Maths problems (interesting ones)

[Denman]

sory i been a bit busy lately.
some new problems coming up later

[Astronouth7303]

Ok, I know this is a little late, but here goes...

The whole 1/2 1/3 debate was bugging me, so I e-mailed her about this with links to a couple of different arguments (in addition to the thread it self). Here's a copy:

...
When I first looked at the problem I quickly thought that it was just a simple case of conditional probability - and that the answer therefore must be 1/3.

However, after giving it some more thought, I'm would side with those who say that it is 1/2. [A friend] agrees.

The problem seems equivalent to: Bob flipped 2 coins. He showed me that one of them came up heads. What's the probability that the other coin is heads?

Essentially, the sex of the 2nd kid is independent of the sex of the
first. Knowing that one of the kids (the door opener) is a girl tells
us nothing about the sex of the second kid.

or to think about it another way. Suppose that there were 1000 doors to be opened. Opening a door and finding a girl there eliminates 500 of them. Of the 500 that are left, 250 of them will be GG. Therefore the probability is 1/2 -[Cool Math Profesor]

She is a profesor at a local college.

[FizMan]

I just wrote my own computer program.


IF you choose a random child to open the door then it's an outcome of 50%

IF you choose a girl to answer the door (I still say the question provides that as a given), then the outcome is 33%


But I don't know... I honestly don't agree with your teacher's logic... yes I agree that the gender of one doesn't influence the gender of the second... but she seems to ignore the fact that there are twice as many families with a boy and a girl than there are with two girls. The gender of the second kid isn't generated when you meet them (i.e. flip the coin); it's already been made... and fact is twice as many "tails" get flipped than "heads."


Also, why does opening the door and getting a girl eliminate 500 doors? Shouldn't it only eliminate 250 doors (b/b) as even in the b/g & g/b families, either one could have answered the door? And wouldn't that leave 250 doors with GG and 500 with some combination of GB?

[Solace]

and that, boys and girls, is the sound of my head exploding.

[Astronouth7303]

Originally Posted by FizMan

But I don't know... I honestly don't agree with your teacher's logic... yes I agree that the gender of one doesn't influence the gender of the second... but she seems to ignore the fact that there are twice as many families with a boy and a girl than there are with two girls.

Of course there is. GG, GB, BG, BB. But that assumes you can identify A and B. But in this instance, you can only identify them by which one opened the door, so you have to account for both instances. A or B may open the door!

A opens:
+GG
+GB
-BG
-BB
B opens:
+GG
-GB
+BG
-BB
It's still 1/2, 50-50.

Also, That list is generated from the combinations of 2 mutually exclusive chances, a table:

  B  G
B BB GB
G BG GG

The bottom line is: THE RESULT OF ONE DOES NOT AFFECT THE RESULT OF THE OTHER.

[FizMan]

I don't understand what your + and - signs are denoting.

[Steve Howland]

Originally Posted by FizMan

I don't understand what your + and - signs are denoting.

the + seems to represent a possible combination and the - shows that it does not work. That post aims to prove that there are 2 ways that the Girl-Girl combination works (because either one could open the door and then the other child would be unknown), which would then make the odds 50-50.

[Denman]

Right, i remembered this thread from a while ago, so i decided to post some new problems up
Number 1

given that the rectangle is 2x4, find the radius

Number 2
Divide $261 (in whole $ increments) into a number of bags so that I can ask for any amount between $1 and $261, and you can give me the proper amount by giving me a certain number of these bags without opening them. What is the minimum number of bags you will require?

When i get around to it i'll add a couple more

[Greg Ross]

Originally Posted by Denman

Number 1

given that the rectangle is 2x4, find the radius

See attached drawing. (The original diagram looked something like the top left picture.)

[DaBruteForceGuy]

Originally Posted by Steve Howland

Very fun puzzles...but here's one that has stumped pretty much everyone I know (similar to marble puzzle but harder)...

You have 12 identical-looking coins. One of the coins is a fake, yet you do not know which one it is nor if it weighs more or less than the other 11. Using 3 turns with a regular balance scale, determine which coin is the fake and if it weighs more or less than the other 11. (props to anyone who gets this in under a day)

AHHHHH my brain hurts ... i need the solution!

[Daniel Brim]

Originally Posted by Steve Howland

Very fun puzzles...but here's one that has stumped pretty much everyone I know (similar to marble puzzle but harder)...

You have 12 identical-looking coins. One of the coins is a fake, yet you do not know which one it is nor if it weighs more or less than the other 11. Using 3 turns with a regular balance scale, determine which coin is the fake and if it weighs more or less than the other 11. (props to anyone who gets this in under a day)

You don't need to use the scale at all.

First, if you need a small graduated cylinder. 11 coins should be exactly the same, so they all have the same density. The fake, although not necessarily weighing the same, will have a different density than the others. Simply put in X mL of water in the cylinder and calculate the displacement of the water for each coin dropped, and the fake coin will be different.

[DaBruteForceGuy]

Well yea i think that is definitly an option if i had to do this in reality... but we're not in "reality", we're in Steve howlands reality... and he descided to give us merley a stupid scale with a bunch of stupid coins... AHHH i need to know the answer dangednabbit! lol

[unapiedra]

ok, here got it from a book some place and i figured out the solution (actually there are two but only two)

--so here it goes: a 9-digit number (123456789; the digits only represent the position for further labeling),
--now each position has to be divideable (english, gramar, I sorry.) by the #of the position (like 1/1=n; 12/2=n; 123/3=n; 1234/4=n and so on; n representing a natural number)
--and finaly, the number consist of the digits 1 through 9, but they can be used only once

So get the solution, and I'm sorry for my terrible english

[Denman]

Right,

For the coins one you can't do it unless you know whether its heavier or lighter.
if you know whether it is heavier or lighter:
put 6 on either side of the scales and weigh
One side will be lighter than the other so you take the side witth the heaveier or lighter one and weigh 3 on 3. One side will be heavier/lighter. Then weigh 1 on 1 of and leave one off. If it is even then the 3rd is fake. If it is not even then the heavier/lighter side is fake

[unapiedra]

I got the solutions for the coins! However I need 4 steps. Maybe I can get it with 3 (I need some time on that)

I'll post the solution with 4 later today (School now).
I know it doesn't count but at least it's something and maybe then someone else can do it then with three...