08 Apr 08
1 edit
Originally posted by eldragonfly[/b]The probability is right, your reasoning is wrong. If you want to continue this, I'll bring out the crayons. Try not to lose too many up your nose.
Wrong! In fact you yourself gave this as the correct solution on page 1 of this thread.Originally posted by LemonJello
[b]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, ther ...[text shortened]... rding to Fabians nomenclature its
2b) he thinks, that he is smart and you are crazy
08 Apr 08
2 edits
eldragonfly - ready to tackle this one, genius?
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?
08 Apr 08
1 edit
Originally posted by pijunEvery day ("everyday" is one word) I do either one of these things (there are 4 things listed, "either" applies to 2 choices only) but NEVER one or more of them in a single day (this means you never do any of them in a single day...how long does it really take?).
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.
1. Brush my teeth
2. Shave
3. Change underwear (yours I hope)
4. Shower
(The fact that it takes you longer than one day to change underwear of some kind is disgusting.)
Today, all I did was shower (phew!)... (you must have started the day before or even earlier...congratulations) what are the odds I showered today?
err I just told you. The others become irrelevant.
Golly, sure is nice to see a properly worded question! About time. Thanks for coming out.
08 Apr 08
2 edits
Originally posted by PBE6You asked him this one earlier. He must be still working on the "did you pass any?" part.
eldragonfly - ready to tackle this one, genius?
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?
If n=3 and k=5, then p=32/33. Am I on the right road?
08 Apr 08
1 edit
Originally posted by luskinThat's what I get as well.
I think that's wrong. Should be p=16/17.
One nice check for the formula: if k = 0 you don't have any additional information, so the probability should be 1/n. And as k -> infinity, it's easy to see that the probability -> 1. If your general formula doesn't satisfy these conditions, it's wrong.
08 Apr 08
2 edits
Originally posted by mtthwYes, mine gives 1/n for k=0, and near enough to certainty for k>=20. That holds for n up to about 10000. It drops slightly below 99% for n=11000 ; k=20.
That's what I get as well.
One nice check for the formula: if k = 0 you don't have any additional information, so the probability should be 1/n. And as k -> infinity, it's easy to see that the probability -> 1. If your general formula doesn't satisfy these conditions, it's wrong.
08 Apr 08
Originally posted by mtthwThat's what I get too for those numbers, and mtthw that is an excellent piece of advice about checking the limits of your equation. I've saved myself on many a math test by checking the reasonableness of my answer before handing in my paper.
That's what I get as well.
One nice check for the formula: if k = 0 you don't have any additional information, so the probability should be 1/n. And as k -> infinity, it's easy to see that the probability -> 1. If your general formula doesn't satisfy these conditions, it's wrong.
08 Apr 08
Originally posted by pijunYOU'RE BOTH WRONG!!!
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
even if you had only 2 cards (silver/silver and silver/gold) the fact that you see the silver side changes the probabiliy from 50/50 increasing
the chances that the card is silver/silver
if you can't see either side than the odss are 50/50
take this into consideration : the odds that a man with 2 cards (silver/silver silver/gold) draws a gold side down are 25 percent and for a silver side down they are 75 percent
08 Apr 08
Originally posted by alexdinoThis is exactly the crux of the question, if you agree then why do you say this if you don't think it's true? you can't have it both ways. And the rest of your "explanation" clearly is wrong. There is no randomness, the gold/gold card is automatically excluded. I'm sorry if this simple observation is rocket science to you and some of the others here.
if you can't see either side than the odss are 50/50