balls in a bag

balls in a bag

Posers and Puzzles

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Quiz Master

RHP Arms

Joined
09 Jun 07
Moves
48793
03 Oct 10

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?

T

ALG

Joined
16 Dec 07
Moves
6190
03 Oct 10

50%

BK

Joined
24 Jan 09
Moves
5514
03 Oct 10

kinda like monty hall problem?...

st johnstone

Joined
14 Nov 09
Moves
417336
03 Oct 10

3 to 2 chance or 66.66%

Quiz Master

RHP Arms

Joined
09 Jun 07
Moves
48793
03 Oct 10

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)

st johnstone

Joined
14 Nov 09
Moves
417336
03 Oct 10

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

T

ALG

Joined
16 Dec 07
Moves
6190
03 Oct 10

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

T

ALG

Joined
16 Dec 07
Moves
6190
03 Oct 10
1 edit

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.

st johnstone

Joined
14 Nov 09
Moves
417336
03 Oct 10

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

Insanity at Masada

tinyurl.com/mw7txe34

Joined
23 Aug 04
Moves
26660
03 Oct 10

1/3

Joined
26 Apr 03
Moves
26771
03 Oct 10
1 edit

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.

BK

Joined
24 Jan 09
Moves
5514
03 Oct 10

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

TY

Joined
12 Sep 10
Moves
6438
04 Oct 10

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%

Read a book!

Joined
23 Sep 06
Moves
18677
04 Oct 10

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.

Joined
26 Apr 03
Moves
26771
04 Oct 10
4 edits

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...