Brain Teaser No 3

Nice description of the tradeoff, Ian. The math is ok too.

It's possible to avoid calculating the sum
sum=1/4 + 2/8 + 3/16 + 4/32 + 5/64 + 6/128 + 7/256 + ...
If you know that
sumb=1/4+ 1/8 + 1/16 + 1/32 ... = 1/2
then you can see that
sum/2+ sumb =sum
so sum=1

A modification of the model where people still adhere strictly to the rule but sometimes settle on having only daughters would yield slightly more girls than boys.

So the proposal offers
- a better "building block" for birth planning policy leading to simplification of the system. The "one child" policy is a bit of a misnomer because it is part of a complicated system that is very dependent on location.
- some freedom of choice
- less abortions and infanticide because the rule puts less strain on the desires of the parents.
- better balanced male/female demography
- slowly declining population(you don't want a fast declining population.
Here is an article about the kind of demographic problems they're struggling with.

Alex,

neat observation adding two simpler sums to get the non-geometric summation - I did it more formally by using a trick taken from integration methods. But it was a dull calculation.

cheers

Ian

I thought I might try to "explain" the Chinese "one boy per family" probability a bit more simply - for my own peace of mind as much as anything. I failed, but felt I would post it anyway.......

The tree diagram below shows the possible routes to a family producing a boy based on a 50/50 probability of a boy at each birth.

The probability of a single boy family (b) = 1/2 (shown in blue in the diagram)
The probability of a girl/boy family (g/b) = 1/2 x 1/2 = 1/4 (part of red then thick black line)
The probability of a girl/girl/boy family = 1/2 x 1/2 x1/2 = 1/8 (red in the diagram)
etc etc

The expectation of a boy in any family = 1/2 + 1/4 + 1/8 ..... = 1
Which is what we expect.

I think most of us know (*) that the sum of the series 1/2 + 1/4 + 1/8 .....=1
This also leads us to know that the sum of the series 1/4 + 1/8 + 1/16..... = 1/2
And the series 1/8 + 1/16 + 1/32.... = 1/4
Etc etc

The probability of a family with one girl = 1/4 (thick black line in diagram)
The probability of a family with two girls = 1/8 (red in diagram)
three girls = 1/16
etc etc

So the probability of a family with at least one girl = 1/4 + 1/8 + 1/16....= 1/2
Which is what we expect. (since half the families have only a boy)

The expectation of a girl in any family depends on two things
First the probability of the family having a girl
Second the number of girls in each family.

Expectation of zero girl = 1/2 x 0 (ie we can ignore this)
expectation of one girl = 1/4 x 1 = 1/4
expectation of two girls = 1/8 x 2 = 2/8
expectation of three girls = 1/16 x 3 = 3/16
etc etc

Hence the expectation of girls = 1/4 + 2/8 + 3/16 + 4/32....
This is a "satisfying" looking series, but how do you find its value?
Summing this series looks easier to me if it is set out as follows

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

Note, there is still one quarter, two eighths, three sixteenths etc
It is now easy to see that the first line sums up to 1/2
the second line sums up to 1/4
the third line sums up to 1/8
etc etc
in other words, adding each line we get 1/2 + 1/4 + 1/8 + 1/16....= 1

This means that the expectation of a girl in a family is 1 ( although many families will have no girls and a fair number will have two or more girls)

The child population (expectation) will be "expectation of a boy = 1" + "expectation of a girl =1"
in other words child population will be 1 + 1 = 2 children per family.

(*) If you aren't sure about this try the following :-
Start with one piece of string and cut it in half
Take one half and cut that into two quarters
Take one quarter and cut that into two eighths
etc

Your original one piece of string is now made up of 1/2 + 1/4 + 1/8 + 1/16 etc

Cheers

Don

Matchsticks

The diagrams below show two ways of forming the perimeter of a polygon using 12 matchsticks (each of unit length). One (red) to form a square of 9 square units, the other (blue) to form an area of 5 square units.

Can you use all 12 matches in a similar way (you must use the full length of each match and you can't break or bend any matches) to form the perimeter of a polygon with an area of exactly four square units ?

Cheers

Don

There are obviously lots of solutions to this.

But one approach might be to construct a 2x2 square, area 4 as required, but perimeter only 8. If we take 2 more matches and extend one side out from the center into some form of triangle (call it 'equilateral'...) and the other 2 to extend a different side inwards in a similar triangle I think it is obvious that the total area remains unchanged at 4 but the perimeter has extended to 12.

A more interesting shape would be to add a single match to each side, two bellying in and two bellying out.

I'd draw a picture but my pen has run out of ink.

Paul

There are obviously lots of solutions to this.

But one approach might be to construct a 2x2 square, area 4 as required, but perimeter only 8. If we take 2 more matches and extend one side out from the center into some form of triangle (call it 'equilateral'...) and the other 2 to extend a different side inwards in a similar triangle .....

.......thereby making an "arrowhead"....(for example) like the one below

Or a right-angled arrow head......

I also like the "bow-out/bow-in" box shape

Cheers

Don

Any more shapes, anybody?

Just so as Paul doesn't have to do all the hard work....

