Posers and Puzzles

Posers and Puzzles

  1. Standard memberabejnood
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    19 Jul '09 23:10
    http://en.wikipedia.org/wiki/Monty_Hall_problem
  2. Standard memberPalynka
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    21 Jul '09 11:29
    Originally posted by abejnood
    http://en.wikipedia.org/wiki/Monty_Hall_problem
    FAIL
  3. Joined
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    11 Aug '09 17:49
    Originally posted by Palynka
    FAIL
    This is so simple.

    At first you have a probability of 1/3 of picking the correct door.
    Then a WRONG door is eliminated.

    This leaves two doors.
    Although you do not know if at first you selected the wrong door.
    The chances are 2/3 you have it wrong. Thus your change of winning the car is 1/3. However you do know that the car is still there. So changing your choice reverses the odds.

    Tadaa.
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  4. Joined
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    11 Aug '09 17:58
    Originally posted by wolfgang59
    Palynka I agree with you entirely ... so why do we get the old chestnut of:-

    A man has two children.
    What is the probability he has one son and one daughter?
    We all agree 0.5

    You hear him on the phone chatting to one of his children called MARY.
    now what is the probabilty he has a son and a daughter?
    0.5 (I hope Palynka agrees) but so many puzzles, quizzes, books and (dare I say it teachers) say 0.67 !!!!!!!!!!
    These things always become easy if you go back to basic.

    If you do know nothing the chanches are

    Girl Girl
    Boy Boy
    Girl Boy
    Boy Girl


    Since you only know the man has one girl, that leaves the options
    Girl Girl 33,333%
    Girl Boy 33.333%
    Boy Girl 33.333%

    So the books and teachers are correct.
    Strangely enough, it would be different if you would know that for example his OLDEST child is a girl. Then your chanches increase because you only have two possibilities left.
    ----------------

    Do not confuse this with the probability of having a Boy after your first is a Daughter, that indeed is 50%

    Chance has no memory
  5. Joined
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    11 Aug '09 22:517 edits
    Originally posted by wolfgang59
    I toss two coins, lets say a £1 coin and a £2 coin they land unseen by me.

    A friend tells me that I have tossed "at least one Head"

    What are the chances that the coins are different? (ie H & T)

    Easy!
    My friend's information tells me that the result is one of these three.

    £1 head and £2 head
    £1 head and £2 tail
    £1 tail and £2 head

    The c ...[text shortened]... l the time) my chances of tossing a head and a tail are 2/3.

    But we all know its 1/2.
    🙄
    Isn't this just straightforward application of Bayes?

    As in:

    P(H&T| at least 1 H) = P(at least 1 H|H&T)*P(H&T)/P(at least 1 H)
    = 1*0.5/0.75 = 2/3.

    In what follows, you are simply not answering the same question. You would be answering P(H&T|at least 1 H or T) = P(at least 1 H or T|H&T)*P(H&T)/P(at least 1 H or T) = 1*0.5/1 = 1/2. So, your mistake must lie somewhere in a faulty equivalence of the problem. In other words, the probablility that you have H&T given that you have at least one H is the same as the probablility that you have H&T given that you have at least one T; but it is NOT the same as the probablility that you have H&T given that you have at least one H or T.

    In your OP, your inference relies on the assumption that the probability of H&T given you have at least "one of something" should be the same as either (1) the probability of H&T given you have at least one H or (2) the probability of H&T given you have at least one T. But there is no reason to think that is true because if I have at least one of something (i.e., I have at least one H or T), that doesn't entail that I have at least one H, nor does it entail that I have at least one T. Why should the probabilities be the same if the information we have is not equivalent? If I tell you that you have at least 1 H (or that you have at least 1 T), then obviously you know that it's also the case that you also have at least 1 H or T. But it doesn't work the other way around, so they don't provide the same amount of information.
  6. Standard memberPalynka
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    14 Aug '09 13:461 edit
    Originally posted by MetBierOp
    This is so simple.

    At first you have a probability of 1/3 of picking the correct door.
    Then a WRONG door is eliminated.

    This leaves two doors.
    Although you do not know if at first you selected the wrong door.
    The chances are 2/3 you have it wrong. Thus your change of winning the car is 1/3. However you do know that the car is still there. So changing your choice reverses the odds.

    Tadaa.
    -----------------
    -----------------
    The Monty Hall problem is NOT the same. The game show host would never open a door with a prize, something which is key. That doesn't apply here. In the Monty hall you're conditioning on the game show host opening a door with a goat (something which happens with probability 1). In this case, you're conditioning on something which doesn't have necessarily probability one.

    And anyway, this is not about memory in probabilities, it's about conditional probabilities.
  7. Standard memberwolfgang59
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    14 Aug '09 13:561 edit
    METBIEROP you are wrong!

    A palynka has demonstarted. (Just substitute heads and tails for boys and girls in original problem)

    Another way of looking at it is this.
    A man has two children.
    One he is speaking to on the phone, the other he isnt.
    What is the probability of the child not on the phone being a boy?
    0.5
    How can knowing the sex of the child on the phone change that?
  8. Standard memberwolfgang59
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    14 Aug '09 14:00
    Originally posted by LemonJello
    Isn't this just straightforward application of Bayes?

