Originally posted by PBE6Arrogant sissyboy! You know nothing about me, your responses are very suspicious and erratic, i have shown complete interest in this topic. Please explain what you are talking about, you appear to be one of those "mathematicians" who can't do the math.
Congratulations, you can cut and paste with the best of them. I'm done trying to help. You're obviously more interested in your own pride than in learning something. Good luck, you're going to need it.
From http://en.wikipedia.org/wiki/Bayes'_theorem
In probability theory, Bayes' theorem relates the conditional and marginal probabilities of two random events. It is often used to compute posterior probabilities given observations. For example, a patient may be observed to have certain symptoms. Bayes' theorem can be used to compute the probability that a proposed diagnosis is correct, given that observation...i got the random event part right, get your head outta' yer hat sir.
Originally posted by eldragonflyi got the random event part right, get your head outta' yer hat sir.[/b]You haven't completed one problem correctly, genius. Get bent.
From http://en.wikipedia.org/wiki/Bayes'_theoremIn probability theory, Bayes' theorem relates the conditional and marginal probabilities of [b]two random events.It is often used to compute posterior probabilities given observations. For example, a patient may be observed to have certain symptoms. Bayes' theorem can be used to compute t ...[text shortened]... t observation...
Originally posted by PBE6??? bad guess. i have taken several high lever mathematic courses in college. But you're right i haven't worked through your examples, but then again... running a simulation on excel is equally suspect. get help you obviously need it.
You haven't completed one problem correctly, genius. Get bent.
Originally posted by eldragonflyDid you pass any?
??? bad guess. i have taken several high lever mathematic courses in college. But you're right i haven't worked through your examples, but then again... running a simulation on excel is equally suspect. get help you obviously need it.
The question of whether you understand conditional probability or not can easily be put to rest. Try this example:
You are given a bag of "n" quarters, one of which is double-headed. A friend randomly selects a quarter from the bag, and proceeds to flip it. On the first k tosses, the quarter comes up heads all k times. Given this information, and without anyone checking the quarter, what is the probability that the double-headed quarter was chosen?
Originally posted by PBE6i was an honors student, and i answered your first question correctly perhaps you missed it.
Did you pass any?
The question of whether you understand conditional probability or not can easily be put to rest. Try this example:
You are given a bag of "n" quarters, one of which is double-headed. A friend randomly selects a quarter from the bag, and proceeds to flip it. On the first k tosses, the quarter comes up heads all k times. Given this inf ...[text shortened]... nyone checking the quarter, what is the probability that the double-headed quarter was chosen?
Originally posted by eldragonflyYou bet I missed it. You began berating us for being stupid on page 2, and after much hemming and hawing, here's was you said on page 5:
i was an honors student, and i answered your first question correctly perhaps you missed it.
"i'll agree with that upon further reflection. Stated another way each card has a 1/3 chance of being randomly selected. But the gold/gold card is excluded, so that leaves p=(1/3) for the silver/silver and p=(1/3) for the silver/gold. But the silver/silver card can be selected twice, therefore for the silver/silver p=(2/3)."
Of course, the silver/silver card can never be selected twice because there is only one draw. Even if it could be, it doesn't follow that the probability of the silver/silver card being drawn once is 2/3.
If you think my question was ill posed, your answer sounds like you spat into a bowl of Alphabets.
Originally posted by PBE6And the point is? Your word problem though interesting was poorly worded, many here have said that. Fact is the way you stated the problem, it is a 50/50% bet, i have tried to explain this simple point to you numerous times.
You bet I missed it. You began berating us for being stupid on page 2, and after much hemming and hawing, here's was you said on page 5:
"i'll agree with that upon further reflection. Stated another way each card has a 1/3 chance of being randomly selected. But the gold/gold card is excluded, so that leaves p=(1/3) for the silver/silver and p=(1/3) for the think my question was ill posed, your answer sounds like you spat into a bowl of Alphabets.
