03 Apr 08
Originally posted by eldragonflyyou know......this kind of games were made for people like you
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 ...[text shortened]... ce. The sample space is Event B, where there are only two possible outcomes that are possible.
P.S.: you should play this with someone and always bet that the face you can see is not the same as the one you can't
GOOD LUCK!
(you'll need it!)
03 Apr 08
Originally posted by mtthwAgain you must be very confused.
I'm not remotely confused. Your three events aren't all actually events though, which suggests you might be.
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 mat ...[text shortened]... 2/3.
If you want P(A | B) = 1/2, you need to manipulate the problem so that P(B) = 1/3.
The placement of 3 cards in a hat is an initial condition. (A)
The selection of a non gold/gold card is an event. (B)
The selection of either a double silver or silver/gold card is an event. (C)
The gold/gold card is *automatically* excluded from the sample space (B) when the selection of a card that has a silver side is chosen.
But thee problem should be re-worded or re-defined as such.
What is the probability that given (A) the selection of a silver/silver card from (C) is .5
04 Apr 08
Originally posted by PBE6i'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).
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.
04 Apr 08
Originally posted by alexdinoWell yes and no, the problem description as given is flawed. And this magically assuming that the silver/gold card is showing the silver side only, and the gold/gold card is somehow not chosen is kind of silly but i see your point.
see my answer and think it over
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!
04 Apr 08
Originally posted by geepamoogleYou simply turn the card over, it has to be either the silver/silver or the silver gold.
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?
04 Apr 08
1 edit
Originally posted by eldragonflyAgain, I'm not confused. Point out exactly which part of my explanation you have a problem with, otherwise we're just wasting our time.
Again you must be very confused.
Or if you won't accept maths, try reality. Try it out - I just have. It doesn't take long to knock up a spreadsheet or program to simulate the game.
Results: win percentage is consistently in the range 65-69%
04 Apr 08
It seems eldragonfly has finally come around, and none too soon.
A note on word problems: I am currently reading a book on language called "The Stuff of Thought" by Steven Pinker, a cognitive scientist. The premise of the book is that our language gives us clues as to the fundamental workings of the brain. Along the way, he gives several instances of ambiguity in language and notes that the brain can construe statements in a myriad of ways depending on the needs and state of the listener.
What does this have to do with word problems? Well, I thought that I had written a clear problem statement, but ambiguity seems to have reared its ugly head (a small one, but ugly nonetheless). However, instead of throwing up your hands and saying "the problem is NON-SENSICAL!!!" (even when it isn't), just ask for clarification. Or make some assumptions and press on. Either solution is preferable to doing your best impression of Alan Dershowitz arguing with a carnie.
04 Apr 08
1 edit
Originally posted by PBE6No 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.
It seems eldragonfly has finally come around, and none too soon.
A note on word problems: I am currently reading a book on language called "The Stuff of Thought" by Steven Pinker, a cognitive scientist. The premise of the book is that our language gives us clues as to the fundamental workings of the brain. Along the way, he gives several instances of ambig ution is preferable to doing your best impression of..
04 Apr 08
Originally posted by mtthwOh but you are. You simply had to reword the problem or admit that is wasn't quite as random as you pretended. 🙁
Again, I'm not confused. Point out exactly which part of my explanation you have a problem with, otherwise we're just wasting our time.
Or if you won't accept maths, try reality. Try it out - I just have. It doesn't take long to knock up a spreadsheet or program to simulate the game.
Results: win percentage is consistently in the range 65-69%