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# Daughters and Sons: CR4 Challenge (10/20/09)

Posted October 18, 2009 5:01 PM

This week's Challenge Question:

You have three colleagues: Arthur, Bill and Charlie. Each has two (and only two) children. Bill mentioned that today is his daughter's birthday and Charlie brought his daughter to work.

Assuming 50:50 male:female distribution and no further information about your colleagues or their families, what are the probabilities that each has a son?

Thanks to ExPat for providing us with this Challenge.

Arthur: 3/4

Bill: 2/3

Charlie: 1/2

Most won't have difficulty with Arthur and Charlie; Bill's case is less intuitive, at least until the problem is broken down.

Explanation:

Designate the two children and Child A and Child B. Each child can be a boy or a girl (B or G). For Arthur, there are four permutations, each with equal probability; three include at least one boy, so the probability that Arthur has a son is 3/4. Child A Child B Has Son?

Child AChild BHas son?
BBY
BGY
GBY
GGN

For Bill, we know that B-B is not an available solution, leaving three permutations, each with equal probability; two include a boy, so the probability that Bill has a son is 2/3. Child A Child B Has Son?

Child AChild BHas son?
BGY
GBY
GGN

For Charlie, we have already met one child (let's designate her Child A), so solutions B-B and B-G are unavailable, leaving two permutations, each with equal probability; one includes a boy, so the probability that Charlie has a son is 1/2. Child A Child B Has Son?

Child AChild BHas son?
GBY
GGN

(We could of course designate the known daughter as Child B, resulting in the same probabilities.)

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Check out these comments that don't yet have enough votes to be "official" good answers and, if you agree with them, rate them!
2
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#1

### Re: Daughters and Sons: CR4 Challenge (10/20/09)

10/19/2009 12:10 PM

There are four combinations for having two children – two sons, one son & one daughter, one daughter & one son, and two daughters. Three of the four combinations have a son, so the probability for Arthur having a son is 3/4 or 0.75. Since Bill and Charlie have a daughter, the combination of two sons is not possible, so there are only three possible combinations. Of these three combinations, only two have sons so the probability for Bill and Charlie to have a son is lower at 2/3 or 0.667. The probability that all three have a son is 0.75*0.667*0.667 or 0.333.

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#2

### Re: Daughters and Sons: CR4 Challenge (10/20/09)

10/19/2009 12:13 PM

Unless Charlie is short for Charlene and she is married to Bill. Where it says Charlie brought his daughter to work they could be saying Charlene brought (his) Bill's daughter to work (for her birthday).

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#95

### Re: Daughters and Sons: CR4 Challenge (10/20/09)

10/25/2009 6:40 AM

And then there are daughters who are not blood relatives and gay couples...

Anonymous Poster
#60

### Re: Daughters and Sons: CR4 Challenge (10/20/09)

10/21/2009 3:20 AM

2/3x1/2x1/2=1/6 (having a daughter and son is the same as having a son and a daughter). Each man only has 3 possibilities with out restrictions. But that only applies to Arthur for this question. Arthur can have 2 boys, or 2 girls or a boy and a girl Bill and Charlie can have a boy and a girl or 2 girls. Therefore Arthur has a 2/3 chance of having a boy, while Bill and Charlie have 1/2 a chance. So again, 2/3x1/2x1/2=2/12=1/6

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#3

### Re: Daughters and Sons: CR4 Challenge (10/20/09)

10/19/2009 12:39 PM

We know Charlie has a daughter. That leaves five children, three sons and two daughters. If we look at the probability that Arthur has no sons, it is 2/5 times 2/5 or 4/25 or 16%. The probability that Arthur has a son is 84%. Same for Bill. The probability that Charlie has a son is 2/5 or 40%. The probability of all three having a son is .84 time .84 time .4 = .282 = 28.2%.

Thanks,

Jim

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#4

### Re: Daughters and Sons: CR4 Challenge (10/20/09)

10/19/2009 1:07 PM

I actually interpreted the "assume 50:50 male:female distribution" to mean equal odds for each child to be male or female, rather than the slightly skewed odds that are observed throughout the world. If instead it means that there are exactly 3 sons and 3 daughters, then the probabilities I posted aren't correct.

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#5

### Re: Daughters and Sons: CR4 Challenge (10/20/09)

10/19/2009 1:32 PM

Your interpretation was right first time (but was your solution?). I totally missed that ambiguity - sorry about that.

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#6

### Re: Daughters and Sons: CR4 Challenge (10/20/09)

10/19/2009 2:58 PM

Hmmm – having a daughter eliminates the possibility of two sons, but does not eliminate it as one of the probabilities. So the probability of having a son and daughter (and thus at least one son) is 2/4 or 0.50, instead of 0.667 as I originally posted. This would change the probability of all having a son to 0.75*0.5*0.5 or 0.188. (Unless perchance Bill and Charlie are married, in which case there are only 4 children and the probability for Arthur & Bill/Charlie both having sons is 0.75*0.50 or 0.375)

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#20

### Re: Daughters and Sons: CR4 Challenge (10/20/09)

10/20/2009 5:39 AM

Yes, his (first) solution was correct also.

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#41

### Re: Daughters and Sons: CR4 Challenge (10/20/09)

10/20/2009 2:49 PM

The more I think about it, the more I also believe my first answer was correct. For two children, there are four combinations - SS, SD, DS & DD with equal probability for each (25%). For ideal distribution, if there are 1,000,000 pairs of children then there would be 250,000 of each combination, with the SD & DS combining to form one group of 500,000. Thus, the distribution for SS/(SD/DS)/DD would be 250K/500K/250K.

Since Bill & Charlie have one daughter, that eliminates the possibility of SS, so they must have one of the three remaining SD, DS or DD. If there is a 50% probability that each has a son, then there would be a 50% probability of a second daughter as well. This would distribute all pairs evenly between SD/DS and DD. If we have the ideal 1,000,000 pairs above and distribute the 750K pairs that have at least one daughter using this probability, there would be 375,000 with DD and 375,000 with (SD/DS), or 250K/375K/375K for SS/(SD/DS)/DD. This does not agree with the 250K/500K/250K distribution above. Therefore, I think the 66.7% probability for Bill and Charlie to have a son is actually correct, since two out of three pairs that have a daughter also have a son.

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#47

### Re: Daughters and Sons: CR4 Challenge (10/20/09)

10/20/2009 4:37 PM

Don't give up on me Dac.

I argue like this--> After revealing a D there can be no SS in the family

Da Db

Dc Sd

Se Df

S S

------------------------------------ Then

Take to work Da: leaves Db at home.

