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Simple gambling problem

Simple gambling problem

Posers and Puzzles

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Originally posted by eldragonfly
Wrong. Then you couldn't perform the card trick as stated in the original, i.e. not the wikipedia page, word problem. You start of with three cards, then the gold/gold card is *magically* eliminated, leaving only two cards to choose from, the SS and the SG, by definition. Indeed this is the foundation of this paradox.
Yes, but to get to the solution, you must keep track of sides, not cards.

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Originally posted by Palynka
There are two possibilities:
- eldragonfly is trolling this thread.
- eldragonfly doesn't have the cognitive abilities to understand why he's wrong.

Two options. Does that mean there's a 50/50 chance? Certainly not.

QED.
One of the best analogies so far.

1 edit
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Originally posted by Palynka
There are two possibilities:
- eldragonfly is trolling this thread.
- eldragonfly doesn't have the cognitive abilities to understand why he's wrong.

Two options. Does that mean there's a 50/50 chance? Certainly not.

QED.
Others have raised the exact same objections and made the same points, you can choose to ignore these facts at your own peril. 😉

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Originally posted by kbaumen
One of the best analogies so far.
😉

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Originally posted by kbaumen
Yes, but to get to the solution, you must keep track of sides, not cards.
For the bayesian solution, that is not the only solution to the problem.

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Originally posted by eldragonfly
For the bayesian solution, that is not the only solution to the problem.
Wow, now you agree that 2/3 is a solution?

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Originally posted by kbaumen
Wow, now you agree that 2/3 is a solution?
kbaumen my bombastic and foolish friend, there is no solution to your poorly worded and errant three card problem.

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Originally posted by eldragonfly
kbaumen my bombastic and foolish friend, there is no solution to your poorly worded and errant three card problem.
The fact that you don't see it doesn't mean it isn't there.

Check the thread - http://www.physicsforums.com/showthread.php?t=229352

It also has some discussion about the problem. You'll have to admit that people there are more knowledgeable.

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kbaumen my bombastic and foolish friend, there is no solution to your poorly worded and errant three card problem.

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Originally posted by Palynka
Look! A three-headed monkey!
How appropriate! You fight like a cow!

4 edits
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Originally posted by PBE6
New problem!

Assume that no person has more than 200,000 hairs on their head.

(a) Prove that at least 2 people in New York City have the same number of hairs on their heads.

(b) Can you prove that this must be the case in Vatican City as well?

(c) Why do people with hair have more hair than people with hairs? 😉
(a) population of new york > 200,000 by inspection
(b) depends on the population of Vatican City, trivial
(c) deliberate nonsense.

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Originally posted by PBE6
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?
This reminds me a lot of the Monty Hall problem.

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Originally posted by eldragonfly
(a) population of new york > 200,000 by inspection
(b) depends on the population of Vatican City, trivial
(c) deliberate nonsense.
(a) You need one more step to prove the proposition.
(b) Look it up!!
(c) Correct!

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Originally posted by PBE6
(a) You need one more step to prove the proposition.
Very well.
n = number of hairs
n must be an integer lower than 200,000. There are therefore 200,000 possible different values of n. (Including baldness!) Assuming then 200,000 people have different amounts of hair then fine, but as there are more then one value of n is used at least more than once. Good enough?

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Originally posted by Bifrost
Very well.
n = number of hairs
n must be an integer lower than 200,000. There are therefore 200,000 possible different values of n. (Including baldness!) Assuming then 200,000 people have different amounts of hair then fine, but as there are more then one value of n is used at least more than once. Good enough?
Yep, that proves it.

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