1. Standard memberwolfgang59
    howling mad
    In the den
    Joined
    09 Jun '07
    Moves
    45641
    03 Oct '10 14:09
    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?
  2. ALG
    Joined
    16 Dec '07
    Moves
    6190
    03 Oct '10 16:14
    50%
  3. Joined
    24 Jan '09
    Moves
    5514
    03 Oct '10 16:27
    kinda like monty hall problem?...
  4. Subscriberroma45online
    st johnstone
    Joined
    14 Nov '09
    Moves
    347600
    03 Oct '10 18:59
    3 to 2 chance or 66.66%
  5. Standard memberwolfgang59
    howling mad
    In the den
    Joined
    09 Jun '07
    Moves
    45641
    03 Oct '10 19:01
    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)
  6. Subscriberroma45online
    st johnstone
    Joined
    14 Nov '09
    Moves
    347600
    03 Oct '10 19:27
    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
  7. ALG
    Joined
    16 Dec '07
    Moves
    6190
    03 Oct '10 20:06
    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
  8. ALG
    Joined
    16 Dec '07
    Moves
    6190
    03 Oct '10 20:091 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.
  9. Subscriberroma45online
    st johnstone
    Joined
    14 Nov '09
    Moves
    347600
    03 Oct '10 20:25
    if you pick a black first then c can be ruled out, that leaves 3 to 2 chance
  10. SubscriberAThousandYoung
    All My Soldiers...
    tinyurl.com/y9ls7wbl
    Joined
    23 Aug '04
    Moves
    24791
    03 Oct '10 21:07
    1/3
  11. Joined
    26 Apr '03
    Moves
    25805
    03 Oct '10 22:301 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.
  12. Joined
    24 Jan '09
    Moves
    5514
    03 Oct '10 22:41
    yes this was my thinking...but now i see this is not completely like monty hall...
  13. Joined
    12 Sep '10
    Moves
    5745
    04 Oct '10 03:46
    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%
  14. Standard memberHandyAndy
    Non sum qualis eram
    At the edge
    Joined
    23 Sep '06
    Moves
    18031
    04 Oct '10 04:15
    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.
  15. Joined
    26 Apr '03
    Moves
    25805
    04 Oct '10 07:424 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...
Back to Top