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#1
08-09-2002, 09:43 AM
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Chinese Children

There are a billion Chinese couples. They won't stop having children until they have a boy. Then they always stop. So five hundred million couples have one child, 250 million have two children etc. If you can sum the appropriate series you can tell me how many children will be born. If you can't there is another way. A logical way. What I like about this problem is that it shows how you can use probability problems to sum infinite series rather than the other way around. I doubt I am the first person to think of this but definitely never saw the idea elswhere.

Any way in this example, what is the answer, what is the series, and how can you solve it without summing the series? (Note. To be rigorous I would have had to state the number couples as approaching infinity.)
#2
08-09-2002, 11:05 AM
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Re: Chinese Children

the series is 500m * [SUM(i=1,n) 1 / 2^(i-1)].

the limit of the series is 2, so at most a billion children will be born.

you can also do this by calculating the expectation of number of children before boy per couple. assuming fair sperm, we've got a series of bernoulli trials with p = 0.5, which means that we can model the number of trials before the first success with the geometric distribution. the basic insight there is that to have a boy on trial k, you need to have k-1 girls first. anyway, the expectation of a geometric random variable is 1/p, or in this case, 2.

btw, such reasoning is common in basic graduate-level stats books, for example, "introduction to probability models" by sheldon ross. my copy of "mathematical statistics and data analysis" by rice also has a nice picture, if not a verbal description, of this reasoning.

good luck

the club
#3
08-09-2002, 11:55 AM
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Re: Chinese Children

Each time a child is born, the probability that it is a boy is 0.5. We know that each family will have exactly 1 boy; therefore the total number of boys is 1 billion, and the total number of children is 2 billion.
#4
08-09-2002, 12:11 PM
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careless error

of course, i screwed up and did the problem as if there were half a billion couples instead of a billion. same methodology, double the answer.

the club
#5
08-09-2002, 02:16 PM
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Re: Chinese Children

what happens if you include twins, triplets, etc (especially identical?)
#6
08-09-2002, 04:13 PM
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Re: Chinese Children

Find the average number of children a couple has.

THis is :

1(1/2) + 2(1/2)^2+ ... n(1/2)^n +...

(rearrange terms) = 1/2 + 1/4 + ...

+ 1/4 +1/8 +....

+...

= 2

now multiply by the number of couples.

All that was used here was the geometric formula and some rearrangement of terms.

#7
08-09-2002, 04:20 PM
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Re: Chinese Children

Right, but the point was to solve it without summing the series.

Of course, as David mentioned the 2 billion figure is not entirely accurate, but if he had specified an infinite number of families the figure of 2 children per family would be correct.
#8
08-09-2002, 08:38 PM
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Re: Chinese Children

Hey guys,

I hate to inject reality here ( I have an MSEE and specialized in Communications Theory, so I know a little about probability), but in China the real problem is that couples are having sonograms, and aborting females. This puts a major bias in birth statistics.
#9
08-09-2002, 08:44 PM
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Re: Chinese Children

1 billion[(1/2)*1 + (1/4)*2 + (1/8)*3 + ...]

From calculus we can identify the series in brackets as one half times the first derivative of the geometric series evaluated at 1/2, so its value is

.5*(1-1/2)^-2 = 2. So the answer is 2 billion.

Suppose there was a benevolent casino with a fair coin flip game which costs nothing to play, and we get paid a dollar for every head, and when we get a tail we also get paid a dollar but then the game terminates. The above series would represent the expected value for this game if we play until we flip a tail since half the time we would win a dollar, 1/4 of the time we win 2 dollars, etc. Now if this game had cost us a dollar per flip and only paid off on a head, and we decided to play until we lost a flip, the game would have an expectation of 0 regardless of how long we play since the expectation on each flip is 0. We would always lose the last bet. But in our benevolent game, instead of losing a dollar on the last bet we win a dollar, so this game has an expectation 2 dollars higher than 0 or 2 dollars.
#10
08-09-2002, 09:11 PM
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Re: Chinese Children

I'm missing something here. The series is

1/2 + 2/4 + 3/8 + 4/16 + 5/32 + ...

What happened to the other 1/2, why are there 2 1/4s, where is the 1/8 coming from, etc?

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