I have been reading a fantastic series of books called Murderous Maths and it's addictive!!! You can fall in love with Maths just by reading these books even if you are normally suspicious of figures and all.

Anyway, I got really stuck yesterday with a very simple-looking sum and I asked my mum (who can't remember) and dad a murderous question, upon which, he shook his head, sighed and said, "That's not really for a 9 yr old, you know ... It's for upper secondary school." Well, he's a physicist after all so he MUST have done university Maths. Anyway, HE tried and he tried ...

How can he not do it!!! Well, he was so beside himself that he just called from the office just now for us to read the problem to him again so he can try it AGAIN(!) at lunchtime. He says he feels confident this time round, that he must have misunderstood the question yesterday and so on... Would you believe that?!

Anyway, it goes like this. Now if someone can help me secretly here *rubbing hands with glee* I can't wait to trounce that physicist tonight! Will tell him my SECRET "weapon" after his BIG surprise.

S has a population of 2015. Men and women totalled 1007. Women and kids totalled 1537. How many KIDS, WOMEN and MEN are there, respectively?

Any ideas?

Violinutter

Three equations, three unknowns, should be okay :

(1) m + w + k = 2015
(2) m + w = 1007
(3) w + k = 1537

Subsitute (2) into (1):

1007 + k = 2015
k = 1008 (4)

Subsitute (4) into (3):

w + 1008 = 1537
w = 529 (5)

Substitute (5) into (2):

m + 529 = 1007
m = 478

So, 1008 kids, 478 men and 529 women.

Whaaaaat! You are amazing! So fast. I'm just about to start my violin practice now so will ask mum to print this out later.

Thanks SO much!!!

VN

I've heard that problem before somewhere...

OK, here's a statistics one...

Imagine a TV game show where the contestant is shown 3 doors. Hidden behind one of them is a fantastic new sports car (substitute other desirable prize if you're not into sports cars). Behind each of the other 2 is a goat. The contestant must choose one of the doors, and will take home whatever is behind that door.

To make the game more interesting, there is a twist. After the contestant has chosen a door (and told the compere their choice), but before it is opened, the compere opens one of the other doors, to reveal a goat. The contestant is now given the opportunity to change their mind if they wish, or stick with their original choice.

Question: should the contestant stick with their original choice, or switch? Justify your decision statistically!

Ignore this, I'm being silly

I hate stats, but here you go- before the door is opened, there is a 1/3 chance of it being the right one. Once the door is opened to reveal a goat, there is a 1/2 chance of it being the right door, and 1/2 of it being the wrong one. So, not much point as each door has the same chance.

That's what I did too. I thought it was quite straightfoward... (and no A level maths for me!)


They should switch. When they pick the door, all the doors are 1/3. However, when one door is opened, this changes. As the original choice is a 1/3 door, the other door inherits the 1/3 from the door that was opened, making it 2/3. So, to have more chance of winning, one should switch.

Just a guess.

Thouston?

VN

The goat one is really famous. When someone put the above answer in a newspaper, hundreds of university professors of Maths wrote in to say he'd got it wrong. But he hadn't. Which just shows that first instincts are often wrong.

You're right - it's a classic, and not straightforward.
Here's the answer and the reasoning behind it.
The contestant SHOULD switch doors and choose the third (unopened) one. Here is the reason:
When they first chose a door they had a 2/3 chance of choosing a goat, and 1/3 that it would be the sports car.
Then the compere revealed a goat behind a second door.
There are 4 possible scenarios:
The contestant has chosen a goat (1), and the compere has revealed the other goat (2). Third door therefore hides sports car.
The contestant has chosen a goat (2), and the compere has revealed the other goat (1). Third door therefore hides sports car.
The contestant has chosen the sports car, and the compere has revealed goat (1). Third door therefore hides the other goat (2).
The contestant has chosen the sports car, and the compere has revealed goat (2). Third door therefore hides the other goat (1).

There is now a 50:50 or 1/2 chance that the third door hides the sports car, rather than the 1/3 chance that the first door has. The probability is different because of the extra information (that the compere revealed a goat behind the second door).

The other classic is the one about children. Assuming there is an equal probability of having a boy or a girl (not quite true in real life, but just for this exercise) and taking all families with 2 children; if one of the children is a boy, what's the probability that the other will also be a boy?

That's another one that resulted in a lot of debate. It's much harder than it sounds, and, even worse, the answer changes if you rephrase the question "...if the older child is a boy..."

VN: yes, well, sorry, but we statisticians are a funny lot, we have to amuse ourselves somehow Don't worry, I'm nearly normal most of the time...

Is it 1/3?

Assuming the non-existence of identical twins, 1/2.

Really?
http://mathforum.org/dr.math/faq/faq.boy.girl.html

Hmm, the way I thought about it is that the gender of one child will have no effect on the gender of the other; thus, the probability of the other child being a boy is 0.5.

Yeah, but that only works if you know the first (older) child is a boy. But since it could be either of them, it means there are 3 different possibilities ({B,B} {B,G} {G,B}), each equally likely.

But we can't have (B,G) and (G, as possible outcomes - they're mutually exclusive. When we know one of the children is male, it either has to be the youngest or the eldest. If it's the eldest, the remaining sample space is (B,G) and (B,; if it's the youngest, the remaining sample space is (B, (G, . In either case, the probability of the other sibling being a boy is 0.5.

Hmm. It apparently depends on the interpretation
http://www.wiskit.com/marilyn/boys.html
Make up your own mind

I agree YAP - doesn't matter what gender the first child is, it's always 50/50 as to whether the second one is male or female.

Now I'm really confused. Can't seem to sort it out in my head

Me too. That's what I first thought after seeing the problem.

