I use it around the lips, eyes, nose, parts of the ear, wrinkles, brow -- and that's just on the head.

I know it's really just a combined draw/pinch, but it works much better than doing the two independently would

If you look at my video in the Sculptris thread in Tools, you'll see I use it a lot!

Jibb

Ok, preasumption, getting a boy or a girl is each 50%

So for the woman we know she does not have 2 girls.
And at least 1 boy.

This makes the woman have these chances for having such kids:

1/3 - B B
1/3 - G B
1/3 - B G
0 - G G (exluded from beginning)

So having 2 Boys is a chance of 1/3

----

For the man, we know he has 2 children.
One of them is a boy (one of the slots), but thats unimportant,
as we only care for the other one. (wich is independant)

And for the other child to be a boy it is 1/2 chance.

So the man has 2 boys with a chance of 1/2

that's no argument, you could also say the woman has at least 1 boy. the chance for the other kid to be a boy is 50%. so she has 50% chance of having two boys.

the problem is that you're mixing slots and the age thing. in my notation the first is the oldest child.

woman:
1/2 - b ?
1/2 - ? b
which gives
1/4 - b g
1/4 - b b
1/4 - b b
1/4 - g b

=> 1/2 chance of having two boys

man:
we know: b ?
which gives
1/2 b g
1/2 b b

=> 1/2 chance of having two boys.

Hahahaaaaaaa!!!
http://www.newgrounds.com/portal/view/437054

No, the information
"has at least 1 boy"
is not the same as
"the older one is a boy"

In the first case, you look at two
people, and state if at least one is a boy
(you look at the probability combination of two events)

In the second, you only look at one person.
(you look at the probability of a single event)

i have two balls, either red or black. at least one of them is red. the probability that the other ball is red is still 50% (why should it be 33%?). so the total probability is 50% for both balls to be red.

no its not 50%

There are 4 outcomes when drawing balls:

(chance - color color)
1/4 - R B
1/4 - B R
1/4 - B B
1/4 - R R

So when saying that at least one is red,
then you sort out the
B B possibility, leaving you with

1/3 R B
1/3 B R
1/3 R R

so now:

1/3 R B (one is Red, the OTHER Black)
1/3 B R (one is Red, the OTHER Black)
1/3 R R (one is Red, the OTHER Red)

gining you a chance to have only 1/3 that
the OTHER is Red

No, you don't have to consider the order of the balls. As I said, you always need Laplace propabilities, and this way you don't get them.

You can arbirarily say that one ball is the "first ball" and the other is the "second ball" (because those are only names). So you have a 100% propability that the "first ball" is red because you already know it's red and a 50% propability that the "second ball" is also red. So you have a 50% propability of having two red balls.



Or let me put it in this way: What if I tell you, there is a ball and it has a 50% propability of being red. Now imagine I tell you there is an other ball that is certainly red and tell you that both balls form a pair. Would you say the propability of this ball being red decreased to 33.3%?

my balls are blue. guess why... (and yeah, that totally takes the ari out of your mathematics schematics...)

I don't know much about this stuff but i would say it is an 50% chance too.

In your ball example:
I wouldn't consider RB and BR as a different possibility.
If you say one ball is red, the other one can be either Red or Blue, which is 50/50. You can't consider the BR case anymore as you know the first one is R so it will be Rx where x is wether B or R.

Many years ago my teacher gave me this probability theory:
You have 3 cups infront of you and in one of those cupse is a marble. Now you pick a cup, and you have a 1:3 chance to get the right cup.
After you chose a cup, the gamemaster is so kind to show you one of the cups that doesn't have the marble and takes it out of the game and asks you if you wanna stick with your cup or choose the other one.

So now it's always smarter to switch to the other cup cause suddenly your winning chances are 1:2. We tried this in many games and your winning chance IS higher when you switch in the second turn.

Maybe some of you (german speaking people) remembers "geh aufs ganze" and there you often has 3 envelopes and two had the Zonk (the bad price). They always let you choose one and sometimes say "ok i give you a hint, this one is a zonk and i'll take it out".

So like inthis example, if you take out one envelope, you take out a possiblity so your chances increase to 1:2.
I think it's the same with the balls. If you say the one is R, then you actually can't consider the case BR and BB.