Take to work Db: leaves Da at home

Take to work Dc: leaves Sd at home

Take to work Df: leaves Se at home

Those are the only 4 D to take to work, and they have left S at home twice and D twice. This is of course a 50% split for a S in the family

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#57

### Re: Daughters and Sons: CR4 Challenge (10/20/09)

10/20/2009 5:59 PM

I won't give up on ya, but I don't think this is quite right.

Take to work Da: leaves Db at home.

Take to work Db: leaves Da at home

These are not independent events that determine the probability of having a son. The establishment of the childrens' sexes was determined long before it was their birthday or before they were brought to work. As far as I see it, the only purpose in stating this information is to establish that both Bill and Charlie have at least one daughter each - nothing about those two statements contributes any additional information as to the probability of having a son. There might be some wordplay there that could change the interpretation of who is whose daughter, but that has no relevance to the probability determination.

Probability can be defined as the ratio of the number of actual occurrences to the number of equally possible occurrences. For any given pair of children, there are four equally possible occurrences - son/son, son/daughter, daughter/son & daughter/daughter. Thus the probability for having two sons is 25%, one son and one daughter is 50%, and two daughters is 25% (not 33% each as some have posted). This higher weighting for one son & daughter is what skews the probability towards Bill and Charlie having a son, since there are two actual occurrences that have a son for three equally possible occurrences (son/son occurrence is not possible).

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#9

### Re: Daughters and Sons: CR4 Challenge (10/20/09)

10/19/2009 4:18 PM

I misread the challenge question (for the umpteenth time this year). I only saw that one daughter was known and not two. Not only that but I made the wrong assumption that there were three sons and three daughters instead of a 50:50 ratio in the general population. Let my try again.

We know Bill and Charlie each has at least one a daughter. That leaves 4 children that we do not know whether they are sons or daughters. If we look at the probability that Arthur has no sons, it is .5 times .5 or .25 or a probability of .75 that he does have at least one son. The probability that Bill has a son is .5 and the probability that Charlie has a son is .5. The probability that all three have sons is .75 time .5 times .5 or .1875 or 18.75%. Close to my previous but wrong answer.

Thanks,

Jim

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#7

### Re: Daughters and Sons: CR4 Challenge (10/20/09)

10/19/2009 3:20 PM

Bill has a daughter, the child left at home will be a brother or sister equally from a choice of two cases .... chance of son 50%.

Charlie has a daughter ...... same logic.... 50%

Arthur is unaffected by the other two .... 75% (interpret as at least one son; not a single son)

We aren't asked the odds for the combination for all three to have a son, but it is 18.75%

.....................................

If however Arthur, Bill and Charlie are Grandfather; Father and Son the chances could be A .... 100%; Bill..... 100%; Charlie ... 75%

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#87

### Re: Daughters and Sons: CR4 Challenge (10/20/09)

10/24/2009 6:34 AM

OK so you say that 50:50 distribution means for example if one has 10 children of which 9 we know are girls for the 10th its 50:50 probability to be boy or girl. Must be a shortcirquit in your brain and others that say similar things. Sorry!

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#88

### Re: Daughters and Sons: CR4 Challenge (10/20/09)

10/24/2009 9:47 AM

So what do you think the probabilities are? Girl more likely, as the couple might have a predisposition to produce girls, or boy more likely as there have been 9 girls? The latter is definitely wrong - if you toss a fair coin and get 9 heads in a row (it could happen and it's easy enough to calculate the odds) the probability of heads on the 10th throw is still 0.5. And what are you sorry about?

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#89

### Re: Daughters and Sons: CR4 Challenge (10/20/09)

10/24/2009 12:49 PM

Predispositions on a math model? OK let's say there are. Did you calculate them and come up with 0.5? My example was about a family with the same distribution as the challenge (50:50) and exagerated to be easier to understand.The possibility of a 10 children family (in a 50:50 boy-girl distribution belonging family) to have 9 girls is of course really slim but not zero. And presuming that that this slim possibility just happened in this family model I'm asking you about possibilities for the next child. The last 9 children have nothing to do with next's possibility? And if not OK if yes how? (Don't tell me predispositions again I'll start shouting).

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#99

### Re: Daughters and Sons: CR4 Challenge (10/20/09)

10/25/2009 1:14 PM

It's not predisposition on a maths model, it's predisposition of a particular couple to be more likely to produce girls than boys.

And you haven't answered the question - do you think the 10th child is more likely to be a girl or a boy?

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#102

### Re: Daughters and Sons: CR4 Challenge (10/20/09)

10/25/2009 5:42 PM

Please read twice my post No 91. Can't explain much better that this question is missing the point. We don't know the first nine born are girls and try to estimate the possibilities for the unborn tenth. We just know nine of the born ten are girls that's all. Can't see the difference? OK.. Thanks for your attention.

Anonymous Poster
#103

### Re: Daughters and Sons: CR4 Challenge (10/20/09)

10/25/2009 6:37 PM

We can only make a reasoned estimate of a specific probability for the tenth if we have some knowledge of the distribution of boy/girl fertility in the population at large. But we certainly have evidence that this can change with time and circumstance; a couple of examples are that there tends to be an increase in the proportion of males conceived following a war, and that female survival rates are higher during times of famine or disease. What we do know is that, if we look at couples who have girls as their first three children, the proportion of girls that are produced as the fourth child is significantly higher than in the population at large. That is not a predisposition in a basic model, but a matter of human variability and observed effects.

Fortunately, the terms of the challenge mean that we are not required to consider such complexities - even if we could find data to support such attempts.

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#90

### Re: Daughters and Sons: CR4 Challenge (10/20/09)

10/24/2009 2:50 PM

If you flip a coin nine times and it comes up tails each time, assuming a perfectly balanced coin, what is the probability it will come up tails the tenth time?

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#91

### Re: Daughters and Sons: CR4 Challenge (10/20/09)

10/24/2009 6:15 PM

Let me rephrase. You're not flipping a coin. you're flipping a coin twice. So the same possibility options are (H,H),(H,T),(T,H),(T,T). If one of two flippings (NOT NECESARILY THE FIRST THAT'S WHERE THINK YOU GET CLOGGED) we know (for example) is T (Tails) it does not eliminate two possibilities it eliminates only one (H,H) The possibility of (H,T) is doubled by (T,H) because results series of happening is irrelevant. Can you see the flaw?

