18 Apr 08
eldragonfly is far from the only person to reach the conclusion on the questions most heavily debated.
I can understand why he thinks as he does, but it doesn't mean I think he's right.
Having posed these questions to some other people, it has become obvious the the problems are tricky in a sense, because the intuitive and simple answer doesn't match the numbers (no matter what eldragonfly says).
It seems to be something that a lot of people cannot easily discern.
Disagree with me? Start by examining everything that COULD happen, and how likely it is. Consider that the mixed color card could be drawn in two visually distinct ways when you do this (and since you can discern the difference with your own eye, they aren't the same..)
THAT was the point of my last questions, that and to see if his numbers would match my own.
18 Apr 08
Originally posted by geepamoogleOne more time:
eldragonfly is far from the only person to reach the conclusion on the questions most heavily debated.
I can understand why he thinks as he does, but it doesn't mean I think he's right.
Having posed these questions to some other people, it has become obvious the the problems are tricky in a sense, because the intuitive and simple answer doesn't mat ...[text shortened]...
THAT was the point of my last questions, that and to see if his numbers would match my own.
The bayesian solution counts sides not cards. The "dealer" is offering bets on either a SS or SG card, provided that the silver side of the SG card is showing. Remember the GG card is excluded in the stated scenario. Now assuming that he picks randomly between these two remaining cards, then it is a fair bet, i.e. he turns over the card he has selected at random and reveals the other side, which can only be silver or gold.
Let's let this represent a silver side of a card ---> 😕
Let's let this represent a gold side of a card ---> 🙂
the dealer shows three cards in his hat :
🙂🙂
😕🙂
😕😕
The 🙂🙂 card is excluded in this example by definition.
Now we are left with two cards, assuming the silver side of the SG is face up. Remember for both cards the silver sides are face up.
😕😕
😕🙂
The dealer then selects randomly between the two cards and makes his even odds wager. He then turns over that card.
only two things can occur:
🙂😕
-or-
😕😕
It is an even money bet in the stated scenario.
18 Apr 08
Originally posted by eldragonflywrong, wrong, WRONG
One more time:
The bayesian solution counts sides not cards. The "dealer" is offering bets on either a SS or SG card, provided that the silver side of the SG card is showing. Remember the GG card is excluded in the stated scenario. Now assuming that he picks randomly between these two remaining cards, then it is a fair bet, i.e. he turns over the card ...[text shortened]... hings can occur:
🙂😕
-or-
😕😕
It is an even money bet in the stated scenario.
The dealer is selecting randomly from the THREE cards. (see original post)
There is a 2/3 chance that the card he picks is S&S or G&G. So whether its gold face up or silver face up he offers any mugs around their 50/50 bet while odds are 2:1 in his favour.
You sure would be poor in the Old West!! 😛
18 Apr 08
2 edits
Originally posted by wolfgang59Wrong. The problem is ambiguously and/or poorly worded, this is not the same as randomly selecting. The dealer is not randomly selecting between the three cards, the gold/gold 🙂🙂 card is automatically excluded from the sample space.
wrong, wrong, WRONG
The dealer is selecting randomly from the THREE cards. (see original post)
There is a 2/3 chance that the card he picks is S&S or G&G. So whether its gold face up or silver face up he offers any mugs around their 50/50 bet while odds are 2:1 in his favour.
18 Apr 08
Originally posted by eldragonflyMy problem was worded just fine. You said so yourself on p.5:
Wrong. The problem is ambiguously and/or poorly worded, this is not the same as randomly selecting.
"No the ambiguity was on my end, it was properly worded. Statistics has its own language, but you did make too many assumptions. It was not in error for me to limit the sample space the the two card experiment, but a little more thought would have yielded the same result."
Now shut up. ðŸ˜
18 Apr 08
Originally posted by PBE6Wrong, your solution is in error. The sides solutions is separate and distinct from the cards solution.
My problem was worded just fine. You said so yourself on p.5:
"No the ambiguity was on my end, it was properly worded. Statistics has its own language, but you did make too many assumptions. It was not in error for me to limit the sample space the the two card experiment, but a little more thought would have yielded the same result."
Now shut up. ðŸ˜
18 Apr 08
1 edit
Originally posted by eldragonflyBTW, this is what a real solution looks like:
Wrong, your solution is in error. The sides solutions is separate and distinct from the cards solution.
********************************************************
Might as well solve this stupid thing one more time, to see if I can make it clearer.
Q1. There are 3 cards are in a hat, one GG, one SG, and one SS. One of the cards is drawn at random (i.e. each card is equally likely to be drawn).
(a) What is the probability that the GG was picked?
(b) What is the probability that the SG was picked?
(c) What is the probability that the SS was picked?
A1. Since there are 3 cards, and each card is equally likely to be drawn, we have:
P(GG) + P(SG) + P(SS) = 1
P(GG) = P(SG) = P(SS)
Therefore P(GG) = P(SG) = P(SS) = 1/3.
