Two kids, with a nod to Teanchi and Beanchi

[bonanova]

Ned, Red, Ted and Zed are identical quadruplets alike in every way except one: the way that they describe the children in families that have two children. 

1. Ned says one is a boy, if that is a true statement; otherwise he says one is a girl.

2. Red says one is a girl, if that is a true statement; otherwise he says one is a boy.

3. If the older child is a boy, Ted says one is a boy; otherwise he says one is a girl.

4. Zed flips a coin and considers the taller (heads) or shorter (tails) child.

If the coin-selected child is a boy, he says one is a boy; otherwise he mentions the sex of the other child.

One of the four men, we don't know which, then tells us:

"Ok, so Teanchi and Beanchi are a married couple (dont ask me whose he and whose she)!  

They have two kids, one of them is a girl."

What is the probability that the other child is a girl?

[bhramarraj]

[1] If Ned said it, then it is not a true statement, and one child is not a girl. So probability for another child being a girl: 0%.

[2] If Red said it, then it is a true statement and one child is a girl. Then probability for other child being a girl: 50%.

[3] If Ted said it, then older child is not a boy but a girl. So probability for other child being a girl: 50%. [4] If Zed said it, and if coin selected child is taller, then the taller child is not a boy but a girl, as well as the other child.. Hence the probability for other child being a girl: 100%.

Similarly if the coin selected child is shorter, then the shorter child is not a boy, but a girl, as well as the other child. Probability for other child being a girl: 100%

[CaptainEd]

Ned only mentions a girl if there is no boy, so if it's Ned, two girls. P(other is a girl) = 1.

Red mentions a girl if there is a girl, so if it's Red, P(other is a girl) is 1/2

Ted names the older child. if he says girl, P(other is a girl) is 1/2

Zed only mentions a girl if the coin-selected child is not a boy AND the other kid is a girl. So, P(other is a girl) = 1

So, assuming chance of speaking to any one of these guys is 1/4, then expected probability that the other kid is a girl is average of 1, 1/2, 1/2, and 1 => 3/4

[Molly Mae]

1. Ned says one is a boy, if that is a true statement; otherwise he says one is a girl.

[1] If Ned said it, then it is not a true statement, and one child is not a girl. So probability for another child being a girl: 0%.

I think your [1] should be 100%. Ned says one is a boy if at least one is a boy. The only time he would say one child is a girl is if both children were girls, leaving 100% that the other was a girl.

Edited July 16, 2012 by Molly Mae

[psykomakia]

Ned says one is a boy, if that is a true statement; otherwise he says one is a girl.

Does this mean that he will say "One is a boy" if and only if there is one boy and one girl, or also if there is one boy and another boy?

[bonanova]

Does this mean that he will say "One is a boy" if and only if there is one boy and one girl, or also if there is one boy and another boy?

If either is a boy.

For our purposes, "One is a boy" and "At least one is a boy" are equivalent statements.

[Yoruichi-san]

I don't think you can say there is equal chance of the guy who said 'girl' being any of the 4 brothers.

The easiest example of why:

You have two bags, one with only red balls, one with only blue balls. You randomly pick a bag and draw out a ball. If it is red (or blue), you cannot say there was equal probability it came from either bag.

4/7

Did this kind of quick, someone should double check my numbers, but, assuming there is equal chance of a child being boy or girl:

1) Ned only says girl if it's GG, so there is 1/4 chance of him saying girl, and then the other has prob of 1 of being girl.

2) Red say girl if GG, GB, BG, so there is 3/4 chance of him saying girl, and there is 1/3 chance of other being girl.

3) Ted says girl if GG, GB, there is 1/2 chance of him saying girl, and 1/2 chance the other is girl.

4) Zed says girl if GG only, there is 1/4 chance of him saying girl, and prob 1 of other being girl.