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#104

### Re: Daughters and Sons: CR4 Challenge (10/20/09)

10/25/2009 7:52 PM

Not true.

For the purpose of this analysis, the order in which the children are born or the coins flipped makes them different. (H,T) is not the same as (T,H).

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#108

### Re: Daughters and Sons: CR4 Challenge (10/20/09)

10/26/2009 4:48 AM

This looks like semantics to me: different, but equivalent. So treating them as a single result but with double the probability is perfectly valid - albeit harder to follow and prone to errors whenever additional constraints need to be added

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#94

### Re: Daughters and Sons: CR4 Challenge (10/20/09)

10/25/2009 6:31 AM

Unfortunately, it would appear that the logic is wrong.

I'll assume for now that the 50:50 of the challenge refers to the general population and that there are no correlations between the sexes of the children within families. Then for a family with two children there are four possibilities - each with equal probability (first-born written first): boy-boy, girl-boy, boy-girl, and girl-girl. You can see that families with at least one daughter have a 2/3 probability of also having a son.

If the 50:50 ratio was intended to mean that the three families had three sons and three daughters in total, the numbers would be slightly different - but the principle stands.

BTW, I can't quite get my mind around your verbal acrobatics - what is the difference between "that each (of Arthur, Bill and Charlie) has a son" and "that all three (of Arthur, Bill and Charlie) have a son"?

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#8

### Re: Daughters and Sons: CR4 Challenge (10/20/09)

10/19/2009 3:51 PM

I think the men's names are obfuscation, as are the mentioned daughters. Since we don't know the children's names or birth order it boils down to:

Each has a one in three chance of having a girl and a boy. (1/3)3=1/9. One chance in nine.

I found it simpler to think of a bag containing an infinite number of both black and white socks.

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#10

### Re: Daughters and Sons: CR4 Challenge (10/20/09)

10/19/2009 9:28 PM

If we know that Bill has 1 daughter and Charlie has 1 daughter then there are 4 children whose gender is unknown. This is a total of 16 combinations with each combination having same probability since sons and daughters are equally likely. Of those 16 combinations only 3 result in at least 1 son for each colleague. So the probability of each having at least 1 son is 3 in 16 or 18.75%. If we are asking for the probability of each having exactly 1 son then there are 2 combination with this result and so the probability of this is 2 in 16 or 12.5%.

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#11

### Re: Daughters and Sons: CR4 Challenge (10/20/09)

10/19/2009 9:44 PM

My first shot was off because I neglected the two known daughters.

Bill and Charlie each have a 50% chance of having a son. There are three combinations available for Arthur 2 boys, 2 girls or 1 of each. Two of the three combinations include a son

0.666 x 0.5 x 0.5 =0.1667

16.67%

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#12

### Re: Daughters and Sons: CR4 Challenge (10/20/09)

10/19/2009 10:15 PM

Bill and Charlie have a daughter each, as there are three sons, one of them must also have a son, this leaves Arthur having either two sons or a daughter and son, and the same for either Bill or Charlie, as only two families are involved the odds are 50/50.

Regards JD.

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#14

### Re: Daughters and Sons: CR4 Challenge (10/20/09)

10/19/2009 11:14 PM

I'm with you on this one, JD. Unless of course:

Bill mentioned that today is his daughter's birthday and Charlie brought his daughter to work.

If Charlie brought Bill's daughter to work (because it was her birthday?), then that would mean that Bill could have two daughters OR Charlie could have two daughters. Between them, they could have only 3 at the most and 1 at the least.

 Dad Sons Daughters Arthur 1 1 Bill 1 1 Charlie 1 1 Dad Sons Daughters Arthur 1 1 Bill 1 1 Charlie 0 2 Dad Sons Daughters Arthur 1 1 Bill 0 2 Charlie 1 1 Dad Sons Daughters Arthur 1 1 Bill 1 1 Charlie 2 0

This looks like 25% to me.

However, if Charlie and Bill each have a daughter, then it becomes:

Either of them might have two daughters as opposed to one, so the probability is 50%.

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#16

### Re: Daughters and Sons: CR4 Challenge (10/20/09)

10/19/2009 11:46 PM

Hi Mikerho

Yes I did not pick up on the fact that Bill's daughter could work at the same place as Charlie, or that Charlie may pass by where she works, meaning we are referring only to one daughter? But I also took 50:50 female/male distribution to mean three sons and three daughters, and as such at least one family has a son and daughter leaving a 50/50 chance there is a son in both of the other two? But you have given me something to think about.

Regards JD.

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#23

### Re: Daughters and Sons: CR4 Challenge (10/20/09)

10/20/2009 8:24 AM

Bill mentioned that today is his daughter's birthday and Charlie brought his daughter to work.

Maybe it meant that Bill has a daughter and Charlie has a daughter. Which would make your chart look like this.

 Dad Sons Daughters Arthur 1 1 Bill 1 1 Charlie 1 1 Dad Sons Daughters Arthur 1 1 Bill 0 2 Charlie 2 0 Dad Sons Daughters Arthur 0 2 Bill 1 1 Charlie 2 0

But, do we take Bill, Arthur, and Charlie into account for the 50:50 male:female ratio?

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#83

### Re: Daughters and Sons: CR4 Challenge (10/20/09)

10/22/2009 6:14 PM

Reading down all the answers given so far, this one is clearly the best. It exactly points out the possibilities. Lets build on that.

There are only three possible scenarios, as per the list.

And as is clearly seen, only the first provides the case that each family has a son. 1 out of three is clearly a 1/3 change that each family has a son. Simple really, once the table is available.

The answer is a 1/3 change.

Your last comment likely does not apply. If it does it changes the picture. However, alone the fact that we would have 9 people, six children and 3 adults there could not be an "even" or 50:50 distribution. I surely trust the question did refer to the 50:50 of the children only.

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#13

### Re: Daughters and Sons: CR4 Challenge (10/20/09)

10/19/2009 11:06 PM

33%

Two of the six children are daughters. That's 33.33% of the total children. Of the remaining children, there is a 50% chance of the remaining 66.66% being daughters, which is 33.33%

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#15

### Re: Daughters and Sons: CR4 Challenge (10/20/09)

10/19/2009 11:22 PM

A B C

SS -- --

SD SD SD

DS DS DS

DD DD DD

3/4 x 2/3 x 2/3 = 1/3

(In each case, both birth orders SD and DS must be counted.) (For B and C, SS is ruled out by the given conditions.)

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#17

### Re: Daughters and Sons: CR4 Challenge (10/20/09)

10/20/2009 3:01 AM

What if Arthur is the father of Bill and Bill and Charlie are married?

