balls in a bag

A different version of a puzzle I put up some time ago.


I have 4 identical bags each containing 2 balls.
1 bag has 2 white balls
1 bag has 2 black balls
2 bags have 1 black and 1 white

I pick a bag at random and take out a ball. It is black.

What is the probability of the other ball being black?

50%

kinda like monty hall problem?...

3 to 2 chance or 66.66%

Originally posted by roma45
3 to 2 chance or 66.66%

I now invite Thomaster and roma45 to fight it out!
(One of you is correct)

my thinking is if you pick a black ball that means the bag with the two white balls is out of the equation. leaving 3 bags with 4 black and 2 white in total, since 1 black is picked that leaves 3 black aginst 2 white, hope i am right

There are four black, and four white balls, so:

p(white) = 0,5
p(black) = 0,5

I name the three bags:
A = two black
B = white and black
C = two white

There are two bags called B (because they are the same).
p(A) = 0,25
P(B) = 0,5
p(C) = 0,25

p(black | A) = 1
p(A | black) = p(black | A) × p(A) / p(black)
p(A | black) = 1 × 0,25 / 0,5 = 0,5

Originally posted by roma45
my thinking is if you pick a black ball that means the bag with the two white balls is out of the equation. leaving 3 bags with 4 black and 2 white in total, since 1 black is picked that leaves 3 black aginst 2 white, hope i am right

There are four ways to pick a black ball.
1. You pick a black ball from bag A.
2. You pick the other black ball from bag A.
3. You pick the black ball from bag B.
4. You pick the black ball from bag B.

if you pick a black first then c can be ruled out, that leaves 3 to 2 chance

1/3

I vote for 50%

There seem to be 4 equally probably ways to get to the current state:

You picked bw bag 1 and got the black ball
You picked bw bag 2 and got the black ball
You picked the bb bag and got the first black ball
You picked the bb bag and got the second black ball.

Of these 4 ways, in two of them the other ball is black.

yes this was my thinking...but now i see this is not completely like monty hall...

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%

The probability of the second ball being black is 0.5, or 50%.

The probability of the first ball being black is also 0.5, which makes the probability of both being black 0.25.

Originally posted by Banana King
yes this was my thinking...but now i see this is not completely like monty hall...

The wording of the question is not quite clear. If Wolfgang *looks* into a random bag, and takes out a black ball *if it is there*, then the odds are completely different to the situation when he randomly takes a ball out of a random bag and it happens to be black.

If it is the former method, then it is very like the Monty Hall problem, Wolfgang has one of the three bags with one or more black balls in, and in two of them there is a white ball left. This gives us 30%

If the extraction is random then it is 50%. This is what I first assumed, and my logic above describes that case, but it only works with random ball picks. We cannot be totally sure, given the wording of the question, that Wolfgang is picking the ball randomly...