Half.

VN

Perhaps it's true to say that you are both right, depending on which theory you follow...?

Having spent the whole afternoon and early evening with this problem bugging me, I am now fairly certain it is actually a half.

It obviously has to be 50/50, 1/2. If there is an equal chance, then it has to be 1/2.



Ben

I appeal to the court of AP and YAP

surely this is nonsense????

The door the contestant first chose had a 1/3 probability of being the right door when he picked it, but at that time the door opened by the host also had a 1/3 probability. As soon as the host revealed the goat behind the second door the probability of that door hiding the car was reduced to zero. That knowledge affects the probabilities of the car being behind either of the other two doors - neither is more likely than the other - each now has a probability of half.

if at the moment the contestant is given the opportunity to switch the probabilities are:

1/3: original choice
1/2: new choice
zero: door with goat

where is the missing 1/6?

I'm also with 50/50 on the boy/girl problem

For the boy girl one, it's a bit more complex than you might think. If you picked a woman at complete random, and found out that at least 1 of her children was a boy, then the probability that the other is also a boy is 1/2. However, if you picked a woman at random from a selection of women who all had 2 children, at least one of which was a boy, it would then be 1/3 that the other was a boy. At least as far as my understanding goes.


Just so it's clear, the host knows he won't be revealing the car.

At the time the door is picked, there is a 1/3 chance that the car is behind that door, 2/3 chance that it is behind one of the other two doors.

One of the other two doors is then shown not to have a car behind.

This means the 2/3 chance that the car is behind one of those two doors, is now a 2/3 chance that the car is behind that one remaining door (the one that hasn't been chosen).

Thus there is a 2/3 probability of winning if you switch.

That make sense?

Any one got any easier puzzle things?? lol

Find two non-distinct numbers, one of which is 2, such that when added together or multiplied the result is 4.

I was with your quote up to this point. But when we rule out the door opened by the host why does that make one of the remaining two doors twice as likely to hide the car and the other door no more likely to hide the car at at all? why is the probability redistributed all to the third door? Surely the 1/3 of the rejected door is redistributed 1/6 each to the two remaining doors. The contestant now has a straight choice between two doors one of which has the car and he has no other information on which to differentiate the probabilities. It's got to be half half !!!!!!!!

PS this is all meant in the tone of "jolly debate" in case that doesn't come across from my words in black and white!



oo eck YAP

give us a mo - I'm half way there already

So does the other number have to be a number other than 2 i.e. you cant have 2 and 2??



I'll have a think

Are you confused as to the meaning of non-distinct?

Of course

This is the kind of thing I'd find easier to explain in person unfortunately, but here goes...

We'll call the doors A, B and C

You pick door A. There has to be probability 1/3 that the car is behind it. Consequently there has to be 2/3 probability it is behind door B or C.

Focusing on doors B and C. We know there is 2/3 probability the car is behind one of them. So if you show that one of them doesn't have a car behind it, there has to be 2/3 probability that the car is behind that one remaining door. The opening of the door does not change the fact that there is a 2/3 probability the car is in the block of doors B and C. It just shows that, within that block, it is definitely not behind one of them, so that 2/3 chance must relate to only one door.

If that doesn't make sense, I recommend http://en.wikipedia.org/wiki/Monty_hall_problem.



Does the other number have to be real, YAP?

Haha, Yeh. I think i understand now though

Sorry for being thick

Ben

I think it does, given the restriction that the two numbers are non-distinct and one of them is two.

S'fairly likely then...

But that's just it - it does! suppose having opened door B the compere then opens door C revealing that the car was behind neither. Would you still maintain the 2/3 probability of being behind doors B and C which you started out with is unchangeable? patently not. The probability is based on the information you have. Each time new information rules an option out reducing its probability to zero, the probabilities of all the remaining options change.


PS: I wonder what the probability of us agreeing on this one is?

I'm not sure how to put this politely, but that's wrong. If the host opened doors B and C and it was behind neither, it would be the 1/3 of the time that it is behind A. The comparable approach for all 3 doors is to say the probability it is behind door A, B or C is 1 (it is definitely there). If you show it is not behind C, that probability of 1 remains for doors A and B. If you then show it is not behind B, that probability of 1 remains for door A, i.e. it is definitely behind door A.

It is the same case for doors B and C, except the starting probability is 2/3. If you know there is a 2/3 probability it is behind doors B and C, and then you show it is not behind C, the 2/3 probability remains for door B. It cannot change the probability that door A has the car.

And we may never agree, some people just can't see it. But what I believe is the established solution, I'm not just making stuff up


To change the subject, does 0.9999 recurring = 1 ?

This question has done my head in for ages...

No, I don't think it does.

Why not?

Because 0.99999- for however long the 9s keep coming does not equal two. It gets closer and closer every time, but it never actually gets there, surely?

I'll agree with that

I think It does !!

Lets say 0.9 (recurring) = x

x = 0.9 (recurring)
10x = 9.9 (recurring)

10x - x = 9x

9.9 (recuring) - 0.9 (recurring) = 9

therefore 9x = 9

9x / 9 = x

therefore x = 1

Does that make Sense??


Ben

Yeah, but that's not as accurate...

Yeah, that's the standard proof for it.

What's not as accurate Oddy?

Whooops, yes it is, misread this line:

9.9 (recuring) - 0.9 (recurring) = 9

How about this one:

Is 0 (zero) an odd or even number??

Even. It divides by two and leaves no remainder.

I think its even aswell, bit for this reason

-5 -4 -3 -2 -1 0 1 2 3 4 5
Odd Even Odd Even Odd Even Odd Even Odd Even Odd

*Edit* Damn, It didnt come out right