Bill and Charlie have 1 daughter together. Is Bill a boy and Charlie a girl, or the other way round? Or are Bill and Charlie both boys or both girls?

Then it is also interesting to know: Is Arthur the father of Bill or the father of Charlie?

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#19

### Re: Daughters and Sons: CR4 Challenge (10/20/09)

10/20/2009 4:05 AM

My wifes sister is named Billy.but today is his daugther birthday.

Regards JD.

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#18

### Re: Daughters and Sons: CR4 Challenge (10/20/09)

10/20/2009 3:35 AM

Hi All

Let daughter = 0 and son =1 Also let Arthur = A , Bill = B and Charlie = C

Then the combinations we can have as follow

A_____B____C

01___01___01

11___00___01

11___01___00

So for A to have at least a son = 100% , For B = 2/6 , For C = 2/6

So for all of them to have at least each a son is (2*2)/(6*6) = 1/9 = 11.111...%

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#37

### Re: Daughters and Sons: CR4 Challenge (10/20/09)

10/20/2009 12:38 PM

Hi all

If there are 3 sons and 3 daughters all together

A has sons 100% of his combinations

B has sons 2/3 of his combinations

C has sons 2/3 of his combinations

So posibillity for all to have at least a son is 1x2/3x2/3 = 4/9 = 44.44%

Hope for the best

Carel

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#21

### Re: Daughters and Sons: CR4 Challenge (10/20/09)

10/20/2009 6:31 AM

Assuming 50:50 male:female distribution and no further information about your colleagues or their families, what are the probabilities that each has a son?

If we are dealing with only the children and not the other members of the the family which have been excluded, and a 50/50 male/female distribution means three sons and three daughters then there are only two combinations, (DD,SD,SS) and (SD,SD,SD), any odds arriving at this distribution are pure speculation.

Regards JD.

PS. and I don't think this is off topic, but sons and daughters will be children.

Anonymous Poster
#24

### Re: Daughters and Sons: CR4 Challenge (10/20/09)

10/20/2009 8:46 AM

66.66 % (DD,SD,SS) is fine; you can't make it higher cause it would imply a probability of 100% wich is impossible to assume just by knowing that two guys have at least one daughter each.

Yahlasit

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#22

### Re: Daughters and Sons: CR4 Challenge (10/20/09)

10/20/2009 6:58 AM

This should be 1/4 ( chances of son out of 3 Male and 1 daughter) x 1/3( Out of 2 male and 1daughter) x 1/1

= 1/12 = 8.33 % chances that each will have a son.

shivaram

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#27

### Re: Daughters and Sons: CR4 Challenge (10/20/09)

10/20/2009 9:22 AM

Should be 3/4 (3 ways to pick a son out of 3 sons and 1 daughter) x 2/3 (2 ways to get a son out of 2 sons and 1 daughter). 3/4 x 2/3 = 6/12 = 50% chance of each having a son.

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#25

### Re: Daughters and Sons: CR4 Challenge (10/20/09)

10/20/2009 8:56 AM

All of the fathers are impotent. Their probability of having either a son or daughter is 0%.

However!

The milkman

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#26

### Re: Daughters and Sons: CR4 Challenge (10/20/09)

10/20/2009 9:12 AM

When Bill mentions that "today is his daughter's birthday" we can't tell whether he says "Today is my daughter's birthday" OR "Today is my daughters' birthday" since both versions are pronounced the same in the English language. If today is the birthday of his daughters (i.e. daughters' birthday - either twins or sharing a birthday) then he obviously has two daughters and no sons, resulting in

chances of having a son A = 75% ; B = 0% and C = 50%

If today is the birthday of his daughter (i.e. daughter's birthday)

chances of having a son A = 75% ; B = 50% and C = 50%; as in my previous post

Anonymous Poster
#44

### Re: Daughters and Sons: CR4 Challenge (10/20/09)

10/20/2009 4:14 PM

Hi SlideRuler;

The question was to be read, not heard and it clearly says his daughter's birthday; if Bill had another girl, then he would have to say: today is one of my daughters' birthday. Likewise, it states that Charlie brought his daughter (not one of his daughters) to work, revealing that he also has one boy, like Bill. So, your other partner also has to have one girl and one boy (to keep the 50/50 sex proportion).

Result: (deducted by how the question was phrased)

The probabilities that each has a son are 100%

Thankyou

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#46

### Re: Daughters and Sons: CR4 Challenge (10/20/09)

10/20/2009 4:34 PM

I started thinking along those lines as well, but I disagree the question read has an exclusivity to each Bill and Charlie. It can be said 'today is my daughter's birthday', without excluding the possibility of another female child. Likewise, if it is known I brought my daughter to work today, that does not preclude another girl child.

I agree, '... his daughter's birthday' has the singular posessive apostrophe. However, the question is about probabilities, not grammar. In a previous post, I disregarded the two possibilities of Arthur having combinations of S/D and D/S, lumping them together as one. The question is not necessarily possibilities, but probabilities.

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#48

### Re: Daughters and Sons: CR4 Challenge (10/20/09)

10/20/2009 4:38 PM

Hope you're right else we're on a kindergarden forum

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#50

### Re: Daughters and Sons: CR4 Challenge (10/20/09)

10/20/2009 4:59 PM

If you "hope", then you're not sure, if you're not sure, then the one on a kindergarden forum should be you.

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#51

### Re: Daughters and Sons: CR4 Challenge (10/20/09)

10/20/2009 5:08 PM

Was kind of sure till I saw comments like yours Yah

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#52

### Re: Daughters and Sons: CR4 Challenge (10/20/09)

10/20/2009 5:20 PM

I have to apologize for confusing your mind, it was not my intent.

Yah

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#64

### Re: Daughters and Sons: CR4 Challenge (10/20/09)

10/21/2009 1:07 PM

Apology accepted Mr H

Anonymous Poster
#53

### Re: Daughters and Sons: CR4 Challenge (10/20/09)

10/20/2009 5:28 PM

Or perhaps me ? I don't know anymore, I'm confused.

Yahlasit

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#65

### Re: Daughters and Sons: CR4 Challenge (10/20/09)

10/21/2009 1:09 PM

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#28

### Re: Daughters and Sons: CR4 Challenge (10/20/09)

10/20/2009 9:31 AM

It should be 3/4( chances of one oboy out of 3 boys and one daughetr) x 2/3( chances of one boy out of 2 boys and one daughter) x 1 ( one and one daiugher) =

1/2

That is 50 % probability that each one will have one son out of two children in the given condition

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#29

### Re: Daughters and Sons: CR4 Challenge (10/20/09)

10/20/2009 10:02 AM

If I read: "Bill mentioned that today is his daughter's birthday and Charlie brought his daughter (Bill's daughter) to work." Then there is only spoken of the fact that Bill has a daughter.

