balls in a bag

so what you say is basically, that first you ask, how likely it was to end up in which bag, having drawn a black ball.

ww: 0/4
bw: 1/4
bw: 1/4
bb: 2/4

meaning that bag bb was much more likely to come up. so in 50% of the cases you will always draw a b, and in 50% of the cases always a w.

Originally posted by tharkesh
Originally posted by mtthw
[b]Out of interest, have you read any of the various previous messages in the thread explaining why that's wrong?

Actually I read all. And all of them neglect the point, that by having drawn already a black ball, the possibilities have been limited to three bags...

The way the riddle was asked, implies, that you front of you: bb, bw or wb. Having drawn b, leaves you in two cases with w and in one with b.[/b]

You need to read this:
http://en.wikipedia.org/wiki/Conditional_probability

From the link
P(A|B) = P(A and B)/P(B)

A: probability that second ball is black
B: probability that first ball is black

P(A and B) = 1/4 (unconditional probability that both balls are black)
P(B) = 1/2 (unconditional probability that the first ball is black)
P(A|B) = 1/2 (probability that the second ball is black given that the first was black)

If you don't like this result you have to accept at least one of the three:
1) Wikipedia's formula is wrong
2) The unconditional probability that both balls are black is not 1/4
3) The unconditional probability that the first ball is black is not 1/2

So which one do you accept?

Originally posted by tharkesh
so what you say is basically, that first you ask, how likely it was to end up in which bag, having drawn a black ball.

ww: 0/4
bw: 1/4
bw: 1/4
bb: 2/4

meaning that bag bb was much more likely to come up. so in 50% of the cases you will always draw a b, and in 50% of the cases always a w.

That's the idea.

Just that, to be precise, that doesn't mean that bag bb was "more likely to come up" just that if you draw a black ball then it's more likely that it came from bag bb than bag bw or wb (note that this is very obvious for bag ww). There's a subtle difference.

i agree that a coin thrown twice is no more likely to get 2 heads then anything else. but its the bag ur looking at not a ball. lets say u have a color changin ball. white and black.
of course it is no more likely to be white then black.
if you have two normal balls, white and black in a bag are they the same as the one? no!
think of it this way. a dice vs if you have six cards Ace to 6 and when u pick a card u put it aside.
throw(or pick) each object(or objects) 6 times. how many times did you get the 6 with the cards? what about the dice?
(phew! i dont think ive ever posted that long before!)

Originally posted by FreakyKBH
Throwing my 50 cents into the mix... a totally non-mathematical mind amongst you formula junkies, so go easy.

WW is eliminated with the first draw of B.

B could only come from BB, BW or BW. It has a 1/3 chance of being from any one of them.

If it came from BB, there is a 3/5's chance of drawing B again, since there are possibly three B's left to since the remaining bags have BB and BW, or 60% chance of being drawn again.

Flame on.

My post above had the chance at 60%, but that can only be correct when identifying the probability of drawing a black ball again from any of the three bags with black balls. I guess I didn't read the OP closely enough to realize we're talking about picking a random bag and then staying with it for the second draw.

In that case, the chance that the second draw will yield another black ball is back to 1/3, since only three of the bags held a black ball in the first place.

Originally posted by FreakyKBH
My post above had the chance at 60%, but that can only be correct when identifying the probability of drawing a black ball again from any of the three bags with black balls. I guess I didn't read the OP closely enough to realize we're talking about picking a random bag and then staying with it for the second draw.

In that case, the chance that t ...[text shortened]... black ball is back to 1/3, since only three of the bags held a black ball in the first place.

So which one is it:
1) Wikipedia's formula is wrong
2) The unconditional probability that both balls are black is not 1/4
3) The unconditional probability that the first ball is black is not 1/2

Originally posted by Palynka
So which one is it:
1) Wikipedia's formula is wrong
2) The unconditional probability that both balls are black is not 1/4
3) The unconditional probability that the first ball is black is not 1/2

I think I may have jacked it all up, but let's try to work thru my mess nonetheless...

Given that the first was established as black--- 3) no longer applies, since we've already drawn. That being said, prior to the draw, with four bags and eight balls--- four of which are black--- the odds were evenly split that the first draw would be black.

WW = 0
BW = .5
BW = .5
BB = 1.00

0 + .5 + .5 + 1.00 = 2. The first pick is coming from one of four bags, and these bags allow for a 100% chance in one bag, a 50% chance in two bags with a 0% chance for the fourth bag.

Because the second pick must come from the same bag as the first pick, the odds are increased in the favor of black (the WW bag has been eliminated), whereas now only three possibilities remain: either you are picking from the BB or either of the two BW's.

If you were to take the bags out of the equation, the first pick of eight balls was decidedly 50/50, since there were just as many whites as black.

Now that you've eliminated two of the previous four whites, but only one black, you're sitting on three blacks and two whites... so my previous idea that it was a 1/3 gambit was wrong.

What I think now is that the odds of pulling another black from that bag is...

5 to 1.

That's my story, and I'll be sticking to it until I am undeniably shown to be the idiot I feel worthy of my station.

Originally posted by Tatanka Yotanka
25%

You picked a black ball and are asking what the chances of the other ball also being black. Since only 1 of the 4 bags had 2 black balls (which must be the case for the second ball to also be black) then the chances of having picked the Black+Black bag are 1 in 4: 25%

I was going to say this, but since Tantanka already did, I endorse this answer.

Originally posted by FreakyKBH
I think I may have jacked it all up, but let's try to work thru my mess nonetheless...

Given that the first was established as black--- 3) no longer applies, since we've already drawn. That being said, prior to the draw, with four bags and eight balls--- four of which are black--- the odds were evenly split that the first draw would be black.

WW = 0 ...[text shortened]... e sticking to it until I am undeniably shown to be the idiot I feel worthy of my station.

🙂

(you are joking, right?)

Originally posted by TwofoldCG
I was going to say this, but since Tantanka already did, I endorse this answer.

Fortunately, the answer does not depend on popular vote. That's the unconditional probability (i.e. before you have the additional information that the first ball is black).