1. Standard membereldragonfly
    leperchaun messiah
    thru a glass onion
    Joined
    19 Apr '03
    Moves
    16870
    03 Apr '08 20:10
    Originally posted by heinzkat
    In the case the man with the hat holds up a gold card, he bets the other side is gold too, simple as that.
    believe it. 😕
  2. Joined
    23 Oct '07
    Moves
    2831
    03 Apr '08 20:11
    Originally posted by eldragonfly
    Drawing a silver sided card is not a conditional probability related to the gold/gold card, according to the problem description, and so Bayes Theorem does not apply.
    see my answer and think it over
    the odds can't be 50/50 because hi draws from 3 cards and even you can exclude the gold/gold card you shouldn't!
  3. Joined
    23 Oct '07
    Moves
    2831
    03 Apr '08 20:13
    Originally posted by alexdino
    it comes down to this the odds are double for him to win because he is beting he drew the silver/silver card (1 assupmtion) while you are beting that he drew the silver/gold card and he drew it with the silver side twards you(2 assumptions!)
    why is this inncorect?
    what is wrong?
  4. Pale Blue Dot
    Joined
    22 Jul '07
    Moves
    21637
    03 Apr '08 20:13
    Originally posted by eldragonfly
    No problem sir let's think sides then. 😳

    There is a 50/50 chance that the other side of the card is silver,

    ...and a 50/50 chance that the other side of the card is gold,

    if a silver side is shown. Period.

    This is the only case described in the original problem. there is no other solution, so it is an error and you don't need to drag in B ...[text shortened]... a valid event [b]in the stated scenario
    .

    http://en.wikipedia.org/wiki/Bayes'_theorem[/b]
    Your reasoning is only correct if the cowpoke is drawing the silver/silver and silver/gold cards an equal percentage of the time. The fact is, he isn't.

    I made the same mistake. 🙂
  5. Standard membereldragonfly
    leperchaun messiah
    thru a glass onion
    Joined
    19 Apr '03
    Moves
    16870
    03 Apr '08 20:15
    Originally posted by alexdino
    see my answer and think it over
    the odds can't be 50/50 because hi draws from 3 cards and even you can exclude the gold/gold card you shouldn't!
    Wrong. The gold/gold card is excluded from the chosen event of showing a silver sided card. 😉
  6. Standard membereldragonfly
    leperchaun messiah
    thru a glass onion
    Joined
    19 Apr '03
    Moves
    16870
    03 Apr '08 20:15
    Originally posted by Green Paladin
    Your reasoning is only correct if the cowpoke is drawing the silver/silver and silver/gold cards an equal percentage of the time. The fact is, he isn't.

    I made the same mistake. 🙂
    As if.
  7. Standard membereldragonfly
    leperchaun messiah
    thru a glass onion
    Joined
    19 Apr '03
    Moves
    16870
    03 Apr '08 20:17
    Originally posted by alexdino
    why is this inncorect?
    what is wrong?
    The event is showing a silver sided card not having three cards in a hat and selecting one with a silver side. You don't need to drag in Bayes Theorem, like really.
  8. Standard memberPBE6
    Bananarama
    False berry
    Joined
    14 Feb '04
    Moves
    28719
    03 Apr '08 20:20
    Originally posted by eldragonfly
    Drawing a silver sided card is not a conditional probability related to the gold/gold card, according to the problem description, and so Bayes Theorem does not apply.
    I don't know what you mean. But here's the definition of conditional probability from the Wikipedia article you referenced:

    Conditional probability is the probability of some event A, given the occurrence of some other event B. Conditional probability is written P(A|B), and is read "the probability of A, given B".

    (http://en.wikipedia.org/wiki/Conditional_probability)

    In this case, the player is trying to guess which card has been drawn at random given some partial information about the card revealed after the draw has been made. I think that fits the definition given above very well.
  9. Standard membereldragonfly
    leperchaun messiah
    thru a glass onion
    Joined
    19 Apr '03
    Moves
    16870
    03 Apr '08 20:221 edit
    Originally posted by PBE6
    I don't know what you mean. But here's the definition of conditional probability from the Wikipedia article you referenced:

    Conditional probability is the probability of some event A, given the occurrence of some other event B. Conditional probability is written P(A|B), and is read "the probability of A, given B".