I assume that all 3 men have a wife.

This means that the following family make up can be possible:

 1 arthur d s bill d s charlie d s
 2 arthur d d bill d s charlie s s
 3 arthur s s bill s d charlie d d
 4 arthur s d bill d d charlie s s
 5 arthur s s bill d d charlie s d

So according to this schedule the chance of all having a son is : 20%

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#84

### Re: Daughters and Sons: CR4 Challenge (10/20/09)

10/22/2009 6:31 PM

In your list only table 1, 3 and 5 are correct possibilities as both Bill and Charlie have a daughter. Table 2 and 4 do not apply.

That leaves only one (table 1) of three possibilities or a 1/3 change. Same answer as above Nr. 23 by cingold.

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#30

### Re: Daughters and Sons: CR4 Challenge (10/20/09)

10/20/2009 11:12 AM

I think many are misreading "Assuming 50:50 male:female distribution" to mean there are exactly three each of daughters and sons, If that were true we would not have to assume it. I read it as saying that the general population is fifty fifty.

The interesting discussion for me is whether we have to count one or two boy/girl combinations. If we consider first and second borns, it appears that there are four possibilities, BB,GG,BG,and GB. On the other hand, if I think of reaching into a bag containing an infinite number each of boy coloured socks and girl coloured socks, and withdrawing just two socks, I find that BB, GG, and BG are equally likely. That was my basis for#11, above.

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#31

### Re: Daughters and Sons: CR4 Challenge (10/20/09)

10/20/2009 11:42 AM

Assuming 50:50 male:female distribution and no further information about your colleagues or their families, what are the probabilities that each has a son?

ExPat has already clarified that the 50:50 male female distribution means that there are three sons.

However, the question asks "what are the probabilities that each has a son?"

I would take that to mean that we are to assess the probablity 1. that Arthur has a son, 2. that Bill has a son and 3. that Charlie has a son. Is that correct?

Also, should we assume that Charlie brought Charlie's daughter to work and that Bill brought Bill's daughter?

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#32

### Re: Daughters and Sons: CR4 Challenge (10/20/09)

10/20/2009 11:48 AM

I agree and disagree.

ExPat clarified that it is the general population that is fifty fifty.

On the rest I agree.

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#35

### Re: Daughters and Sons: CR4 Challenge (10/20/09)

10/20/2009 12:25 PM

ExPat clarified that it is the general population that is fifty fifty.

Right you are!

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#33

### Re: Daughters and Sons: CR4 Challenge (10/20/09)

10/20/2009 12:09 PM

It is 50-50. There are 4 children left, 3 boys and one girl. The possible combinations of these taken two by two are six. B1B2, B1B3, B2B3, B1G, B2G AND B3G. The probability of Arthur getting a BG combination is 50-50, which also means the other two get the same.

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#34

### Re: Daughters and Sons: CR4 Challenge (10/20/09)

10/20/2009 12:24 PM

Assuming Charlie brought his own daughter to work (not Bill's daughter, the sentence is ambiguous)

Arthur 0.75 (= 1 - prob of 2 daughters = 1 - 0.52)

Bill 0.5 (we know he has a daughter and probability that his other child is a son = 0.5)

Charlie - same as Bill

Cheers..........Codey

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#36

### Re: Daughters and Sons: CR4 Challenge (10/20/09)

10/20/2009 12:30 PM

I agree, and have awarded you the cherished GA vote.

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#38

### Re: Daughters and Sons: CR4 Challenge (10/20/09)

10/20/2009 12:40 PM

When I first read the challenge I thought it might mean each in the sense of any one of them (the way various posters have answered it), but when I re-read it I couldn't see it that way. And it would have been "probability", whereas the challenge said "probabilities".

Cheers............Codey

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#85

### Re: Daughters and Sons: CR4 Challenge (10/20/09)

10/22/2009 6:45 PM

You give out a GA without knowing if the answer is correct and therefore to deserve one.

I beg to differ and thrust the answer is a 1/3 change as pointed out above. But ok. I will not be mean and take the GA away as to you the answer was good. We all are happy to receive and stay tuned as we should be.

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#39

### Re: Daughters and Sons: CR4 Challenge (10/20/09)

10/20/2009 12:59 PM

Since we know Authur must have a boy (because only one girl was not stated); you then have 3 unaccounted for children. 2 boys and a girl the chance of Authur having a girl is 1/3 or 33.3333%.

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#40

### Re: Daughters and Sons: CR4 Challenge (10/20/09)

10/20/2009 2:19 PM

As I see it, unless the writer is really bad (deliberately bad) at writing English:

Arthur Bill and Charlie are colleagues, so they know each other, and as such it would not be unusual for them to know each others families.

"Bill mentioned that it was his daughters birthday today", rather than calling her by name or begging the question "Which one?" The logical conclusion is that he has only one daughter.

The same follows for Charlie, who did not specify that he brought "one" of his daughters to work, but his daughter, implying only one daughter.

The 50:50 male:female distribution can only be interpreted as meaning that of the 6 childern 3 are boys and 3 are girls. As Bill and Charlie have only 1 daughter each, the third daughter must be Arthurs, and therefore Arthur Bill and Charlie also have a son each. So the probability that each of the 3 colleagues has a son is 100%.

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#42

### Re: Daughters and Sons: CR4 Challenge (10/20/09)

10/20/2009 3:22 PM

All right, here we go:

Arthur cannot have combination D/D, as this makes 4 daughters; thus Arthur has 100% probability of having at least one son.

Now, IF Art has combo D/S, then all three have 100% probability of one son; but IF Art has S/S, then Bill and Chuck each have 50% probability of one son... Thus:

Arthur: 100% probability of (at least) one son

Bill: [100% + 50%] / 2 opportunities = 75% probability of one son

Charlie: [100% + 50%] / 2 opportunities = 75% probability of one son

I hope some of this baloney is right...

