Originally posted by PBE6Given the question in the way it was asked, the yes it is a fair bet:
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?
It is only a one case scenario. If there were additional questions after that, ie. If they continue playing and the question is always "even money that the other side is silver." regardless of what side is showing OR the bet changes each time after to say that it's even money for whatever side is showing will show up on the reverse, the odds will be different.
But, in the question below, it is a one time event and the bet is merely, Do I have the S-S bar or the S-G bar?
I get the point that the fact that he is showing a silver side increases the chances of it being S-S, but there was just as much chance of showing a gold side than it was a silver side if it was the S-G bar and then the question would be "even money that the other side is gold"
saying that repeating it would yield a 2/3 ratio would be like arguing that the trick was already done 2 times prior and each time the other side was silver, so this time it MUST be the S-G bar. It's information that is not referenced or alluded to in the original question.
It's a one time event. And the rephrase of the question is, "do I have the S-S bar or the S-G bar?"
Originally posted by broblutoWhat's probability then? Doesn't 50% probability mean that in 50 from 100 cases the other side will silver? The fact is that it's a bigger chance that the other side will be silver because from the three possible silver sides you see one of them. And to only one of these silver sides, the other side is gold. Read the boy/girl question in previous pages, it's similar. Or check here - http://en.wikipedia.org/wiki/Three_cards_problem.
Given the question in the way it was asked, the yes it is a fair bet:
It is only a one case scenario. If there were additional questions after that, ie. If they continue playing and the question is always "even money that the other side is silver." regardless of what side is showing OR the bet changes each time after to say that it's even money for whate ...[text shortened]... ime event. And the rephrase of the question is, "do I have the S-S bar or the S-G bar?"
Originally posted by kbaumenYes. I read it, and I get it. But the fact still remains, that for a one time shot, it's either S-S or S-G. Probability goes out the door in a one time event between A or B because in reality, I am either going to lose or I am going to win. The trick would have to be done at least three times to REALIZE the odds. If the odds are 2 in 3, then I would I have to play 3 times in order to win once. That's a bad bet. However, if I play once, I'm not going to be a 1/3 winner. I'm either a winner or a loser. There's no "cosmic" design that says because I'd lose 2 out of 3 times, I would lose the first time or even the second. The theory says that the closer you get to infinity for the number of times played, the odds of winning converge to 33%. However, the first time played in that series will either be 0% or 100% winning.
What's probability then? Doesn't 50% probability mean that in 50 from 100 cases the other side will silver? The fact is that it's a bigger [b]chance that the other side will be silver because from the three possible silver sides you see one of them. And to only one of these silver sides, the other side is gold. Read the boy/girl question in previous pages, it's similar. Or check here - http://en.wikipedia.org/wiki/Three_cards_problem.[/b]
My contention isn't the math or the theory. My contention is that theory goes out the window in a real life one time event.
And technically, it is only a fair bet if both parties agree it is fair. Kind of like fair market value for homes or stock. It's only worth what the selling party and the purchasing party agree to what it's worth, regardless of what any third party says.
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Originally posted by broblutoThe fact that there are two possible outcomes, doesn't make them equally probable.
Yes. I read it, and I get it. But the fact still remains, that for a one time shot, it's either S-S or S-G. Probability goes out the door in a one time event between A or B because in reality, I am either going to lose or I am going to win. The trick would have to be done at least three times to REALIZE the odds. If the odds are 2 in 3, then I would I have e purchasing party agree to what it's worth, regardless of what any third party says.
EDIT: For example a 6-sided dice. What's the odds that you'll roll 6? 1 to 6, no? Two outcomes - you get six, or you don't get six but you can't look at it like this: I'll either get six or won't get six, so it's 50-50.
Originally posted by kbaumenOr a lottery ticket. It's either winning or it's not but more likely the latter.
The fact that there are two possible outcomes, doesn't make them equally probable.
EDIT: For example a 6-sided dice. What's the odds that you'll roll 6? 1 to 6, no? Two outcomes - you get six, or you don't get six but you can't look at it like this: I'll either get six or won't get six, so it's 50-50.
Originally posted by broblutoNo, it doesn't. Say you're given a jar of jelly beans, 350 of which are red and 1 of which is white. You are to pick 1 jelly bean at random (you will be blindfolded). Before you pick the jelly bean, you're given the chance to make a one-time even money bet on which colour you'll select. Which would you bet on?
My contention isn't the math or the theory. My contention is that theory goes out the window in a real life one time event.
What would you do in this case? Disregard the theory that predicts red is 350 times more likely to be chosen than white? As you say, in a one-time draw it's either a 100% win or a 100% loss, so what would you do?