Q2. The chosen card is now placed on a table in a random manner such that one side is exposed and one is covered (i.e. either side of the card is equally likely to be exposed). Without looking, what is the probability that the exposed side is:
(a) silver?
(b) gold?
A2. There are 3 cards with 2 sides each, so there are 6 possibilities:
P(GG card chosen, G1 face up) = P(GG card chosen)*P(G1 face up) = (1/3)*(1/2) = 1/6
P(GG card chosen, G2 face up) = P(GG card chosen)*P(G2 face up) = (1/3)*(1/2) = 1/6
P(SS card chosen, S1 face up) = P(SS card chosen)*P(S1 face up) = (1/3)*(1/2) = 1/6
P(SS card chosen, S2 face up) = P(SS card chosen)*P(S2 face up) = (1/3)*(1/2) = 1/6
P(SG card chosen, G face up) = P(SG card chosen)*P(G face up) = (1/3)*(1/2) = 1/6
P(SG card chosen, S face up) = P(SG card chosen)*P(S face up) = (1/3)*(1/2) = 1/6
(a) Of the above possibilities, 3 meet the criteria of having a silver side exposed. Therefore, the probability that the exposed side is silver is 3/6 = 1/2.
(b) Of the above possibilities, 3 meet the criteria of having a gold side exposed. Therefore, the probability that the exposed side is gold is 3/6 = 1/2.
Q3. The exposed side is revealed to be silver. Given this information:
(a) what is the probability that the chosen card was SS?
(b) what is the probability that the chosen card was SG?
(c) what is the probability that the chosen card was GG?
A3. For brevity, let's rename the probabilities given in Q2 above as follows:
P(GG card chosen, G1 face up) = P(GG,G1) = 1/6
P(GG card chosen, G2 face up) = P(GG,G2) = 1/6
P(SS card chosen, S1 face up) = P(SS,S1) = 1/6
P(SS card chosen, S2 face up) = P(SS,S2) = 1/6
P(SG card chosen, G face up) = P(SG,G) = 1/6
P(SG card chosen, S face up) = P(SG,S) = 1/6
(a) Of the above possibilities, only P(SS,S1), P(SS,S2) and P(SG,S) meet the criteria of having a silver side exposed. Of those, only P(SS,S1) and P(SS,S2) meet the criteria of being the SS card. Therefore, the probability that the card chosen was SS given that a silver side has already been exposed is given as:
P(SS chosen given silver side exposed) = (P(SS,S1) + P(SS,S2)) / (P(SS,S1) + P(SS,S2) + P(SG,S))
= (1/6 + 1/6) / (1/6 + 1/6 + 1/6)
= 2/3.
(b) Of the above possibilities, only P(SS,S1), P(SS,S2) and P(SG,S) meet the criteria of having a silver side exposed. Of those, only P(SG,S) meets the criteria of being the SG card. Therefore, the probability that the card chosen was SG given that a silver side has already been exposed is given as:
P(SG chosen given silver side exposed) = P(SG,S) / (P(SS,S1) + P(SS,S2) + P(SG,S))
= (1/6) / (1/6 + 1/6 + 1/6)
= 1/3.
(c) Of the above possibilities, only P(SS,S1), P(SS,S2) and P(SG,S) meet the criteria of having a silver side exposed. Of those, none of them meet the criteria of being the GG card. Therefore, the probability that the card chosen was GG given that a silver side has already been exposed is given as:
P(GG chosen given silver side exposed) = 0 / (P(SS,S1) + P(SS,S2) + P(SG,S))
= 0 / (1/6 + 1/6 + 1/6)
= 0
18 Apr 08
1 edit
Originally posted by eldragonflyIt's exactly the same problem, you retard.
Wrong. you are showing the probability of this:from: http://en.wikipedia.org/wiki/Three_cards_problem#Card_versionnot your cards in a hat even money bet.
"..the side facing up is black, what are the odds that the other side is also black?"
How can you obviously take the time to search out a solution, post a link to it, yet not read the solution!?
18 Apr 08
Originally posted by eldragonfly"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"
Wrong. The problem is ambiguously and/or poorly worded, this is not the same as randomly selecting. The dealer is not randomly selecting between the three cards, the gold/gold 🙂🙂 card is automatically excluded from the sample space.
How can you say the dealer is not picking from 3 cards?
READ THE ORIGINAL PROBLEM !!!!!!!!!!!!!
If you still do not understand the problem do it experimentally. After 20 trials or so you will see you are wrong.
18 Apr 08
Originally posted by PBE6The answer is yes, this is a fair bet, the gold/gold card is excluded from the sample space.
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?
18 Apr 08
1 edit
Originally posted by eldragonflyNobody cares.
[quote]Originally posted by PBE6
[b]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. No ...[text shortened]... The answer is yes, this is a fair bet, the gold/gold card is excluded from the sample space.[/b]