Your magic dealer shows someone that he has two silver cards in his hat. He draws one and then says :
..Now he says "I'll bet you even money that the other side of this card is silver too...whaddaya say, partner?"Now as stated this is an even bet, one shot one play.
The reasoning here is rather direct.
He could reveal the other side of the card which is either silver or gold ---- by problem definition.
So he could turn card 1 over revealing, let's say another silver side.
Therefore this is the silver/silver card.
Or he could have carefully replaced card1 into the hat and then picked the other silver card, which of course is the silver/gold card, showing the other side to be gold. This is card2.
eldragonfly is right it's 50/50, if it's any other answer then you've worded your question wrong. The fact that you mention the gold/gold card is irrelevant.
This question would make as much sense as saying
"a magician places a gold/gold card, a silver/gold card and a silver/silver card and a flourescent pink/neon green card, and a severed thumb, and a piece of sticky tack, and his underwear in a hat....... then proceeds to take out a card where one side is silver..................."
as soon as we get that info the sticky tack and severed thumb become irrelevant as does the gold/gold card. In order to include them in any kind of probability calculations your question would have to be worded as such: "funny looking magician in a hat picks out an object out of the hat.. what are the odds it is "silver/silveR" or a severed thumb or a ........"
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makes as much sense as this following problem I'm making on the fly.
Every day I do either one of these things but NEVER one or more of them in a single day.
1. Brush my teeth
2. Shave
3. Change underwear
4. Shower
Today, all I did was shower (phew!)... what are the odds I showered today?
err I just told you. The others become irrelevant.
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or if we just go back to the original question.. Once one side is shown to be silver.. We know the card is either silver/silver or silver/gold.
once one side is silver the other side must either be silver or must be gold. Only 2 choices.. 50/50.. Had you said that there were only 2 CARDS (since 2 is all we're dealing with)!! being silver/silver and silver/gold... and if one card is chosen at random and 1 side is silver, what are the odds the other side is silver too? Once you tell us the fact that one side is silver, we know the other side has a 1 out of 2 chance of being silver and a 1 out of 2 chance of being gold. There are no mathematical formulas needed. This is a case where people start plugging in formulas before using common sense (which isn't so common, yes we know that joke).
Maybe a bad diagram will help
||-------------||
silver/silver silver/gold
we are told one side is silver.. 2 sides left. either silver or gold. You can even choose the other silver, same result.. the fact that one card is double silver one single silver is irrelevant to your formula since you told us one side is silver already!
----------||------||
silver/silver silver/gold
same deal, 2 sides/possibilities left.. 1/2 chance silver, or a 1/2 chance it's gold.
Not sure how to make this any more elementary
edit: My state of the art diagram won't display as wanted
Originally posted by PBE6Wrong! In fact you yourself gave this as the correct solution on page 1 of this thread.
{quoting eldragonfly:} "i'll agree with that upon further reflection. Stated another way each card has a 1/3 chance of being randomly selected. But the gold/gold card is excluded, so that leaves p=(1/3) for the silver/silver and p=(1/3) for the silver/gold. But the silver/silver card can be selected twice, therefore for the silver/silver p=(2/3)."
Of course ...[text shortened]... e, it doesn't follow that the probability of the silver/silver card being drawn once is 2/3.
Originally posted by LemonJello
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.
Originally posted by PBE6
Correct!
and on page 2:
Originally posted by PBE6
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.
afx gave the same answer 2:1
Originally posted by afx
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
Originally posted by pijunthank you sir.
eldragonfly is right it's 50/50, if it's any other answer then you've worded your question wrong. The fact that you mention the gold/gold card is irrelevant.
This question would make as much sense as saying
"a magician places a gold/gold card, a silver/gold card and a silver/silver card and a flourescent pink/neon green card, and a severed thumb, and a pi ...[text shortened]... s any more elementary
edit: My state of the art diagram won't display as wanted