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#43

### Re: Daughters and Sons: CR4 Challenge (10/20/09)

10/20/2009 3:49 PM

Concidering the 50:50 distribution Arthur has 25% chance to have 2 boys, 25%chance to have 2 girls and 50% chance to have one of each. Bill and Charlie equally have 2/3*100% chance to have 1 boy and 1 girl and 1/3*100% chance to have 2 girls (no chance to have 2 boys) So the chance for all tree to have (at least) ONE son each is 75%* 2/3*100%*2/3*100% =1/3*100% ~ 33.333333..... % and the chance for all three to have ONLY ONE son each is 50%*2/3*100%*2/3*100% =2/9*100% ~22.222222....%

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#45

### Re: Daughters and Sons: CR4 Challenge (10/20/09)

10/20/2009 4:24 PM

It's a classic problem, a variation of the Monty Hall riddle. The trick to understanding it is to identify each child in some way; typically this is done by age. This sounds irrelevant, but it isn't. Using this understanding, a distinction is made between "1 son 1 daughter" and "1 daughter 1 son" as two different permutations.

Each man has two children for a total of four possibilities:

Older.......S.....S.....D.....D

Younger...S.....D.....S.....D

Without further information, we could presume that each possibility is equally likely. In three of the four permutations, there is at least one son. Thus for Arthur, there is a 75% probability that he has a son.

For the other two men, one of the permutations has a probability of zero, and the other three are equally likely. Thus for Bill, the probability that he has a son is 67%. Likewise for Charlie.

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#49

### Re: Daughters and Sons: CR4 Challenge (10/20/09)

10/20/2009 4:40 PM

Hi Mitsurati;

Please don't take offense, but I've always believed that there is no knowledge or tool that replaces common sense, you should take a closer look to how the question is formulated, the naughty boy who wrote it concealed the answer within.

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#67

### Re: Daughters and Sons: CR4 Challenge (10/20/09)

10/21/2009 2:56 PM

Absolutely right on the Monty Hall reference. However (as implied by your use of the word "typically"), age isn't the only way of identifying the daughters.

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#72

### Re: Daughters and Sons: CR4 Challenge (10/20/09)

10/21/2009 7:07 PM

Nah, this don't fit the Monty Hall problem. The key to that is Monty knowing which door is correct. To match that, you'd have to have the guy deliberately bringing his daughter to work, for example.

Are you using the word "each" correctly? To say "..the odds of each having a son..." means every one having a son.

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#73

### Re: Daughters and Sons: CR4 Challenge (10/20/09)

10/22/2009 4:27 AM

Hello TVP45

If the challenge said (as your #72) "....what are the odds of each having a son?..." it could mean the odds for each (3 numbers) or the the odds of every one having a son (1 number). Because "odds" is both singular and plural in this context.

But it says "....what are the probabilities that each has a son?" which implies to me 3 separate probabilities.

Cheers......Codey

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#74

### Re: Daughters and Sons: CR4 Challenge (10/20/09)

10/22/2009 6:32 AM

Good point! I missed that ambiguity altogether. I will reread the question, get out my dice (no, not that pair - the fair ones), throw another eye of newt in the statistics pot, and try again. Thanks.

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#75

### Re: Daughters and Sons: CR4 Challenge (10/20/09)

10/22/2009 6:37 AM

And, having reread it, I see your point clearly, although there's the additional problem that I don't know whose daughter Charlie brought to work. However, skipping over such nit-picking, the probabilities for each are thus: 0.75, 0.75, 0.75.

Cheers.

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#54

### Re: Daughters and Sons: CR4 Challenge (10/20/09)

10/20/2009 5:32 PM

I just gave a GA to SlideRuler for #7.

I see my mistakes.

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#55

### Re: Daughters and Sons: CR4 Challenge (10/20/09)

10/20/2009 5:33 PM

The semantics of the statement provide two clues. Bill mentioned that today is his daughter's birthday. If he had more than one daugher he would have said that today was the birthday of one of his daughters. Charlie brought his daughter to work. If he had more than one daughter, then he would have brought one of his daughters to work. The implication is that Bill and Charlie each have one, and only one daughter. Therefore, Arthur has the third daughter. Therefore, there is a 100% chance that each has one daughter and one son.

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#56

### Re: Daughters and Sons: CR4 Challenge (10/20/09)

10/20/2009 5:37 PM

I don't agree. I have three daughters but I wouldn't bother to identify which one had a birthday unless the person I was speaking to knew the girls.

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#61

### Re: Daughters and Sons: CR4 Challenge (10/20/09)

10/21/2009 5:09 AM

Neither do I. When the challenge says "Assuming 50:50 male:female distribution....." I think it means the probability of each is 50:50, not that this particular sample of 6 children is split 50:50. The latter interpretation would give a trivial problem.

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#58

### Re: Daughters and Sons: CR4 Challenge (10/20/09)

10/20/2009 7:37 PM

Arthur, Bill and Charlie could be Son, Father and Grandfather (or even Father and 2 sons). Hey, it could happen !

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#59

### Re: Daughters and Sons: CR4 Challenge (10/20/09)

10/20/2009 9:00 PM

I couldn't figure out why there were so many different answers. Then I looked a little harder and I think we are solving many different problems. Even though the challenge question is short, it is subject to interpretation.

"You have three colleagues: Arthur, Bill and Charlie. Each has two (and only two) children".

I think this simply means there are six children, two for each colleague. Or does it?

"Bill mentioned that today is his daughter's birthday and Charlie brought his daughter to work."

This might mean Bill (father) and Charlie (mother) have a daughter and Charlie brought their daughter to work on her birthday. Or it might mean that Bill and Charlie are not related and Charlie brought Bill's daughter to work. Or it might mean Bill and Charlie are not related and they each have a daughter. So we might be considering four or six children. And we might know the sex of one or two children.

"Assuming 50:50 male:female distribution and no further information about your colleagues or their families"

I originally thought this meant that for each child whose sex is unknown, there was a 50:50 chance each was a son or a daughter. But it seems this might mean alternately mean that half are sons and half are daughters.

"What are the probabilities that each has a son?"

Does this mean that each person has exactly one son or at least one son?

I think we might have any of the following nine cases:

1) Six kids – three sons and three daughters – Bill and Charlie each have at least one daughter

2) Six kids – three sons and three daughters – Bill has at least one daughter

3) Six kids – four unknown and two known to be daughters - Bill and Charlie each have at least one daughter – each family has at least one son

4) Six kids – four unknown and two known to be daughters - Bill and Charlie each have at least one daughter – each family has exactly one son

5) Six kids – five unknown and one known to be a daughter - Bill has at least one daughter – each family has at least one son

6) Six kids – five unknown and one known to be a daughter - Bill has at least one daughter– each family has only one son

7) Four kids – two sons and two daughters – Bill and Charlie have at least one daughter as a couple

8) Four kids - three unknown and one known to be a daughter– Bill and Charlie have at least one daughter as a couple – each family has at least one son

9) Four kids - three unknown and one known to be a daughter– Bill and Charlie have at least one daughter as a couple – each family has only one son

Case 1) There is only one daughter who's parent is unknown. The empty slots are two for Arthur and one each for Bill and Charlie. Half the empty slots are for Arthur, either of which would result in a son for each parent. 50% chance all three have a son (and only one son).