(3 and 4 have slightly more complicated explanations to be entirely accurate, but I think that's good enough to give the drift)

So there is 1/4*(1/4+3/4+1/2+1/4) = 1/4*(7/4)=7/16 chance of saying girl. If the person does say girl, then (omitting or 'factoring out' the X1/4 for simplicity):

1) There is (1/4)/(7/4)=1/7 chance it was Ned.

2) There is (3/4)/(7/4)=3/7 chance it was Red.

3) There is (1/2)/(7/4)=2/7 chance it was Ted.

4) There is (1/4)/(7/4)=1/7 chance it was Zed.

So the expectation for the other being a girl is:

1/7*1+3/7*1/3+2/7*1/2+1/7*1=4/7

Edit: Bah, formatting...why does it ignore my spacing, etc after I click okay in the spoiler screen?

Edited July 16, 2012 by Yoruichi-san

[bushindo]

Thought I had a question, but it turned out to be unnecessary. Agree with Yoruichi-San on the answer. Nice puzzle!

[CaptainEd]

Durn, bit by Bayes again...

[bonanova]

The OP intended to deny bias among reporters, since each can comment on any two-child family,

Even tho some statements are denied to some reporters for some families.

However ...

The greater intent of this puzzle was to cool the debate somewhat, housed in the Teanchi-Beanchi thread.

We don't know the algorithm used by the reporter, and now we understand that it matters.

Most of us "1/3" zealots assume the reporter was like Red.

But there is no basis for that [or any other] assumption.

Ned:

BB BG GB one is a boy

GG one is a girl

1

Red: 

GG GB BG one is a girl

BB one is a boy

1/3

Ted: first-mentioned is older child. 

BB BG one is a boy

GB GG one is a girl

1/2

Zed: first-mentioned is coin-selected child. 

BB BG one is a boy

GB one is a boy

GG one is a girl

1

Avg = (6+2+3+6)/6/4 = 17/24 = .7083

[bushindo]

The OP intended to deny bias among reporters, since each can comment on any two-child family,

Even tho some statements are denied to some reporters for some families.

However ...

The greater intent of this puzzle was to cool the debate somewhat, housed in the Teanchi-Beanchi thread.

We don't know the algorithm used by the reporter, and now we understand that it matters.

Most of us "1/3" zealots assume the reporter was like Red.

But there is no basis for that [or any other] assumption.

Ned:

BB BG GB one is a boy

GG one is a girl

1

Red:

GG GB BG one is a girl

BB one is a boy

1/3

Ted: first-mentioned is older child.

BB BG one is a boy

GB GG one is a girl

1/2

Zed: first-mentioned is coin-selected child.

BB BG one is a boy

GB one is a boy

GG one is a girl

1

Avg = (6+2+3+6)/6/4 = 17/24 = .7083

Some comments

Let A be the spoken statement "One of the children is a girl",

Let S stand for speaker,

The calculation highlighted in red (see quoted post) implicitly assumes that given the statement A, all speakers have the same chance of being the one who said it. In other words, the implicit assumption is

P( S = Ned | A ) == P( S = Red | A ) == P( S = Ted | A ) == P( S = Zed | A )

If we put in the proper conditional probabilities, the equation becomes

1 * P( S = Ned | A ) + (1/3) * P( S = Red | A ) + (1/2) * P( S = Ted | A ) + 1* P( S = Zed | A )

= 1 * (1/7) + (1/3)* (3/7) + (1/2)*(2/7) + 1 * 1/7 = 4/7

[bonanova]

Did I forget to mention that uncle Jed had put their names in a hat and drew one of them to speak?

More interestingly, does the "One Boy, One Girl" thread any longer have a definitive answer?

[Rockmond]

If Ned said it, probability is 1 that it is GG;

If Red said it, it is either BG, GB, or GG, probability is 1/3 it is GG

If Ted said it, you know the older child is G, probability is 1/2 it is GG

If Zed said it, you know the coin selected child is G. He mentions sex of other child (G) so probability of GG is 1

1/4*1+1/4*1/3+1/4*1/2+1/4*1=17/24 = probability that it is GG.