    As in:

    P(H&T| at least 1 H) = P(at least 1 H|H&T)*P(H&T)/P(at least 1 H)
    = 1*0.5/0.75 = 2/3.

    In what follows, you are simply not answering the same question. You would be answering P(H&T|at least 1 H or T) = P(at least 1 H or T|H&T)*P(H&T)/P(at least 1 H or T) = 1*0.5/1 = 1/2. So, your mis ...[text shortened]... doesn't work the other way around, so they don't provide the same amount of information.
    So you are saying its 2/3?
  9. Joined
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    14 Aug '09 14:011 edit
    -------------------------------------
    The Monty Hall problem is NOT the same. The game show host would never open a door with a prize, something which is key. That doesn't apply here. In the Monty hall you're conditioning on the game show host opening a door with a goat (something which happens with probability 1). In this case, you're conditioning on something which doesn't have necessarily probability one.

    And anyway, this is not about memory in probabilities, it's about conditional probabilities.
    --------------------------------------

    I was not replying to the 2 coins paradox but to the Monthy Hall problem. I was stating: The Monthy Hall problem is simple.

    However you are also correct when you state the influence of the gameshow-host is causing this.

    I was not replying to the coins.
  10. Joined
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    14 Aug '09 15:22
    Originally posted by wolfgang59
    METBIEROP you are wrong!

    A palynka has demonstarted. (Just substitute heads and tails for boys and girls in original problem)

    Another way of looking at it is this.
    A man has two children.
    One he is speaking to on the phone, the other he isnt.
    What is the probability of the child not on the phone being a boy?
    0.5
    How can knowing the sex of the child on the phone change that?
    Hello Wolfgang.

    No need to shout :-)

    I am quite wright really.
    And believe me, I know what I am talking about.

    First lets me take your statement:

    When you do know nothing the probability is 50%. That is totally correct.


    Now you ask, how does knowing the gender of the other child change that.

    Let me explain through an example.
    Given = They have 2 children. No more no less.

    If we take 1.000 random family's with 2 children, then we would get the following (or something close).

    250 x Girl:Girls
    250 x Girl:Boy
    250 x Boy:Girl
    250 x Boy:Boy

    Stop for a minute and please agree on this point or disagree and explain
    that to me.

    Now we also class the man with his boy on the phone in this group.
    you only can exclude the Girls:Girls group.

    Now do the math and you'll have to agree.

    -----------
    The same goes for flipping coins.
    Replace the groups of family with people tossing coins, reveal one toss and the compairison is the same.

    But once again, if you allready have flipped a coin, (lets say heads) then the possibility of the next coin beeing heads is 50%.
    It simply is not the same question.
  11. Joined
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    14 Aug '09 16:283 edits
    Originally posted by wolfgang59
    So you are saying its 2/3?
    Yes, I think in the problem as you laid it out, the probability of H&T given you have at least 1 H is 2/3. Same would go for the probability of H&T given you have at least 1 T. Like I mentioned, isn't this what an application of Bayes would tell us?

    The reason why you won't accept that it is 2/3 is because you keep swaying yourself with a bad argument. Your argument is the following:

    (1) The probability of H&T given I have at least 1 H should be the same as the probability of H&T given that I have at least "one of something".
    (2) Obviously, the probability of H&T given that I have at least one of something is 1/2.
    (3) So, the probability of H&T given I have at least 1 H should also be 1/2.

    Your premise (1) is false (and as a result, so is the conclusion 3). I don't understand why you keep thinking (1) is true because there are absolutely no good reasons to think it is true. That you have at least 1 H; that you have at least 1 H or T; these two statements obviously do not provide the same information. Why would the two conditional probabilities be the same if the information we have in the two cases is different?

    EDIT: Actually, what you did is you presented something like a reductio, where you say that an obvious contradiction arises from the conjunction of (1) P(H&T| at least 1 H) = 2/3; (2) P(H&T| at least one of something) = 1/2; and (3) P(H&T| at least 1 H) = P(H&T| at least one of something). Yes, an obvious contradiction does result from that. But I think you did not correctly identify which of the 3 to reject.
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    14 Aug '09 16:33
    OH DEAR!!
  13. Standard memberwolfgang59
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    15 Aug '09 11:30
    Metbierop you are quite wrong.
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    17 Aug '09 11:40
    If you hear a daughter on the phone, chances are better than even that the other is a boy.

    Why? Simple - girls are far more likely to spend time on the phone 🙂. That increases the relative probability of the BG/GB combination compared to GG.
  15. Standard memberPalynka
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    18 Aug '09 13:432 edits
    Originally posted by LemonJello
    Yes, I think in the problem as you laid it out, the probability of H&T given you have at least 1 H is 2/3. Same would go for the probability of H&T given you have at least 1 T. Like I mentioned, isn't this what an application of Bayes would tell us?
    It's a subtle point, but what you say is not the same as the stated problem. It assumes that the friend would always say that there's at least one H when the coins are different.

    If you don't assume that, then conditioning on "at least 1T" is not the same as conditioning on "the friend tells you that there's at least 1T"!

    Please see my second post on the first page for more details, along with the math.

    ---------------------
    His "at least one of something" is not a good way to explain what he means to say. Let's forget about that for now.
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