Case 2) There are two daughters who's parents are unknown. The empty slots are two for Arthur, one for Bill, and two for Charlie. There is a 60% chance Bill will have a son (3 of 5). Given that Bill has a son and a daughter, this leaves two daughters and two sons. There is a 50% chance Arthur's first child is a son (2 of 4). If Arthur's first child is a son, there is 66.7% chance his second child is a daughter (2 of 3). Therefore, there is a 33.3% chance (50% times 66.7%) that his first child is a son and the second is a daughter. The same odds exist for the first child being a daughter and the second being a son. Summing these together, there is a 66.7% chance Arthur will have a single son if Bill has a single son. The product of these probabilities is 50% times 66.7% or 33.3% chance all three will have a (single) son.

Case 3) The probability that Arthur has no sons is 50% times 50% or 25%. This means there is a 75% chance Arthur has a son. For both Bill and Charlie, there is a 50% chance each unknown child is a son. That means the probability that all three has at least one son is 75% time 50% time 50% or 18.75%.

Case 4) There is a 50% chance that Arthur's first child is a son and a 50% chance his second child is a son. If the first child is a son, there is a 50% chance Arthur has two sons. Therefore, if Arthur's first child is a son, there is a 25% chance he has only one child. If Arthur's first child is a daughter (50%), there is a 50% chance his second child is a son. The probability that Arthur has only one son is 25% plus 25% or 50%. The chance Bill has one son is 50%. Same for Charlie. Probability that all three have only one son is 50% time 50% time 50% = 12.5%.

Case 5) Similar to Case 3) except the probabilities are 75% time 75% times 50% or 28.125%.

Case 6) Similar to Case 4), the chance each family has only one son is 50%. The chance all three families have only one son is 50% time 50% time 50% = 12.5%.

Case 7) This follows the same logic as the second part of case 2). If Arthur's first child is a son, there is 66.7% chance his second child is a daughter (2 of 3). Therefore, there is a 33.3% chance (50% times 66.7%) that his first child is a son and the second is a daughter. The same odds exist for the first child being a daughter and the second being a son. Summing these together, there is a 66.7% chance Arthur will have a single son leaving the other son for Bill and Charlie.

Case 8) The probability that Bill and Charlie's other child is a son is 50%. The chance that at least one of Arthur's two children is a son is 75% as in Case 3) above. The probability that each of Arthur, Bill, and Charlie has at least one son is 50% time 75% or 37.5%.

Case 9) The probability that Bill and Charlie's other child is a son is 50%. The chance that at exactly one of Arthur's two children is a son is 50% as in Case 4) above. The probability that each of Arthur, Bill, and Charlie has at least one son is 50% time 50% or 25%.

So the answer appears to be 66.7%, 50%, 37.5%, 33.3%, 28.125%, 25%, 18.75%, or 12.5% depending on how you interpret the challenge question.

Thanks,

Jim

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#63

### Re: Daughters and Sons: CR4 Challenge (10/20/09)

10/21/2009 9:56 AM

I notice that we collectively try to make just about all of these challenge questions into a Saint Ives type of puzzle, trying to insert hidden meaning into a grammatical slip-up.

I venture this is why there are so many different answers... Is there really a conspiracy out there?

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#62

### Re: Daughters and Sons: CR4 Challenge (10/20/09)

10/21/2009 7:35 AM

0.422

Cheers, and see ya in Vegas.

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#66

### Re: Daughters and Sons: CR4 Challenge (10/20/09)

10/21/2009 2:45 PM

I never knew how many different answers this would generate! What is undoubtedly true is that the answer depends on exactly what information is given or inferred. Different answers can be valid, depending on the inferences made from information conveyed, intentionally or otherwise, in the problem statement.

I would like to emphasize that this is not a trick question: no crucial facts are withheld (to do so would violate "assume ... no further information"). Furthermore, wording should be taken purely at face value: e.g. usages such as "his daughter" convey nothing about being an only daughter or otherwise, even though in everyday life one might make some inference about the choice of words.

Having re-read my question I realize that there is one ambiguity that I don't think anyone has picked up. My intent in mentioning her birthday was to indicate that Bill had a daughter, and if it hadn't been her birthday, I could have used some other tidbit of information to convey the same information. However, if it is interpreted as meaning "Bill has a daughter whose Birthday is today" the odds of Bill having a son are significantly changed.

So, following up on that thought, here's a bonus question: Dick also has exactly two children and his daughter* has a birthday in October. What are the odds that Dick has a son? Assume the probability of a child (boy or girl) having a birthday in any specified month is exactly 1/12. I used month rather than day just to keep the numbers manageable. I'll post the answer after the main solution is posted.

One further piece of information: all four answers are unique (at least for my interpretation of the problem statements).

By the way: Happy 95th Birthday to Martin Gardner, to whom I'm indebted; and to whom I hope I'm not doing too much of a disservice. (There's a lead there people, if not an actual clue.)

* I refer you again to the 2nd para, final sentence of this post.

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#68

### Re: Daughters and Sons: CR4 Challenge (10/20/09)

10/21/2009 4:24 PM

I think you just made things worse. I don't know how to square:

"Furthermore, wording should be taken purely at face value: e.g. usages such as "his daughter" convey nothing about being an only daughter or otherwise, even though in everyday life one might make some inference about the choice of words."

with:

"One further piece of information: all four answers are unique (at least for my interpretation of the problem statements)."

p.s. Martin Gardner is God, I join you in your birthday wish.

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#69

### Re: Daughters and Sons: CR4 Challenge (10/20/09)

10/21/2009 5:08 PM

FWIW, this relates to what we are trying to figure out here.

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#70

### Re: Daughters and Sons: CR4 Challenge (10/20/09)

10/21/2009 5:24 PM

Bingo! You found me out!

I didn't actually see that specific article till after I submitted the question, but I did later. It confirmed my nagging doubt about seemingly superfluous information altering the possible outcomes, and gave me the basis for the bonus question.