[bushindo]

Did I forget to mention that uncle Jed had put their names in a hat and drew one of them to speak?

More interestingly, does the "One Boy, One Girl" thread any longer have a definitive answer?

If Ned said it, probability is 1 that it is GG;

If Red said it, it is either BG, GB, or GG, probability is 1/3 it is GG

If Ted said it, you know the older child is G, probability is 1/2 it is GG

If Zed said it, you know the coin selected child is G. He mentions sex of other child (G) so probability of GG is 1

1/4*1+1/4*1/3+1/4*1/2+1/4*1=17/24 = probability that it is GG.

Regarding these answers

[Yoruichi-san]

Lol...settled one debate and started another...

We were assuming the original choice was random, just like in my previous extreme example, the bag I drew the ball out of was randomly chosen. But after the result was observed, the probability of it coming from either bag was not equal.

In this case, the 4 brothers do not contribute equally to the probability that the "girl" is reported, hence when "girl" is reported, they are not equally likely to have spoken.

There are 16 possible cases, with (we assume) equal probability, each of the four brothers being picked to speak and each of the four possible child scenarios:

Ned: GG GB BG BB

Red: GG GB BG BB

Ted: GG GB BG BB

Zed: GG GB BG BB

There are 7 cases in which "girl" will be reported, those in cyan. Out of those 7 equally likely cases, Ned contributes 1, Red contributes 3, Ted contributes 2, and Zed contributes 1.

Hence when "girl" is spoken, there is 1/7 chance it was Ned, 3/7 chance it was Red, 2/7 chance it was Ted, and 1/7 chance it was Zed.

I do really like how the coin toss in number 4 is a "red herring"...it doesn't matter if you always pick tallest, use an unfair coin, or use some voodoo ritual...in the end it amounts to the same thing .

Edited July 17, 2012 by Yoruichi-san

[bonanova]

I totally get that the likelihood, or ease, of saying girl depends on the reporter's algorithm.

But if Jed drew their name from a hat, the four are equally likely to have been the reporter.

And then, likely or not, the OP tells us that the reporter said girl.

Repeating, we are not given that the reporter said girl and then asked which reporter is the most likely to have spoken.

The OP saying we don't know which one spoke was meant to be even stronger: we don't even have a clue.

So I need to understand better Bushindo's analysis.

[Yoruichi-san]

When the reporter drew the name out of the hat, it was equally likely to have been any of the 4 brothers, and it was equally likely to have been any of the 4 child pairings, hence it was equally likely to have been any of the 16 possibilities I outlined above.

But after it was announced girl, you can eliminate the cases where "boy" would have been reported. Therefore the remaining 7 cases are equally likely, but not equally distributed among the 4 brothers.

Even if we are not asked which reporter is likely to have spoken, it is still part of the problem of figuring out the likelihood of the other child being a girl.

The key is that the cases were "girl" is reported are equally likely, not the reporter.

Take my example with the balls and the bags above. Let me make it more similar to this problem by asking "If a draw a blue ball, what is the likelihood that the next ball will be blue?" Bag A has all red balls, bag B has all blue balls. We are told that we are drawn a blue ball, likely or not, but after you know you drew a blue ball, it is definitely NOT equally likely to have come from either bag. Saying it would would give a prob of (1/2)*0+(1/2)*1=1/2 for the next ball being blue, which is clearly erroneous. Bag B contributes 100% of the prob of getting blue balls, so if the ball was blue, it is 100% likely to have come from bag B, and hence the next ball drawn from the bag will 100% likely be blue.

Edited July 18, 2012 by Yoruichi-san

[Yoruichi-san]

Doh...had a revelation. Let's bypass the whole "which brother spoke" thing and just say...

