03 Apr 08
Originally posted by eldragonflysee my answer and think it over
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.
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!
03 Apr 08
Originally posted by alexdinowhy is this inncorect?
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!)
what is wrong?
03 Apr 08
Originally posted by eldragonflyYour 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.
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]
I made the same mistake. 🙂
03 Apr 08
Originally posted by eldragonflyI don't know what you mean. But here's the definition of conditional probability from the Wikipedia article you referenced:
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.
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.
03 Apr 08
1 edit
Originally posted by PBE6apples and oranges.
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.
************************
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.
03 Apr 08
Originally posted by eldragonflyAs I noted above, event A = the other side of the drawn card is silver, and event B = the side facing up is silver.
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.
03 Apr 08
1 edit
Originally posted by PBE6Originally posted by alexdino
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?
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?
03 Apr 08
Originally posted by alexdinoIt'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.
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?