# Puzzling Adventures: A Kingdom for My Child -- August 2008 Solution

Solutions:

1. Let's consider again the three children that a family could have and associate the number at which they would stop. (As a thought experiment, we imagine that they have three children in a certain order, but "undo," or send away, the extras once they have a boy.):

000—3
001—3
010—2
011—2
100—1
101—1
110—1
111—1

So, the average number is under two (14/8).

2. Of the eight scenarios above, 6/8 (or 3/4) would have one boy and one girl.

3. We write down the number of children needed to achieve completeness in each case:

000—3
001—3
010—2
011—2
100—2
101—2
110—3
111—3

This gave us an average of 20/8 children per family, or 2.5.

4. If the youngest happened to be a girl, then the others would either both be boys or both girls (otherwise their parents would have stopped already). Therefore, the probability of at least one boy is 1/2.

5. Once again we are, with equal likelihood, in one of these eight family situations. There are 10 girls available (because some families stop before three) and in 7/10 of the cases, there is a boy, so the likelihood is 7/10.

000—3
001—3
01x—2
01x—2
10x—2
10x—2
110—3
111—3

For more on this subject, see Statistics is Easy! by Dennis Shasha and Manda Wilson (Morgan & Claypool, July 2008).

Back to Puzzling Adventures main page

Dennis Shasha is at the Courant Institute of Mathematical Sciences, New York University. His most recent puzzle book, Puzzles for Programmers and Pros, was published in May 2007 by John Wiley and Sons/Wrox.

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1. 1. jpender 11:56 AM 8/1/08

For answer number 5, shouldn't it be 6/7? There are 7 out of 8 families that have girls. 6 of those 7 also have boys. To be Chiwachi-Complete, the family can either have 3 children, or a boy and a girl, and you state that the family is Chiwachi-Complete.

2. 2. Gemini in reply to jpender 02:35 PM 8/1/08

I agree with you!

3. 3. Gemini 02:41 PM 8/1/08

I agree with you. We might not consider the two girls in a 001-family as two situation, and so forth. So I believe the answer should be 6/7.

4. 4. trickydicky in reply to Gemini 09:35 PM 8/1/08

I thought it was 6/7 too, but now I think it is 7/10. If you see a girl, then she could be any one of the pool of 10 girls in the six girl-containing families. For each girl, does she have a brother? Only the 3 of 10 girls that are in an all-girl family would have no brother. The other 7/10 do have a brother

5. 5. Gemini 11:52 PM 8/1/08

I change my opinion. Before see trickydicky's comment, I realized that in the Chiwachi-Complete society, the ratio of male and female isn't approximate to 1:1. Considering a group which include 10,000 girls, 3,000 girls should come from the 000-family. So if you see a random girl, she has 3/10 possibility of coming from 000-family.Hence, the answer should be 1-3/10=7/10.

6. 6. btvdan 02:44 PM 8/4/08

I never understand how to count these problems. Maybe someone can explain it better for me. The question asked, "...then what was the likelihood that her family also had a boy?" We are interested in counting the total number of families that the girl could belong to and then selecting the number of families from the selection space that have the interesting characteristic (the family has a boy). This leads me to the 6/7 result. It seems to me that 7/10 answers a different question (the likelihood that the girl has a brother). It seems the fact that the girl comes to school only serves to eliminate 3 boy families from consideration. This also asks the question... can the family be more likely to have a boy than a girl have a brother (when considering both within this same framework of rules).

Is there a better way to explain this? Considering the 10,000 girls type thinking doesn't help me because it sounds like the other question to me.

Thanks.

7. 7. felipecarvajal 04:53 PM 8/4/08

Given the faulty and contradictory premises set forth in the wording of the puzzles, I consider them to be a "wash", that is, unworthy of consideration. Respectfully, ><)))> Felipe

8. 8. t 03:33 PM 8/5/08

I think #5 is wrong and should be 6/7. A girl as described has an equal likelihood of being from any of the 7 possible family arrangements (01X, 01X, 001, 000, 10X, 10X, 110). She's not more likely to be in family 000 than family 110 simply because there are more girls in that family arrangement.

Sure, 70% of girls from "Chi-wac" complete families have brothers, but that's not what is being asked.

Despite enjoying math, this isn't a book that I'll be buying anytime soon.

9. 9. go2gokul 02:13 PM 11/21/08

I think for #1, the probability is 9/4. The family will stop getting children the moment they get a baby boy so if we consider 1 o be boy and 0 to be girl, the sample space would be
000 = 3
001=3
01=2 (No more children as they got a baby boy)
1=1 (No more children)
So the average value would be 9/4 = 2.25

I think it is not appropriate to consider 111, 110 or 101 kind of families in the sample space. Because as per the problem statement there will be no such families as they stop getting children after they see a baby boy

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