There are 16 possible cases, with (we assume) equal probability, each of the four brothers being picked to speak and each of the four possible child scenarios:

Ned: GG GB BG BB

Red: GG GB BG BB

Ted: GG GB BG BB

Zed: GG GB BG BB

Since "girl" was spoken, we know it has to be one of the 7 cases in cyan above. 4 of these have the other child being a girl. Hence the probability of the other child being a girl is 4/7.

*Takes a deep breath* There, much simpler .

[bonanova]

I'm trying to see where my thinking has gone astray:

p(other is girl) =

p(other is girl if Ned spoke) x p(Ned spoke) +

p(other is girl if Red spoke) x p(Red spoke) +

p(other is girl if Ted spoke) x p(Ted spoke) +

p(other is girl if Zed spoke) x p(Zed spoke).

If the reporter had worn a name tag, we're done: we know p(other is girl if x spoke) for each x.

We're only uncertain about the name tag, it seems. Jed makes each tag probability 25.

Bushindo, I perceive an extra level of conditionals.

[bushindo]

I'm trying to see where my thinking has gone astray:

p(other is girl) =

p(other is girl if Ned spoke) x p(Ned spoke) +

p(other is girl if Red spoke) x p(Red spoke) +

p(other is girl if Ted spoke) x p(Ted spoke) +

p(other is girl if Zed spoke) x p(Zed spoke).

If the reporter had worn a name tag, we're done: we know p(other is girl if x spoke) for each x.

We're only uncertain about the name tag, it seems. Jed makes each tag probability 25.

Bushindo, I perceive an extra level of conditionals.

You're right, there is an extra layer of conditionals. We need to incorporate the information about what was said ("One of the kids is a girl") into the calculations as well.

The equation that you wrote below in green is correct mathematically. (I replaced your 'If' with 'Given' since it fits better into the Bayesian framework)

p(Other is girl) =

p(Other is girl Given Ned spoke) x p(Ned spoke) +

p(Other is girl Given Red spoke) x p(Red spoke) +

p(Other is girl Given Ted spoke) x p(Ted spoke) +

p(Other is girl Given Zed spoke) x p(Zed spoke).

However, the left hand side, P(Other is a girl), is not the solution to this puzzle. The probability that we want is the probability that "Other is a girl" given that a reporter (we don't know which one) said 'One kid is a girl'. That is, we want P( 'Other is a girl' Given 'One kid is a girl').

From my derivations in we can see that P( 'Other is a girl' Given 'One kid is a girl') can be expressed as

P('Other is a girl' Given 'One kid is a girl' ) =

p('Other is girl' Given [ Ned spoke and 'One kid is a girl' ]) x p(Ned spoke Given 'One kid is a girl')

p('Other is girl' Given [ Red spoke and 'One kid is a girl' ]) x p(Red spoke Given 'One kid is a girl')

p('Other is girl' Given [ Ted spoke and 'One kid is a girl' ]) x p(Ted spoke Given 'One kid is a girl')

p('Other is girl' Given [ Zed spoke and 'One kid is a girl' ]) x p(Zed spoke Given 'One kid is a girl').

Essentially, we don't know which reporter is the speaker, so take a weighted average over all reporters based on the conditional probabilities P( Reporter Given 'One kid is a girl'). Now let's examine values of these probabilities

Since uncle Jed drew the names out of the hat, we have

p(Ned spoke) = p(Red spoke) = p(Ted spoke) = p(Zed spoke) = 1/4

From Yoruichi-san's analysis, we have

p(Ned spoke Given 'One kid is a girl') = 1/7

p(Red spoke Given 'One kid is a girl') = 3/7

p(Ted spoke Given 'One kid is a girl') = 2/7

p(Zed spoke Given 'One kid is a girl') = 1/7

From the equation in green above, it seems that was solving for p(Other is girl Given Ned spoke) and p(Other is girl Given Red spoke) and so on. However, those probabilities are actually different conditionals. They actually are

