote in message
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Casey Hawthorne · science forum beginner Joined: 19 Jun 2005 Posts: 27

The correct answer is 2/3, given that a boy has already been picked.

Conditional probabilities can be counter intuitive, which is why we
have the machinery of probability to work them out!

--
Regards,
Casey

Mike · science forum addict Joined: 17 Sep 2005 Posts: 74

Better check your machinery! I know that problems often have counter
intuitive answers but that doesn't apply here. When one approaches the
King and asks about his sibling you either care about birth order or you
don't.

If you don't care the sample points are simply (b,g) and (b,b).

If you do then the sample points are (b,g) [because he is the older
sibling], (g,b) [because he is the younger sibling], (b,b) [because he is
the older sibling] or (b,b) [because he is the younger sibling]. It's
reckless of me to call (b,b) a single sample point but the sample "point"
(b,b) must be counted twice because these cases represent two different
situations. Got it?

Be careful not to dismiss your intuition either. In this case it should be
obvious to your intuition that the sex of one child doesn't depend on the
sex of the other which is all that this problem is asking. Whether or not a
person is King or Queen or anything else his/her sibling's gender is not
affected.

The error people make is that in listing the sample points they list two of
them as if birth order matters [ (g,b) and (b,g) ] and then (b,b) as if
birth order doesn't.

If you want to stretch your intuition do a google search of Simpson's
Paradox or the Monte Hall paradox. The latter is particularly interesting
in that the correct answer to the problem (provided by a woman [Marilyn vos
Savant] I might add) was lambasted by the mathematical establishment as a
symbol of all that is wrong with math education today. Hah!

Regards,
Mike
"Casey Hawthorne" <caseyhHAMMER_TIME@istar.ca> wrote in message
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Mike · science forum addict Joined: 17 Sep 2005 Posts: 74

There are (at least) two errors in my last post indicated by double brackets
in the copy below. In the first, "or" should be replaced by "and" and in
the second "these cases represent" should be replaced by "it represents".
(For clarity and accuracy, grammer is important, you know!)

"Mike" <MDC@nomail.com> wrote in message
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Henry · science forum addict Joined: 15 May 2005 Posts: 58

On 24 Sep 2005 00:03:08 -0700, "Footer" <footer105@yahoo.com> wrote:

Nigel · science forum beginner Joined: 03 Jun 2005 Posts: 37

Henry wrote:

Jon Haugsand · science forum beginner Joined: 03 May 2005 Posts: 37

* useweb@nospam.com

Casey Hawthorne · science forum beginner Joined: 19 Jun 2005 Posts: 27

It doesn't matter whether you think the King is the older brother or
not!

With out knowing anything (my usual case or casey):

The sample space is:
b,b
b,g
g,b
g,g

But, you are told the King is one of two children, so the sample space
is now truncated to the following:
b,b
b,g
g,b

The above all have at least one boy.

There are two cases out of the three that have girls.
--
Regards,
Casey

Jon Haugsand · science forum beginner Joined: 03 May 2005 Posts: 37

* Casey Hawthorne

Mike · science forum addict Joined: 17 Sep 2005 Posts: 74

Of course it matters. The only way that 2/3 is the answer is if you assume
that the King can't / doesn't have an older brother (as in primogeniture).
Otherwise, the answer is most definitely 1/2 as your intuition should tell
you.
If you don't agree would love to hear your answer to Jon Haugsand's (and the
original poser's [Footer]) question.

"Casey Hawthorne" <caseyhHAMMER_TIME@istar.ca> wrote in message
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Rusty · science forum beginner Joined: 07 Sep 2005 Posts: 46

"Casey Hawthorne" <caseyhHAMMER_TIME@istar.ca> wrote in message
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Dan Akers · science forum addict Joined: 19 Jul 2005 Posts: 56

Footer wrote:
"The king comes from a family of two children. What is the probability
that the other child is his sister?"
The answer in the back of the book is 2/3. Now I understand the
reasoning behind this (sample space is {(b,g)(g,b)(b,b)} and two of
those have 1 boy/1 girl), but I am having trouble wrapping my head
around it. It seems to me that the possible outcomes should be of the
form (king,sibling) and thus be limited to (b,g) and (b,b) giving the
more intuitive answer 1/2.
I am picturing another scenario also. Imagine a crowd of people
consisting only of sibling pairs with equal amounts of the four possible
groups present. If one were to walk up to a guy at random and ask if he
was with his brother he would say yes half the time, correct? Because
although the (b,b) pair occurs only 1 out of 4 times, there are two guys
representing it as opposed to one with either of the mixed pairs. Is
this not analogous to the problem above, or is my logic faulty?"
_____________________________________
Re;
The book is correct. The probability of his sibling being a girl is 2/3
given that there is a 50% probability of a person being a girl (or boy).
I think your logic is indeed in error and so is your "sample space".
Try looking at the problem in this way with the sample space being all
the possible pair permutations of girl/boy; BB, GG, BG, and GB with each
permutation being equally likely. We are interested in knowing the
probability of BG or GB given the fact that B has been revealed. Your
intuition tells you that since B is revealed, GG is eliminated and there
then must be a 50% chance of the sibling being a girl since, in your
mind, only BG and BB are possible. Well, you're right, the only
possible combinatorial outcomes are BG and BB and that is where
intuition breaks down. What if, behind a blind, I flipped two coins. I
then showed you one of the coins was on heads; what is the probability
that the other is tails? Since the other coin could have been the first
or the second coin flipped, the unique possible permutations are HH, HT,
and TH. Thus, there is a 2/3 probability of the other being tails; that
is to say, there is a 2/3 probability of the permutation TH or HT being
flipped. The problem is concerned with the probability of the BG and GB
permutations; not the probability of an individual being a B or a G.
This problem is seen in almost every probability text book under the
heading, "Conditional Probability". The counter to it is: Given that in
a two children family, one of the children is a boy, what is the
probability that the other child is a boy? The answer here is 1/3.
How's that for counter-intuitive!! I hope that helps...

-Dan Akers

Dan Akers · science forum addict Joined: 19 Jul 2005 Posts: 56

Jon wrote:
"If you meet a random person on the street and somehow you are told that
he has a sibling. What is the chance that this sibling is a woman? With
your reasoning you can conclude that is 2/3. Isn't it odd?"
______________________________________
Re;
Not a random "person"; but a random "man" (condition 1) who has ONE
(condition 2) other sibling; then there is indeed a 2/3 probability of
that man's sibling being female and 1/3 being male.

-Dan Akers

Dan Akers · science forum addict Joined: 19 Jul 2005 Posts: 56

Footer wrote;
"I am picturing another scenario also. Imagine a crowd of people
consisting only of sibling pairs with equal amounts of the four possible
groups present. If one were to walk up to a guy at random and ask if he
was with his brother he would say yes half the time, correct? Because
although the (b,b) pair occurs only 1 out of 4 times, there are two guys
representing it as opposed to one with either of the mixed pairs. Is
this not analogous to the problem above, or is my logic faulty?"
_____________________________________
Re;
The first condition on the group, to make it an analogy of the "king"
problem, would be to make an announcement at the gathering that all
girl-girl or GG pairs are to please leave the gathering. Since this
eliminates one of the four unique possible permutational possibilities,
only BG, GB, and BB remain. Thus, on a random selection of boys, 2/3
would say they were with a sister and 1/3 with a brother.

-Dan Akers

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