    (http://en.wikipedia.org/wiki/Conditiona ...[text shortened]... ealed after the draw has been made. I think that fits the definition given above very well.
    apples and oranges.
    ************************
    Conditional probability is
    the probability of some event A,
    given the occurrence of some other event B.
    Conditional probability is written P(A|B), and is read "the probability of A, given B".
    ************************

    Show me your event B, show me your other event, show me the kwon.
  10. Joined
    07 Sep '05
    Moves
    35068
    03 Apr '08 20:23
    This thread is still going? It's not that complicated, and it was answered correctly very early on.
  11. Standard membereldragonfly
    leperchaun messiah
    thru a glass onion
    Joined
    19 Apr '03
    Moves
    16870
    03 Apr '08 20:24
    Originally posted by mtthw
    This thread is still going? It's not that complicated, and it was answered correctly very early on.
    No it wasn't.

    Show me the kwon.

    😛
  12. Standard memberPBE6
    Bananarama
    False berry
    Joined
    14 Feb '04
    Moves
    28719
    03 Apr '08 20:24
    Originally posted by eldragonfly
    apples and oranges.
    We should play. 😉
  13. Standard memberPBE6
    Bananarama
    False berry
    Joined
    14 Feb '04
    Moves
    28719
    03 Apr '08 20:31
    Originally posted by eldragonfly
    apples and oranges.
    ************************
    Conditional probability is
    the probability of some event A,
    given the occurrence of some other event B.
    Conditional probability is written P(A|B), and is read "the probability of A, given B".
    ************************

    Show me your event B, show me your other event, show me the kwon.
    As I noted above, event A = the other side of the drawn card is silver, and event B = the side facing up is silver.
  14. Joined
    23 Oct '07
    Moves
    2831
    03 Apr '08 20:471 edit
    Originally posted by PBE6
    This one comes from the Old West, apparently.

    A man with a hat shows you 3 cards, one completely gold, one completely silver, and one gold on one side and silver on the other. He puts them in his hat, and picks one at random. He then shows you one side of the card he picked, which happens to be silver. Now he says "I'll bet you even money that the other side of this card is silver too...whaddaya say, partner?"

    Is this bet a fair one?
    Originally posted by alexdino
    it comes down to this the odds are double for him to win because he is beting he drew the silver/silver card (1 assupmtion) while you are beting that he drew the silver/gold card and he drew it with the silver side twards you(2 assumptions!)


    why is this inncorect?
    what is wrong?


    ''..................................................................................................So there are 2. Similarly, to calculate P(B) we count the number of times silver is face up (B). The outcomes that satisfy this condition are:

    gold/silver with silver face up
    silver/silver with silver(1) face up
    silver/silver with silver(2) face up

    So there are 3. Therefore, the probability that the other side is silver (A), given that the side face up is silver (B), is:

    P(A g B) = P(A U B) / P(B) = 2/3 ''
    ????????????????????????


    first you say ''not quite'' than you give the same explination only overcomplicated!!!
    i'm sorry but i stick to my afirmation




    P.S.: you have to see the easy way too not only your point of view!


    PBE6 PLEASE ANSWER WHY IN GODS NAME IS THIS WRONG?
  15. Standard memberPBE6
    Bananarama
    False berry
    Joined
    14 Feb '04
    Moves
    28719
    03 Apr '08 20:54
    Originally posted by alexdino
    Originally posted by alexdino
    it comes down to this the odds are double for him to win because he is beting he drew the silver/silver card (1 assupmtion) while you are beting that he drew the silver/gold card and he drew it with the silver side twards you(2 assumptions!)


    why is this inncorect?
    what is wrong?


    ''................................. ...[text shortened]... oo not only your point of view!


    PBE6 PLEASE ANSWER WHY IN GODS NAME IS THIS WRONG?
    It's wrong because you counted the number of assumptions you had to make, as opposed to counting the useful outcomes. For example, you could reword the dealer's 1 assumption that "he drew the silver/silver card" as 2, "he drew the silver/silver card and one silver side was face up" and "he drew the silver/silver card and the other silver side was face up". This would make the bet seem fair, when it is not.
Back to Top

Cookies help us deliver our Services. By using our Services or clicking I agree, you agree to our use of cookies. Learn More.I Agree