View Full Version : Chinese Children

08-09-2002, 09:43 AM
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.)

08-09-2002, 11:05 AM
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

08-09-2002, 11:55 AM
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.

08-09-2002, 12:11 PM
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

08-09-2002, 02:16 PM
what happens if you include twins, triplets, etc (especially identical?)

08-09-2002, 04:13 PM
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.

08-09-2002, 04:20 PM
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.

08-09-2002, 08:38 PM
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.

08-09-2002, 08:44 PM
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.

08-09-2002, 09:11 PM
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?

08-10-2002, 01:22 AM
I don't think this problem is to be taken literally (that is the females are aborted, what about twins triplets etc).

I thought it was fair to assume a couple has a child and continues to have children until they have a boy, and they only have one child at a time.

08-10-2002, 10:34 PM
I'll try the clever way first.

First a lemma: If a family has exactly one boy, then the expected number of children in the family is 2.

general concept of proof: if that were not the case then half of the chilren born would not be boys, which violates our assumption.

So we have 1 billion families and they all will have exactly 1 boy. Therefore on average a family has 2 children. The answer is then 2 billion children.

I can also sum the infinite series, but that is boring.

1B * (1*(.5)+2*(.5)^2+...)


1B * x * Sum(k*x^k-1) for x=.5


=1B*x*d/dx[1/(1-x)] - geometric series


=2B when you plug in .5 for x

08-13-2002, 05:20 PM
Apparently, 1 in 5 people in the world are Chinese.

There are 5 people in my family, so it must be one of them.

It's either my mom or my dad. Or my older brother

Colin. Or my younger brother Ho Cha Chu.

But I think it's Colin.

08-13-2002, 10:45 PM
Half the total children born will be boys, and each couple will have one boy, so there will be one billion children.

This problem is very similar to the "bee and trains" problem I heard a while ago.

Two trains start 100 miles apart and approach each other at 20 miles per hour. A bee, starting at the first train, flies towards the second train at 100 miles per hour. Upon reaching the second train, it turns around and flies back towards the first train also at 100 miles per hour. It continues to do this back and forth until the trains meet. When the trains meet, how far has the bee traveled?

The word on the street is that John von Neumann was asked this question and immediately gave the correct answer. The guy who asked him the question said, "wow... you sure found the trick quickly," to which von Neumann responded, "what trick? I summed the series..."

08-14-2002, 02:23 AM
pretty funny

08-16-2002, 05:09 AM
It seems to me that if we use the geometric distribution, we haven't really gotten away from summing the series, we've just hidden it inside some very intuitive and easy to understand probability concept. You can say 'geometric distribution', but finding the expectation of that is basically summing a series. (At least, I can't think of another way.) Of course, I don't think there's anything wrong with this. All of math (good math) is an attempt to put difficult or intimidating problems into a previously known context so the solution seems natural, and the geometric distribution seems to do that in this case.