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What's the Difference?

You meet a woman in the grocery store and you ask her how many children she has. She tells you that she has two children. You ask if any of them are boys, and she replies "yes." What is the likelihood that both her children are boys?

The following day you meet a different woman, who introduces you to her son. You ask if she has any other children and she replies "yes." What is the likelihood that both her children are boys?

Ham Fisted
Saturday, August 21, 2004

In the first case

the possibilities are


bb
bg
gb
gg

so probability of having a boy is 3/4

in the second case it is
bg
bb

(since one of them is boy)
There for the probability is 1/2

Sundar Rajan.G.S
Sunday, August 22, 2004

In the first case she already admitted to having a boy, which excludes the case 'gg'.  The remaining cases - bg, gb, bb - are equally likely and their probabilities must add up to 1, hence:

P(bb | gg excluded) = 1/3

This is the 'painfully easy' problem in different clothing. :)

Chuck Boyer
Sunday, August 22, 2004

Well like Chuck said, it's the "painfully easy" problem in not-very different clothing :)
But I just had a thought, that second problem doesn't state that the children are 2... That is unless the statement "What is the chance that both her children" implies they are 2, but I am not sure.....

Christian Kamel
Monday, August 23, 2004

Another thought, why do we consider here in this problem the solutions gb and bg as different ones? Is it different having a boy and a girl or a girl and a boy? I think order doesn't matter here so the whole solution space should have been (bb, bg, gg) and excluding gg for the first problem we should come up with the probability being half....
We could argue that the order stands for the age of the children but then again they could be twins.... so why don't we account for this fact too....
If my thoughts are true that would make it pretty different than the "painfully easy" problem stated earlier.

Christian Kamel
Monday, August 23, 2004

Yes, I meant to specify that the woman has 2 children in both problems.

Ham Fisted
Monday, August 23, 2004

Birth order doesn't matter, per se.  What does matter is that there are two distinct children (a 2-tuple), which provides two ways (permutations) to generate a boy/girl pair (combination). 

Case 1 excludes only the 'gg' outcome, leaving both boy/girl permutations in the sample space.

Case 2, OTOH, excludes both 'gb' and 'gg' outcomes.  This excludes one of the boy/girl permutations from the sample space.

~~~~~~~

It might be easier to visualize things if you map the sample space to points on a cartesian plane:  Let 0 = boy, 1 = girl.  Case 1 says the possible outcomes correspond to the points (0,0), (0,1) and (1,0).  Note that (0,1) and (1,0) are two distinct points on the plane!

Case 2 says the outcome is either the point (0,0) or the point (0,1).  One of the boy,girl points - (1,0) - is excluded.

Changing the order of the coordinates just swaps ordinate and abscissa; there remain two distinct points that map to a boy/girl *combination*

Chuck Boyer
Tuesday, August 24, 2004

Why is gb excluded in case 2 ? :S

Christian Kamel
Tuesday, August 24, 2004

> Why is gb excluded in case 2 ? :S

Hello again, Christian! :)

I was (arbitrarily) assigning the visible child -- a known 'b' --  to the left member of the 2-tuple, and the unseen child to the right member.  That leaves one degree of freedom in the problem, in which case only the outcomes bb and bg fit the given circumstances.  One could as easily assign the visible child to the right member of the 2-tuple, in which case the only outcomes would have been bb and gb. Either way, one of the boy/girl permutations gets excluded. Although {b, g} and {g, b} are the same _set_, the outcome 2-tuples (b,g) and (g,b) -- which we'd abbreviated to bg and gb -- are decidedly _not_ the same. 

My point is that order may be arbitrary, but it _is_ significant.  Here's an example in which we wouldn't usually question the significance of order: consider how many ways there are to combine a pair of binary digits to make a two-digit binary number: { 00, 01, 10, 11 }. 

There are two ways to pair a '1' and a '0', but  '01' and '10' are definitely _not_ the same number.  Which one represents a decimal '2' depends on whether we read the bit positions from right to left or from left to right. Although it's customary to put the least significant digit on the right, that custom is arbitrary.

The key issue here is the distinction between permutations and combinations; it's the permutations that define the basic elements of the sample space. 

Probability problems can be tricky because it's easy to get confused about the basic sample space and the "sigma algebra" defined over that space.  Sigma algebras don't get covered in the math courses that introduced most of us to probability concepts, so if anyone's interested, here are two URLs that lead to the formal world of "probability spaces":
http://encyclopedia.thefreedictionary.com/probability%20space

http://en.wikipedia.org/wiki/Probability_theory

Cheers! :)

Chuck Boyer
Tuesday, August 24, 2004

One more URL: http://mathworld.wolfram.com/SampleSpace.html

:)

Chuck Boyer
Tuesday, August 24, 2004

As always you bring something interesting to the table Chuck :)

Well I got what you said about solution spaces, I still have to read those links but in the meantime, I still don't think the second case is any different from the first one.
The fact that you'd seen (the/one of) boy children in case 2 doesn't make any difference if you 100% trust the lady's words in case 1 which we have no reason to doubt.

If I understand correctly your argument is that you'd seen 1 digit from the {0,1} number, but obviously you don't know if you'd seen it starting the right digit or the left digit, so you can't exclude 01 or 10 and have to include them both in the solution space, which renders the case 2 problem exactly similar to the case 1 problem.
I hope I'm explaining myself well, English is my second language :)

Christian Kamel
Tuesday, August 24, 2004

> As always you bring something interesting to the table Chuck :)

Thanks, Christian! So do you. :)

...and we both seem to have too much time on our hands! ;)

If I understand what you've been saying, it appears that you see 'bg' and 'gb' as identical outcomes; I'm trying (unsuccessfully? ;) to say why they are not. That seems to me to be our point of departure, and I probably muddied the water with too many illustrations.  Maybe we can chat off-line some about those URLs I posted. :)

> I hope I'm explaining myself well, English is my second language :)

You use the language very well indeed, and better than many of us who speak it natively! :)

Chuck Boyer
Wednesday, August 25, 2004

First, I'd like to argue, that both cases given above are identical. It should not make a difference if the woman _tells_ us she has at least one boy or if she _shows_ us the same fact.

But the boy/girl question get's more complicated if we take in consideration, that  51% of all newborn are female ;-).

We get this diagram:

g (51%) b (49%) = 24,99%
g (51%) g (51%) = 26,01%
b (49%) g (51%) = 24,99%
b (49%) b (49%) = 24,01%

bb out of (bg/gb/bb) = 24,01 out of 73,99 = 32,45%

Actually I think, that it indeed makes a difference, if in the second question we are informed if the son presented is the older or younger one, that reducing the possibilities

bb/bg/gb/gg to bb/bg

(if the son is the older one..), where the probability would be 49%.

Given  the "painfully easy" example, it would make a difference if the information given is "One of the coins shows head" or "The penny shows head". In the second case, the likelihood for the dime showing head would be 50%.

PS: Maybe we also should take care of the fact, that a certain percentage of all childs are identical twins, which also have the same sex. ;-)

Dagobert
Monday, September 06, 2004

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