03 Apr 08
2 edits
Originally posted by PBE6that does not meet the definition of a conditional probability. Those are independent events. Show me how your gold/gold card fits in, i can assure you that it doesn't.
As I noted above, event A = the other side of the drawn card is silver, and event B = the side facing up is silver.
03 Apr 08
Originally posted by eldragonflyYes it was.
No it wasn't.
The only possible dispute is over the definition of the problem. I think it's pretty obvious what it means, but when in doubt I'd defer to the person who asked the question in the first place.
Who is the person you are currently arguing with. Pointless.
03 Apr 08
Be it noted here that the man only makes the bet once the randomly chosen side is revealed.
Had the side facing up been gold, his bet would have been even odds the other side was gold.
Here is the simplest way to think of it, seriously.
There are three cards in the hat, one of which has differing sides. A card is picked at random from the hat and placed with one side showing, and the other unknown by either you or him. His wager is that the back side of the card matches the front side, and he is paying even odds.
So you double your money if the card chosen is the silver/gold, regardless of which side turns up, but lose your bet if one of the other two cards shows up.
It just so happens that in this particular case, the side which comes up is silver rather than gold.
03 Apr 08
1 edit
Originally posted by mtthwYou're super-duper exactly wrong, the problem is poorly described, the sample space is not defined/redefined, and the magical dragging in of Bayes Theorem is wrong.
Yes it was.
The only possible dispute is over the definition of the problem. I think it's pretty obvious what it means, but when in doubt I'd defer to the person who asked the question in the first place.
Who is the person you are currently arguing with. Pointless.
http://en.wikipedia.org/wiki/Event_(probability_theory) :
"In probability theory, an event is a set of outcomes (a subset of the sample space) to which a probability is assigned. Typically, when the sample space is finite, any subset of the sample space is an event (i.e. all elements of the power set of the sample space are defined as events)."
03 Apr 08
1 edit
Originally posted by geepamoogleThat's a redefinition of the problem you are now changing things around to suit the faulty and incorrect solution proffered my man.
Be it noted here that the man only makes the bet once the randomly chosen side is revealed.
Had the side facing up been gold, his bet would have been even odds the other side was gold.
Here is the simplest way to think of it, seriously.
There are three cards in the hat, one of which has differing sides. A card is picked at random from the hat a ...[text shortened]... ust so happens that in this particular case, the side which comes up is silver rather than gold.
03 Apr 08
Originally posted by eldragonflyThe question could be more precisely defined, but there's only one sensible interpretation. But, as I said, I'd be prepared to be corrected by the person who set the question in the first place.
You're super-duper exactly wrong, the problem is poorly described, the sample space is not defined/redefined, and the magical dragging in of Bayes Theorem is wrong.
And don't bother quoting definitions of conditional probability to me. I understand it fine, thanks. Have done for decades.
03 Apr 08
Originally posted by mtthwBFD.
The question could be more precisely defined, but there's only one sensible interpretation. But, as I said, I'd be prepared to be corrected by the person who set the question in the first place.
And don't bother quoting definitions of conditional probability to me. I understand it fine, thanks. Have done for decades.
The question is IMPROPERLY defined, yet there are only two outcomes to the event of showing a silver card: either
1) the other side is silver
or
2) the other side is gold.
No more no less.
03 Apr 08
1 edit
Originally posted by eldragonflyTwo events with different probabilities. Your statement tells us nothing.
The question is IMPROPERLY defined, yet there are only two outcomes to the event of showing a silver card: either
1) the other side is silver
or
2) the other side is gold.
No more no less.
Whereas the question says "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."
It could, ideally, state that he randomly chooses one side of the card to show you. That would be better. But it's reasonable to interpret it that way.
03 Apr 08
1 edit
Originally posted by mtthwAgain that does not _satisfy_ the definition of conditional probability.
Two events with different probabilities. Your statement tells us nothing.
Whereas the question says "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."
It could, ideally, state that he randomly chooses one side of the card to show you. That would be better. But it's reasonable to interpret it that way.
Let's look at this another way, since you're so confused here.
Event A = 3 cards in hat, 1 silver/silver, 1 silver/gold, 1 gold/gold
Event B = the showing of a silver card, which could be only one of two cards.
Event C = the other side of the silver sided card that has been displayed.
Event A does not belong, the selection of the gold/gold card is impossible, because Event C is not dependent upon Event A. The selection of the gold/gold card is excluded from the sample space. The sample space is Event B, where there are only two possible outcomes that are possible.
03 Apr 08
2 edits
Originally posted by eldragonflyI'm not remotely confused. Your three events aren't all actually events though, which suggests you might be.
Again that does not _satisfy_ the definition of conditional probability.
