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Dear Long Live Yorke


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you have a 50% chance of picking BIG if you switch you get .5BIGyou have 50% of picking small if you switch you get BIG.5*.5 BIG + .5*BIG= .75 BIGnot switching.5 BIG + .5 (.5BIG) = .75 BIGeither way your expected value is the middle of big and small .75 BIGswitching doesn't matterthe mistake in logic is thinking that your first choice tells you something about how much the BIG might be but that's already been decided

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for some of us, every month is STD awareness month.

When you have absolutely no ****ing clue what you're talking about, just keep your mouth shut. If it helps, think about losing $2k in equity in a free roll, the amount of absolute money we're talking about is irrelevant as our bank accounts are clearly not very similar. Thanks for chiming in with your words of wisdom.
Oh damn, you're BKiCe too? I'mpressed.
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Once I pick an envelope, I fix a number, so by switching I can get either double or half that number:So, I get either 2X or .5X. There's a mistake in my logic somewhere, but it's pretty subtle I think.
I can't tell if you'd seen my responses when you wrote this.... was my response wrong?Basically, your handling of X here is incorrect. X, at this point, is what? The value of the envelope you have chosen. So you can't say you are either going to double it or halve it because, depending on which one you've chosen, you only get one of those, and you've already chosen the envelope, so you can't just add those probabilities together. You are either going to (halve it or double it), or (keep it). If you keep it, your expected value is 3/4 of the max value, because the choice was made before. If you trade it, you have to use the formula I provided with an X and Y, with the same expected value. It's been too long since I've had a stats class to remember all the terminology that explains this mathematically.
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Once I pick an envelope, I fix a number, so by switching I can get either double or half that number:So, I get either 2X or .5X.There's a mistake in my logic somewhere, but it's pretty subtle I think.
If that is, in fact, you in your avatar, I think you look smarter in the old one and I am more likely to trust you with your old avatar because for some reason you now remind me of the lead singer of Everclear. Just throwing that out there.
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If that is, in fact, you in your avatar, I think you look smarter in the old one and I am more likely to trust you with your old avatar because for some reason you now remind me of the lead singer of Everclear. Just throwing that out there.
Is this a joke? Yeah, it must be.
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<br />If that is, in fact, you in your avatar, I think you look smarter in the old one and I am more likely to trust you with your old avatar because for some reason you now remind me of the lead singer of Everclear. Just throwing that out there.<br />
Yeah, LLY is famous:http://www.thom-yorke.info/biography.phpHe's a physicist AND a musician.
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Dammit. But seriously, his avatar leads me to believe him less just the same.
To be fair, I wasn't sure if his last avatar was him or not. I hoped it was, but suspected it wasn't.
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Dammit. But seriously, his avatar leads me to believe him less just the same.
To be fair, I wasn't sure if his last avatar was him or not. I hoped it was, but suspected it wasn't.
I'll admit it, I'm an idiot. I thought the old av was him too until a couple months ago when I read something about radio head on here and that the lead singers name was Thomas Yorke. Then it all made sense.
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I can't tell if you'd seen my responses when you wrote this.... was my response wrong?Basically, your handling of X here is incorrect. X, at this point, is what? The value of the envelope you have chosen. So you can't say you are either going to double it or halve it because, depending on which one you've chosen, you only get one of those, and you've already chosen the envelope, so you can't just add those probabilities together. You are either going to (halve it or double it), or (keep it). If you keep it, your expected value is 3/4 of the max value, because the choice was made before. If you trade it, you have to use the formula I provided with an X and Y, with the same expected value.
X is whatever value we see in the envelope we open. Half the time it's the larger value and by switching we get .5X, and half the time it's the smaller envelope and by switching we get 2X. So, X is a fixed number and doesn't depend on which envelope we chose first. If I keep it, my expected value is certainly X, because that's how much I'll get every time. If I change, I get either double X or half X, which has a higher expected value. So, why don't I switch? Clearly switching is the wrong answer.
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If that is, in fact, you in your avatar, I think you look smarter in the old one and I am more likely to trust you with your old avatar because for some reason you now remind me of the lead singer of Everclear. Just throwing that out there.
LOL at everything about this. I had the old avatar since day 1. I'm digging the new one.
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X is whatever value we see in the envelope we open. Half the time it's the larger value and by switching we get .5X, and half the time it's the smaller envelope and by switching we get 2X. So, X is a fixed number and doesn't depend on which envelope we chose first. If I keep it, my expected value is certainly X, because that's how much I'll get every time. If I change, I get either double X or half X, which has a higher expected value. So, why don't I switch? Clearly switching is the wrong answer.
see my answer switching and not switching have the same EV
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X is whatever value we see in the envelope we open. Half the time it's the larger value and by switching we get .5X, and half the time it's the smaller envelope and by switching we get 2X. So, X is a fixed number and doesn't depend on which envelope we chose first. If I keep it, my expected value is certainly X, because that's how much I'll get every time. If I change, I get either double X or half X, which has a higher expected value. So, why don't I switch? Clearly switching is the wrong answer
I think the problem is that, since you have an envelope in your hand, you can't say that it's either 2X or .5X, because those two X's are not the same -- they each refer to a different situation, so using the same X for both and then mixing them is where the error comes in. The situation you are describing is for this situation:I give you an envelope with 1 dollar in it. There are two other envelopes, one with 0.5 dollars and one with 2 dollars. You can pick either of those two or keep your own envelope. What should you do? Then, your formula that leads to always switching gets the correct answer, because on average you end up with 1.25.
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I think the problem is that, since you have an envelope in your hand, you can't say that it's either 2X or .5X, because those two X's are not the same -- they each refer to a different situation, so using the same X for both and then mixing them is where the error comes in.
But X is always what we find in the envelope, that's how it's defined. No matter whether we picked the smaller one or the larger one, it's always X, never 2X or .5X.
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Anyway, here's what I think the resolution of the problem is so we can all move on:When we look into the envelope, we see exactly how much money is in it. We have figured out what X is. I claim that it is therefore wrong to say that we have picked the smaller one and the larger one with a 50/50 chance. By knowing how much money is in the envelope, we have been given some information as to which one we more likely picked. The problem is that there's really no way to fill the envelopes with all pairs of a number and it's double with equal probability. This sort of distribution could never be normalized and would never add up to 1. We need to fill the envelopes with a real probability distribution.In other words, there must be a nearly 0% chance that we fill the envelope with $1,000,000,000,000 and $2,000,000,000,000, so if I open the envelope and find $1,000,000,000,000, it's more likely that the two envelopes were filled with 500,000,000,000 and 1,000,000,000,000 than $1,000,000,000,000 and $2,000,000,000,000.I think it's neat and quite subtle. So, problem is really ill posed. We have to apply some workable probability to how the filler of the envelopes decides to fill them, and when we do so, the problem becomes solvable and the paradox is eliminated.Imagine that he only fills it with 1, 2, 4, 8, 16, 32, 64, and 128. Then if you get 128, you know you can't switch because the two pairs are 64 and 128, so you can only lose.