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Originally posted by PBE6That analogy is wrong.
No, it doesn't. Say you're given a jar of jelly beans, 350 of which are red and 1 of which is white. You are to pick 1 jelly bean at random (you will be blindfolded). Before you pick the jelly bean, you're given the chance to make a one-time even money bet on which colour you'll select. Which would you bet on?
What would you do in this case? Disregard t ...[text shortened]... As you say, in a one-time draw it's either a 100% win or a 100% loss, so what would you do?
There are only 1 of 2 possible outcomes in the question asked. Not 1 out of 350.
I have a 50/50 chance of it being SS or SG, not a .3/99.7.
Originally posted by broblutoThe analogy is fair, though. You are using the fact that there are only two outcomes as an argument that they have equal probability, when they don't. All the analogies mentioned show clearly that you have have only two outcomes without it being 50-50. The only difference is a matter of degree, which means that in the analogies your intuition agrees with the theory.
That analogy is wrong.
There are only 1 of 2 possible outcomes in the question asked. Not 1 out of 350.
I have a 50/50 chance of it being SS or SG, not a .3/99.7.
Once you accept that SS & SG are not necessarily of equal probability you can move on and calculate the correct value.
Originally posted by broblutoNo, it's not a false analogy. It's simply a more extreme example. You're betting on red or white, and not a particular red bean versus the white bean, so there are only 2 possible outcomes (red or white). However, the red outcome is 350 times more likely than the white outcome.
That analogy is wrong.
There are only 1 of 2 possible outcomes in the question asked. Not 1 out of 350.
I have a 50/50 chance of it being SS or SG, not a .3/99.7.
In the question I posed at the beginning of this thread, you're betting on silver or gold on the reverse, and not a particular silver or gold side, so there are only 2 possible outcomes (silver or gold). However, the silver outcome is twice as likely as the gold outcome.
Originally posted by PBE6No, it's wrong, because I know that it's either A or B, not A or B or C or D etc...
No, it's not a false analogy. It's simply a more extreme example. You're betting on red or white, and not a particular red bean versus the white bean, so there are only 2 possible outcomes (red or white). However, the red outcome is 350 times more likely than the white outcome.
In the question I posed at the beginning of this thread, you're betting on sil ...[text shortened]... comes (silver or gold). However, the silver outcome is twice as likely as the gold outcome.
There are only 2 possible cards it could be. The jelly bean anology is there are 350 beans it could be. Not analogous.
If you agree that I have a 1 in 350 chance of picking the white bean, then you MUST agree, that I have a 1 in 2 chance of the card being the SG card.
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Originally posted by broblutoWhy must I? It's wrong. 😕
No, it's wrong, because I know that it's either A or B, not A or B or C or D etc...
There are only 2 possible cards it could be. The jelly bean anology is there are 350 beans it could be. Not analogous.
If you agree that I have a 1 in 350 chance of picking the white bean, then you MUST agree, that I have a 1 in 2 chance of the card being the SG card.
Originally posted by broblutoDid you read the explanation here - http://en.wikipedia.org/wiki/three_cards_problem. It should make everything clear. 50% would be correct if all four sides were silver. Then there is an even number of cases.
No, it's wrong, because I know that it's either A or B, not A or B or C or D etc...
There are only 2 possible cards it could be. The jelly bean anology is there are 350 beans it could be. Not analogous.
If you agree that I have a 1 in 350 chance of picking the white bean, then you MUST agree, that I have a 1 in 2 chance of the card being the SG card.
Now here the cards are different and one of the cards has silver side on both sides but the other only on one side. If we randomly choose a silver side, it 66% that it's the completely silver card, because it has both sides silver.
Here the analogy with beans goes quite well. Consider 2 red beans as the silver sides from completely silver card and 1 white bean as the silver side from the mixed card. Now if we choose a bean randomly, the probability to choose a red one is twice as big as the probability to choose the white one, so there is 66% probability that we will get the red one, hence 66% that we see a silver side from the completely silver side.
Still wrong?
Originally posted by PBE6Irrelevant personal attack PBE6. Both questions are the same, you can't really be that stoopid, but obviously you are.
mtthw, geepamoogle, LemonJello, Aethereal, deriver69...I love all you guys. Even if eventually, one of you think I'm stupid, or I think you're stupid, I'll respect your opinions and be glad that I've known you.
eldragonfly, you're the biggest laughing stock since ACME Novelties went public. To be honest, I'm surprised you're kicking up such a fuss, because ...[text shortened]... days you're going to have to tell me how you grew so thick on a diet gruel and bologna.