Let's look at this another way, since you're so confused here.
Event A = 3 cards in hat, 1 silver/silver, 1 silver/gold, 1 gold/gold
Event B = the showing of a silver card, which could be only one of two cards.
Event C = the other side of the silver sided card that has been displayed.
Doing it the Bayes' Theorem way:
Statement A: side 1 is silver
Statement B: side 2 is silver
P (A | B) = P (A & B) / P(B)
This is always true - there are no conditions to satisfy. Conditional probability always applies. It's a matter of whether you can calculate those.
P(A & B) = P(the all silver card is chosen). In my interpretation of the question this is 1/3, as the cards are chosen randomly. I don't see how the question can be read any other way.
So the only questionable area is P(B). What is the probability side 2 is silver if we know nothing? In my interpretation of the question this is 1/2, as side 2 is equally likely to be any of the six sides.
=> P(A | B) = 2/3.
If you want P(A | B) = 1/2, you need to manipulate the problem so that P(B) = 1/3.
03 Apr 08
1 edit
Originally posted by PBE6If my way of thinking is redefining the problem, let let me revisit the problem as stated.
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?
The issue here seems to be some vagueness in the details of how it works in the wording. But the problem follows a well-known pattern, and it makes sense to assume that the problem is identical.
However, let me look at what questions are not answered which may affect the problem at hand.
According to the problem, the selection of the actual card is random, meaning the dealer does not have pre-knowledge of which card will be selected, and any of the three cards have an equal chance of being selected, even the one that is gold-gold, even though later on it is stated the one which was selected had a silver side, because that is what you see.
The main vagueness which occurs in my mind is whether or not the dealer knows which card it is before making the offer.
My assumption here is that he does not, and the analysis I gave works off that assumption.
However, what if I am wrong on this count? Then the dealer has full knowledge of whether the bet he makes is good or bad, and this ceases to be a probability problem at all. And the answer most likely is that the bet is a bad one.
So it is reasonable to assume the dealer has not seen the other side of the selected either for purposes of analyzing the problem.
That also means that it could just as easily been a gold side showing. In fact, given the setup, Silver and Gold had equal chances of coming up.
Now, the fact that it is a silver side showing provides us with information we did not have with the setup alone.
Before the card was drawn and placed in front of us, we knew one of six possible events would happen. One of the three cards would be drawn, and one of the two sides on that card turned face up. The card and side selection is assumed to be fair in the absence of information to the contrary, which means any of the 6 possibilities are equally likely.
Now the drawing occurs, and we see a silver side. This eliminates the three gold sides from consideration. It could have just as easily been a gold side selected, and we would then be eliminating the silver sides.
The important thing is that any of the 3 silver sides are equally likely to have been the silver side picked.
We are then offer even money on the other side being silver (i.e. the same color).
To examine the chances of this, we can examine how many of the possible sides match their opposite side.
And the answer is that TWO of the silver sides are on the silver/silver card, while only ONE is on th silver/gold card, which means the dealer has a 2-in-3 chance of being correct.
If you were to repeat this process until silver turned up 30 times, and then tally the results, you would expect to see the silver /silver card 20 times, and the silver/gold only 10 times. The other 30 times, it would be a gold side showing up.
You would lose twice as often as you win.
Or think of it this way.
When the Silver/Silver card is picked, you'll ALWAYS see a silver side.
When the Silver/Gold card is picked, you'll only see silver HALF the time. (The other half of the time, you'll see the gold side).
Since either card is equally likely to be picked, then half the time the latter is chosen, it will come up on the gold side. This means of the cases in which a silver side is showing, the Silver/Gold card is half as likely to have been chosen as the Silver/Silver.
03 Apr 08
2 edits
Also a note on a fatal assumption as well.
There being only n number of possibilities does not by necessity mean that each possibility has an equal chance of occurring.
Suppose I toss 2 normal, fair 6-sided dice. There are 11 possible sums which could come up. However, none of the possible throws have a 1-in-11 chance of coming up. Some of the rolls, like 7, have a greater chance, while others, like 2 or 12, have a smaller chance.
Therefore, the statement "there are only two possibilities" is not proof the possibilities are equal. Rather, you have to show that they are equal given the assumptions. (Assuming the card/side selection is fair is reasonable and allows the problem to be analyzed. Should the selection be skewed, that would by necessity be an explicitly stated part of the problem.)
So now, the question I would ask is this.
You see a silver side.
It could either be the Silver card or the Mixed card.
What is your proof that both possibilities are equally likely, given that any of the silver sides are equally likely to have been picked?
Or are you suggesting that the silver side on the Mixed card is twice as likely to be picked as either of the two sides of the Silver card individually?