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#71

### Re: Daughters and Sons: CR4 Challenge (10/20/09)

10/21/2009 5:28 PM

Now I see why I couldn't make up my mind about the answer

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#76

### Re: Daughters and Sons: CR4 Challenge (10/20/09)

10/22/2009 10:59 AM

Kudos !

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#77

### Re: Daughters and Sons: CR4 Challenge (10/20/09)

10/22/2009 11:30 AM

GA for the find DAC!

Now all that's left is to solve the problem (and the Bonus one)!

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#78

### Re: Daughters and Sons: CR4 Challenge (10/20/09)

10/22/2009 1:33 PM

After reading the boy-girl paradox, I can see some of the reasons we are all getting different answers. Some consider that revealing one daughter essentially "fixes" one of the children, so the probability of the other child being male is even (0.50). Others (myself included) consider that the daughter could apply to either child, which expands the combinations that could include a son and increases the probability of the other child being a son to 0.67.

However, by providing additional information about Bill's daughter (birthday is today) and Dick's daughter (birthday is in October) you have pushed this into a Frequentist approach. I had a hard time understanding how the birthday or birth month of the daughter can affect the probability of the gender of the other child, but I finally realized that this approach is just proportioning the probability between 0.50 and 0.67 based on the probability of the additionally supplied information. In the case of "birthday is today", the probability for this is fairly low (1/365) which provides more constraint ("fixes") to the daughter and pushes the probability of the other child being a son towards 0.50. In the case of "birthday in October", this is a more probable event (1/12) so there is less of a constraint on the daughter, which pushes the probability of the other child being a son up towards 0.67.

Based on my interpretation of the Frequentist approach (I think there are errors in the example, so hopefully I did this right), the probability for Bill's other child to be a son is 0.5003 (365+365)/(365+365+1+364+364) while the probability of Dick's other child to be a son is 0.5106 (12+12)/(12+12+1+11+11). The probability for Arthur has been well-established to be 0.75, and I am going to stick with 0.67 for Charlie.

 Bill Total Child Pairs 532900 S S 133225 S D w/BD 365 S D wo/BD 132860 D w/BD S 365 D wo/BD S 132860 D w/BD D w/BD 1 D w/BD D wo/BD 364 D wo/BD D w/BD 364 D wo/BD D wo/BD 132496
 Dick Total Child Pairs 576 S S 144 S D w/BD 12 S D wo/BD 132 D w/BD S 12 D wo/BD S 132 D w/BD D w/BD 1 D w/BD D wo/BD 11 D wo/BD D w/BD 11 D wo/BD D wo/BD 121
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#79

### Re: Daughters and Sons: CR4 Challenge (10/20/09)

10/22/2009 2:06 PM

(I think there are errors in the example, so hopefully I did this right)

Upon further review, I think the error was in my interpretation of the problem statement - gotta pay attention to wording on problems like these

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#80

### Re: Daughters and Sons: CR4 Challenge (10/20/09)

10/22/2009 3:10 PM

DAC: Excellent analysis BUT

"and I am going to stick with 0.67 for Charlie"

Can't all daughters be split into "Daughters which were brought ro work" (Dbtw) and "Daughters not brought to work" (Dnbtw) in just the same way as the Birthday split. You solve for sratio 1/12 yields .5106: 1/365 yields .5003 and now we have 1/VeryLargeNumber yields ?

Following the tables the answer will be (VLN+VLN)/(VLN+VLN+!+(VLN-1)+(VLN-1)) which obviously tends to .5000

This gives a final answer A =.75; B =.5003; C =.5000 (D =.5106)

SLR

Associate

Join Date: Oct 2009
Posts: 46
#81

### Re: Daughters and Sons: CR4 Challenge (10/20/09)

10/22/2009 3:32 PM

Good Answer from me for actually improving on the "official" solution. I think my first - and intended - interpretation of Bill's case (Post #66) is just about defensible, but I have to confess that your choice of the second is easier to justify.

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#82

### Re: Daughters and Sons: CR4 Challenge (10/20/09)

10/22/2009 5:44 PM

Yeah - I suppose that's possible ... good catch!

Guru

Join Date: Apr 2007
Posts: 3531
#126

### Re: Daughters and Sons: CR4 Challenge (10/20/09)

10/27/2009 6:11 PM

Agreed - except that EXPAT specified in an addendum that the birthday was post-hoc, which avoids the problem. (I have at least one daughter followed by "when was your older daughter born" must give some date - but does not change the probabilities).

there was more on the requirements in post #97...

Guru

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Posts: 3531
#127

### Re: Daughters and Sons: CR4 Challenge (10/20/09)

10/27/2009 6:18 PM

EXPAT specified in an addendum that the birthday was post-hoc, which avoids the problem. (I have at least one daughter followed by "when was your older daughter born" must give some date - but does not change the probabilities).

The frequentist approach is only justified if there is some preconstraint that sets the child's birthdate - such as "you may only tell us the sex of a child of is the child's birthday".

It turns out this is a much more subtle challenge than first appeared - I've (retrospectively I suppose) given the thread a 5* rating.

Guru

Join Date: Apr 2007
Posts: 3531
#97

### Re: Daughters and Sons: CR4 Challenge (10/20/09)

10/25/2009 11:25 AM

Assuming independence of outcomes, and the only information obtained was that each couple has two children and that Bill and Charlie each have at least one daughter (i.e. we didn't check Arthur's family and then list all families that had a daughter) DAC's first answer (2/3) was what you require for the main challenge.

The bonus question as written remains ambiguous*, but a perfectly natural interpretation would give the answer 2/3:
If we had asked: "Do you have exactly two children", followed by "Do you have any daughters?" and "What is the birth-month of your oldest daughter (if more than one)" and October was the outcome then the probability would be unchanged at 2/3.

However, if we had asked a large number of families "Do you have exactly two children, and is any of them an October-born girl" and Dick was randomly selected from the fathers who replied yes, then we can see a reason that the probability should be different - i.e. that if we ask this same question about each month in turn, then families with a one daughter or with two daughters born in the same month would be counted once, whereas those with two daughters born in a single month would be counted twice. For the present case this gives
Probability = (B&OG+OG&B)/((B+G)&OG+OG&(B+G)-OG&OG)
. = (12+12)/(24+24-1) = 24/47
(Ignoring leap-years, the answer for a pre-specified birth-date would have been
(2x365)/(4x365-1) = 730/1439
These gives the same result for 1/50 as calculated in Wikiipedia's "Frequentist Approach".

I think I may have missed some of your versions? Ouch! anyway.

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