03 Apr 08
There are 3 cards, that makes 6 sides.
He showed you a silver side.
there are three possibilities:
a) the mixed card
b) the first side of the silver card
c) the second side of the silver card
so, if anything is fair, the chance, that the other
side is silver too, is 2:1.
According to Fabians nomenclature its
2b) he thinks, that he is smart and you are crazy
03 Apr 08
1 edit
Originally posted by LemonJellosomething like that. But if it is required to show the silver side of a card, he then has only 2 choices :
Not a fair bet. We should really be keeping track of sides, not cards. If, as supposed, you see a silver side, then that eliminates it from being the gold/gold card. So counting the remaining possible sides, there are two silvers, one gold -- each of which should be equally likely to be on the other side given what we know.
In other words, one who accepts the bet should on average lose 2 out of every 3 times.
1) the silver/silver -
2) the silver/gold -
and since he does appear to pick from the 2 cards at *random*, it is a therefore 50/50 bet.
03 Apr 08
1 edit
Originally posted by eldragonflyThe choice of card is random (each card has a 1/3 chance of being drawn) and the choice of side is random too (each side has 1/2 chance of being shown face up). It just so happens that in this example, silver was face up.
something like that. But if it is required to show the silver side of a card, he then has only 2 choices :
1) the silver/silver -
2) the silver/gold -
and since he does appear to pick from the 2 cards at *random*, it is a therefore 50/50 bet.
EDIT: LemonJello and afx both gave the correct answer.
03 Apr 08
Originally posted by eldragonflyyes and no. There are your two cases, but they don't have the same propability.
something like that. But if it is required to show the silver side of a card, he then has only 2 choices :
1) the silver/silver -
2) the silver/gold -
and since he does appear to pick from the 2 cards at *random*, it is a therefore 50/50 bet.
you have
1a) silver/silver-card front side
1b) silver/silver-card back side
2) the silver-gold-card silver-side
so its 2:1 for having the silver-silver-card
03 Apr 08
9 edits
Originally posted by PBE6Reword the problem then, this is nonsensical.
The choice of card is random (each card has a 1/3 chance of being drawn) and the choice of side is random too (each side has 1/2 chance of being shown face up). It just so happens that in this example, silver was face up.
EDIT: LemonJello and afx both gave the correct answer.
"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?"
If he held up a sliver card then there are only two choices, hence it is a 50/50 bet that the other side of a card showing a sliver side is silver.
He can't make this bet if he draws the gold/gold card. It is an even bet.
There is no other solution.
You don't make the case for holding up a gold sided card, so your "demonstration" is incorrect.
03 Apr 08
Originally posted by eldragonflyCompletely wrong. This is a simple application of Bayes Theorem, as follows:
Reword the problem then, this is nonsensical.
"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 o ...[text shortened]... 't make the case for holding up a gold sided card, so your "demonstration" is incorrect.
P(A g B) = P(A U B) / P(B)
where P(A g B) = probability of A given B
P(A U B) = probability of A and B happening together
P(B) = probability of B
In this case:
A = the other side is silver
B = the silver side is face up
Counting the sides, there are 6 possible outcomes in this game:
gold/gold with gold(1) face up
gold/gold with gold(2) face up
gold/silver with gold face up
gold/silver with silver face up
silver/silver with silver(1) face up
silver/silver with silver(2) face up
To calculate P(A U B), we simply count the number of outcomes where silver is face up (B) and the other side is silver (A). The outcomes that satisfy these conditions are:
silver/silver with silver(1) face up
silver/silver with silver(2) face up
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
Since the dealer is only offering even money, or 1:1, the bet is not a fair one. If you still don't believe me, try playing this game and keeping track of the result, or setting up an Excel spreadsheet to run the game for you.
03 Apr 08
Originally posted by eldragonflyAs LemonJello said, think in terms of sides rather than cards. There is a 2/3 chance of the underside of the card being the same colour as the side showing.
Reword the problem then, this is nonsensical.
"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 o ...[text shortened]... 't make the case for holding up a gold sided card, so your "demonstration" is incorrect.
03 Apr 08
Originally posted by PBE6Thanks PBE6 but you are still wrong and/or the problem is poorly worded.. your application of the Bayes Theorem is incorrect. 😕
Completely wrong. This is a simple application of Bayes Theorem, as follows:
P(A g B) = P(A U B) / P(B)
where P(A g B) = probability of A given B
P(A U B) = probability of A and B happening together
P(B) = probability of B
In this case:
A = the other side is silver
B = the silver side is face up
Counting the sides, there are 6 possible outcomes ...[text shortened]... me and keeping track of the result, or setting up an Excel spreadsheet to run the game for you.
03 Apr 08
Originally posted by Green PaladinNo problem sir let's think sides then. 😳
As LemonJello said, think in terms of sides rather than cards. There is a 2/3 chance of the underside of the card being the same colour as the side showing.
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 Bayes Theorem, or start magically counting sides of a card that is not included. The probability of drawing the gold/gold card is zero.
http://en.wikipedia.org/wiki/Conditional_probability
As i say reword the problem, the double gold card should is not a valid event in the stated scenario.
http://en.wikipedia.org/wiki/Bayes'_theorem
03 Apr 08
2 edits
Originally posted by alexdino''..................................................................................................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:
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!)
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!
03 Apr 08
Originally posted by alexdinoThat is incorrect.
''..................................................................................................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
...[text shortened]... afirmation
P.S.: you have to see the easy way too not only your point of view!