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Posers and Puzzles

Posers and Puzzles

  1. Standard member wolfgang59
    Infidel
    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. 03 Oct '10 16:14
    50%
  3. 03 Oct '10 16:27
    kinda like monty hall problem?...
  4. Subscriber roma45
    st johnstone
    03 Oct '10 18:59
    3 to 2 chance or 66.66%
  5. Standard member wolfgang59
    Infidel
    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. Subscriber roma45
    st johnstone
    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. 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. 03 Oct '10 20:09 / 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.
  9. Subscriber roma45
    st johnstone
    03 Oct '10 20:25
    if you pick a black first then c can be ruled out, that leaves 3 to 2 chance
  10. Subscriber AThousandYoung
    It's about respect
    03 Oct '10 21:07
    1/3
  11. 03 Oct '10 22:30 / 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.
  12. 03 Oct '10 22:41
    yes this was my thinking...but now i see this is not completely like monty hall...
  13. 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 member HandyAndy
    Non sum qualis eram
    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. 04 Oct '10 07:42 / 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...