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Anyway, here's what I think the resolution of the problem is so we can all move on:When we look into the envelope, we see exactly how much money is in it. We have figured out what X is. I claim that it is therefore wrong to say that we have picked the smaller one and the larger one with a 50/50 chance. By knowing how much money is in the envelope, we have been given some information as to which one we more likely picked. The problem is that there's really no way to fill the envelopes with all pairs of a number and it's double with equal probability. This sort of distribution could never be normalized and would never add up to 1. We need to fill the envelopes with a real probability distribution.In other words, there must be a nearly 0% chance that we fill the envelope with $1,000,000,000,000 and $2,000,000,000,000, so if I open the envelope and find $1,000,000,000,000, it's more likely that the two envelopes were filled with 500,000,000,000 and 1,000,000,000,000 than $1,000,000,000,000 and $2,000,000,000,000.I think it's neat and quite subtle. So, problem is really ill posed. We have to apply some workable probability to how the filler of the envelopes decides to fill them, and when we do so, the problem becomes solvable and the paradox is eliminated.Imagine that he only fills it with 1, 2, 4, 8, 16, 32, 64, and 128. Then if you get 128, you know you can't switch because the two pairs are 64 and 128, so you can only lose.
that's new info you never said there was a known max or a possible max
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that's new info you never said there was a known max or a possible max
Right. I'm saying the original question was ill posed and therefore impossible. To ask a question that makes sense, there needs to be some probability distribution that sums to 1, so in a sense there is a limit higher numbers, for instance.
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Right. I'm saying the original question was ill posed and therefore impossible. To ask a question that makes sense, there needs to be some probability distribution that sums to 1, so in a sense there is a limit higher numbers, for instance.
I still think you're smarter with your old avatar.
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Right. I'm saying the original question was ill posed and therefore impossible. To ask a question that makes sense, there needs to be some probability distribution that sums to 1, so in a sense there is a limit higher numbers, for instance.
but what if it was a check??? AAAAAHHHHHH.....
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Right. I'm saying the original question was ill posed and therefore impossible. To ask a question that makes sense, there needs to be some probability distribution that sums to 1, so in a sense there is a limit higher numbers, for instance.
nah it can be theoretical money and any number can be doubled
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nah it can be theoretical money and any number can be doubled
It can be theoretical money, but what determines how much money is put in the envelope? How can you come up with a system that determines what goes into the envelopes? At some point, a person or a machine or something must actually put the money in the envelope, or decide the value, or whatever. If you make a machine that does this, it at some point must have been given some sort of probability distribution that controls what money it puts in the envelope. If it's a human, then we internally have some sort of distribution. Either way, there's no way we can have a flat probability over the entire number line, since it would integrate to infinity.
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