Originally posted by PBE6I get the probability to be 1/[1+(n-1)/(2^k)]. It seems to yield the proper results for the cases k = 0 and k --> infinity.
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?
I get the feeling this thread has degenerated into a poo-fling competition.
As far as the original three problem goes, I agree that the answer is you only have a 1-in-3 chance of winning, because only 1 of the three equally likely silver sides have a gold side opposite it.
This thing has been explained at least a half-dozen ways it would seem.
However, some do not accept this answer insisting that because it's one of two cards, the answer HAS to be 1-in-2 and a fair bet.
So does someone wish to propose a NEW problem, because debating the old one seems to be pointless at this point in time. (I do urge you to try this experiment out some time though.
Try to make an equivalent setup and experiment to see what happens in real life. Find a way to place three tokens in a bag, one each of two colors and one marked with a combination of the colors. Randomly draw a token and look at one side.
If it matches your chosen color, note what the other side is.
If it doesn't match, place it back into the bag and ignore the result.
Half the time when you pick the token with two colors, you'll see the wrong side first and will have to place in into the bag without noting the result.
Alternatively, label 6 tokens with the following.
* Gold, Gold/Gold A
* Gold, Gold/Gold B
* Gold, Gold/Silv
* Silver, Gold/Silv
* Silver, Silv/Silv A
* Silver, Silv/Silv B
Draw one randomly and note the result. Do this until you you have silver show up a certain number of times and note the ratio of Silver/Silver to Gold/Silver.
Originally posted by pijunI'm getting so sick of pansies whining and demanding threads be closed.
can we end this thread pls?
too many childish remarks, and the point is the original poster while having the right idea wrote it out all wrong.
Here's a better suggestion: Once you realize a thread, according to your standards, has 'degenerated' into childish remarks, or whatever else it is you're afraid of...
...STOP CLICKING ON IT! DON'T READ IT!!
Originally posted by LemonJelloYep that's it.
I get the probability to be 1/[1+(n-1)/(2^k)]. It seems to yield the proper results for the cases k = 0 and k --> infinity.
Nice to see some calculations rather than people going out of their way to fail to get the point.