1. Joined
    24 Apr '05
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    3061
    02 Oct '11 22:392 edits
    Originally posted by wolfgang59
    WRONG

    If that were true it would mean that every 2 child family with a daughter had a 2/3 chance of having a son and daughter.

    Switch genders and we can then say every 2 child family with a son has a 2/3 chance of having a son and a daughter.

    So therefore EVERY 2 child faamhaving a son and daughter - which is clearly false.
    I cannot understand your final conclusion (the one you state is clearly false).

    Regardless, your anlaysis here is flawed. Thomaster had basically asserted (before later retracting it) that P[family F has brother & sister | family F has a daughter named Mary] = 2/3. Regardless of whether or not he was right, your analysis assumes that P[F has brother & sister | F has a daughter named Mary] equates to P[F has brother & sister | F has at least one daughter]. But that turns out to be false, just like it is generally false that P[X | I1] equates to P[X | I2] when information I1 happens to entail I2 but I2 does not entail I1.
  2. Standard memberwolfgang59
    Quiz Master
    RHP Arms
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    09 Jun '07
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    48793
    03 Oct '11 07:22
    Originally posted by wolfgang59
    WRONG

    If that were true it would mean that every 2 child family with a daughter had a 2/3 chance of having a son and daughter.

    Switch genders and we can then say every 2 child family with a son has a 2/3 chance of having a son and a daughter.

    So therefore EVERY 2 child faamhaving a son and daughter - which is clearly false.
    Originally posted in much haste!

    WRONG

    If that were true it would mean that every 2 child family with a daughter had a 2/3 chance of having a son and daughter.

    Switch genders and we can then say every 2 child family with a son has a 2/3 chance of having a son and a daughter.

    So therefore EVERY 2 child family having either a son or daughter has a 2/3 chance of having a son and a daughter- which is clearly false.

    We debated this about 18 months ago (but I cannot find it). I first posed the question as a coin toss problem.
  3. Joined
    24 Apr '05
    Moves
    3061
    03 Oct '11 18:571 edit
    Originally posted by wolfgang59
    Originally posted in much haste!

    WRONG

    If that were true it would mean that every 2 child family with a daughter had a 2/3 chance of having a son and daughter.

    Switch genders and we can then say every 2 child family with a son has a 2/3 chance of having a son and a daughter.

    So therefore EVERY 2 child family having either a son or daughter ha ...[text shortened]... s about 18 months ago (but I cannot find it). I first posed the question as a coin toss problem.
    In that case, I have another objection against your analysis in addition to my previous one. Your analysis seems to rest on the following assumption (call it A), which I think is generally false:

    A: If P[X | I1] = m; and if P[X | I2] = m; then P[X | I1 OR I2] = m.

    (Here in this case, X = F has a son & a daughter; and I1 = F has at least one daughter; and I2 = F has at least one son, and m=2/3).

    Am I correct that your analysis relies on A? If so, do you have some justification for using this? Because as far as I can tell, it is generally false and only holds for certain cases. Consider:

    P[X | I1 OR I2] = P[X & (I1 OR I2)]/P[I1 OR I2]
    ... = P[(X & I1) OR (X & I2)]/P[I1 OR I2]
    ... = {P[X & I1] + P[X & I2] - P[(X & I1) & (X & I2)]}/P[I1 OR I2]
    ... = {P[X & I1] + P[X & I2] - P[X & (I1 & I2)]}/{P[I1] + P[I2] - P[I1 & I2]}

    By assumption here P[X & I1]/P[I1] = P[X & I2]/P[I2] = m. So it should follow that P[X | I1 OR I2] = m if and only if the following is satisfied:

    P[X & (I1 & I2)] = m*P[I1 & I2].

    But as far as I can tell, this only holds for the following cases:
    (a) It holds if I1 and I2 are mutually exclusive.
    (b) Else, it holds if X and [I1 & I2] are independent and P[X] =m; or if P[X | (I1 & I2)] = m (= P[X | (I1 OR I2)]).

    But since none of these should hold in this case, your reductio doesn't work, since one should simply reject A, rather than the premises that P[X | I1] = P[X | I2] = m.

    What am I missing?
  4. Standard memberPalynka
    Upward Spiral
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    02 Aug '04
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    8702
    04 Oct '11 15:122 edits
    I agree with iamatiger. It depends on the prior beliefs on the process on how Mary's name was selected.
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