Cheers

Don

One can take Don's bow-out/bow-in shape and flip the points back making something like

--
\ \
/ /
\ \
/ /
--

if you see what I mean.

All in-out permutations are also allowed e.g.

.--
.\/
./\
/..\
\../
.--

etc ( dots added to get round forum formatting )

ian

I see at least two groups of solutions that can easily be constructed

one group is based on a square 0,0 > 0,2 > 2,2 > 2,0 > 0,0
Take any wobbly string of 4 sticks such that the distance between the two endpoints is 2 units, and make a copy
Use two straight edges and two wobbly edges to connect the four corners.
Wobbly edges can be on opposite sides or on joining sides.
A wobbly edge can be rotated and mirrored yielding 4 possibilities. Two increase surface and two decrease it.

Another group of solutions is based on a parallellogram: 0,0 > 0,2 > 2+x,2 > x,2 > 0,0 (x<2* sqrt(3)).
The top and bottom lines are straight and horizontal,
You can again make any wobbly string(and duplicate it) as long as the distance between the beginning and end is between 2 and 4.
Combinations of rotation and mirroring are again possible.
If x=2* sqrt(3) then the length of the wobbly lines is 4 (which makes them unwobblified) and you have a parallellogram.

Ah, and then there are squares and parallellograms consisting of pairs of matching 3-stick wobblies.

Alexander, Don, Ian and Paul, huge Kudos for being able to do this stuff.

I started to read the thread, only to be resuscitated by my other half after drifting into unconsciousness, (Algebra has the same effect) while I was under, a couple of questions have become apparent...do you lads enjoy papering irregular shaped rooms and secondly, do you put the plug in while having a bath?

Beano fluked an O-level in maths.

Here are my three favourite solutions to the 12 matchsticks.

I like the star and the Maltese Cross because of their satisfying symmetry and the fact that the star can be adjusted to give any area from zero units to just under 12 units. However, working out the precise angles for these two, needed to give 4 units, takes a little longer than the very clear-cut 3/4/5 triangle, modified to reduce the basic area from 6 units to 4 units precisely.

Cheers

Don

quote:

.do you lads enjoy papering irregular shaped rooms and secondly, do you put the plug in while having a bath?

Beano,

First, glad to hear that you survived.........

As for wallpapering, I always like to manage with two rolls less than the retailer recommends...

....and stagnant water in a bath with a plug...no-way....although I do cheat on this one and use the plug until the bath is filled, but as soon as I get in I pull the plug and turn on the t........

I think I'll stop here just in case you have nobody around to revive you this time...

Cheers

Don

I'm the opposite on the wallpaper front Don, I'll take extra rolls just to be on the 'safe side' of my calculations.

Beano

I always peel my apple in the form of a rhombicosidodecahedron,
and the plughole of my bath has been designed specially to maximize vorticity, Beano.
Actually, no, you've witnessed all my efforts on brain teasers in many years.

Don, I especially like the solution on the right,
because it eluded me completely and still it's so simple and obvious in hindsight.

B]The Blue-eyed Green Sisters[/B]

If you met two of the Green sisters, (assuming the two are perfectly random selections from all the girls in the Green family), it's exactly a 50/50 chance that both girls will have blue eyes.

What's your best guess as to the total number of blue-eyed sisters in the Green family?

Cheers

Don

quote:

Don, I especially like the solution on the right,
because it eluded me completely and still it's so simple and obvious in hindsight.

Thanks Alex - glad you liked it!

Cheers

Don

Three.

quote:

Originally posted by AlexanderVH:
Three.

and one sister with non-blue eyes

Ian

quote:

quote:
Originally posted by AlexanderVH:
Three.


and one sister with non-blue eyes

Ian

Superb, both of you.

Who would like to post an explanation for the benefit of others?

Cheers

Don

Oh!

I nearly forgot. But the next set of figures would be 15 and 21.

But the probability of so many girls in one family is quite frightening.....

Cheers

Don

If there are m blue eyed girls(which is the number we are looking for) and the Green family has n girls
then the chance of selecting two blue eyed girls is
m/n * (m-1)/(n-1), which we know should be exactly 1/2
This leads to the equation 2m(m-1)=n(n-1)
Because m and n are integers there will be few solutions.
let's see what the product x(x-1)looks like for increasing x:
X | X*(X-1)
1 | 0
2 | 2
3 | 6
4 | 12
5 | 20
6 | 30
7 | 42
...
15| 210
...
21| 420
...
So the first solution is m=3 and n=4
When you select two girls for this solution,
the chance of the first girl being blue eyed is 3/4
The chance of the next girl being blue eyed is 2/3

For the next solution
the chance of the first girl being blue eyed is 15/21
the chance of the next girl being blue eyed is 14/20

Neat solution, to which I can't add anything

Cheers

Don

Freight Train, Freight Train going so fast.....

Two track-workers are walking towards each other alongside a railway. A freight train overtakes one of them (ie the train is going the same way as the track-worker) in 20 seconds.

Exactly 10 minutes later it meets the other trackworker comming in the opposite direction and passes him in 18 seconds.

How long after the train has passed the second trackworker will the two workers meet.

Assume constant speeds throughout for the trackworkers and the freight train.

Cheers

Don

90 mins ?