p('Other is girl' Given [ Ned spoke and 'One kid is a girl' ]) = 1

p('Other is girl' Given [ Red spoke and 'One kid is a girl' ]) = 1/3

p('Other is girl' Given [ Ted spoke and 'One kid is a girl' ]) = 1/2

p('Other is girl' Given [ Zed spoke and 'One kid is a girl' ]) = 1

So I believe the mistakes in are as follows

1) It solves for a different quantity, P(Other is a girl) versus P(Other is a girl Given 'One kid is a girl'), with respect to the solution

2) It actually computed p('Other is girl' Given [ Ned spoke and 'One kid is a girl' ]) instead of p('Other is girl' Given Ned spoke) and so on.

[bonanova]

You're right, there is an extra layer of conditionals. We need to incorporate the information about what was said ("One of the kids is a girl") into the calculations as well.

The equation that you wrote below in green is correct mathematically. (I replaced your 'If' with 'Given' since it fits better into the Bayesian framework)

p(Other is girl) =

p(Other is girl Given Ned spoke) x p(Ned spoke) +

p(Other is girl Given Red spoke) x p(Red spoke) +

p(Other is girl Given Ted spoke) x p(Ted spoke) +

p(Other is girl Given Zed spoke) x p(Zed spoke).

However, the left hand side, P(Other is a girl), is not the solution to this puzzle. The probability that we want is the probability that "Other is a girl" given that a reporter (we don't know which one) said 'One kid is a girl'. That is, we want P( 'Other is a girl' Given 'One kid is a girl').

From my derivations in we can see that P( 'Other is a girl' Given 'One kid is a girl') can be expressed as

P('Other is a girl' Given 'One kid is a girl' ) =

p('Other is girl' Given [ Ned spoke and 'One kid is a girl' ]) x p(Ned spoke Given 'One kid is a girl')

p('Other is girl' Given [ Red spoke and 'One kid is a girl' ]) x p(Red spoke Given 'One kid is a girl')

p('Other is girl' Given [ Ted spoke and 'One kid is a girl' ]) x p(Ted spoke Given 'One kid is a girl')

p('Other is girl' Given [ Zed spoke and 'One kid is a girl' ]) x p(Zed spoke Given 'One kid is a girl').

Essentially, we don't know which reporter is the speaker, so take a weighted average over all reporters based on the conditional probabilities P( Reporter Given 'One kid is a girl'). Now let's examine values of these probabilities

Since uncle Jed drew the names out of the hat, we have

p(Ned spoke) = p(Red spoke) = p(Ted spoke) = p(Zed spoke) = 1/4

From Yoruichi-san's analysis, we have

p(Ned spoke Given 'One kid is a girl') = 1/7

p(Red spoke Given 'One kid is a girl') = 3/7

p(Ted spoke Given 'One kid is a girl') = 2/7

p(Zed spoke Given 'One kid is a girl') = 1/7

From the equation in green above, it seems that was solving for p(Other is girl Given Ned spoke) and p(Other is girl Given Red spoke) and so on. However, those probabilities are actually different conditionals. They actually are

p('Other is girl' Given [ Ned spoke and 'One kid is a girl' ]) = 1

p('Other is girl' Given [ Red spoke and 'One kid is a girl' ]) = 1/3

p('Other is girl' Given [ Ted spoke and 'One kid is a girl' ]) = 1/2

p('Other is girl' Given [ Zed spoke and 'One kid is a girl' ]) = 1

So I believe the mistakes in are as follows

1) It solves for a different quantity, P(Other is a girl) versus P(Other is a girl Given 'One kid is a girl'), with respect to the solution

2) It actually computed p('Other is girl' Given [ Ned spoke and 'One kid is a girl' ]) instead of p('Other is girl' Given Ned spoke) and so on.

Thanks. That makes sense now.