171–180 of 318 posts
https://www.youtube.com/watch?v=_NGPncypY68 this video explains the paradox properly, with a very reasonable explanation.
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Exactly! The logic of "the average of +100% and -50% is positive" only applies if the denominator is the same. In this case it isn't. The relative percentage "rule of thumb" doesn't even feel like a primitive in math. It seems wrong to reason from that.
why isn't this listed in wikipedia as an answer.
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Try thinking of it this way: The information is in which door the host does not open. 2/3rds of the time, your initial choice will be a goat. In those cases the host deliberately avoids opening the remaining door with a car - thereby telling you where it is. If you chose the car door initially, then indeed you get no new information. But that only happens 1/3rd of the time. I am curious if that changes your opinion.
Read their second paragraph. That is exactly how they think of it, and they do think it makes it obvious you should switch.
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that's an excellent treatment. I'm curious: this video picks a particular example of a distribution of the amounts and shows that the expected profit from switching is not defined. Can one somehow prove that for all possible distributions, either the expected profit from switching is 0 or it is undefined?
the point of the video is that given a distribution, the total expected profit cannot be defined, because the infinite sum of the probabilities of each case don't have an order, and also adds up to positive infinity and negative infinity. it doesn't really matter what the distribution of the amounts are, as long as there's an infinite number of possibilities in the distribution (ie., it's not a finite amount of possi…
For example if instead of halving the probability of each value as he does in the video we take 1/20 then we get the expected value as the sum of 9/20^2 - 9/20^2 + 90/20^3 - 9/20^3 + … Here the series is absolutely convergent since the positive terms sum to 9/10 and the negative terms sum to 9/10.
A simple non-mathematical resolution is that no information has been gained by selecting the first envelope. Since no information about the contents was available before selecting the first envelope switching envelopes is a neutral act. While this sounds like just a lazy intuitive explanation, it calls out the key component - a lack of information about the outcome - that can help avoid the paradox in more complex si…
There's an amusing amount of probability puzzles with seemingly paradoxical solutions out there, to the point that it's almost more surprising to see one where the intuitive answer is correct: - https://en.wikipedia.org/wiki/Boy_or_Girl_paradox - https://en.wikipedia.org/wiki/Penney%27s_game - Waiting time paradox - Honorary mention: https://en.wikipedia.org/wiki/St._Petersburg_paradox (feel free to add more)
A simple non-mathematical resolution is that no information has been gained by selecting the first envelope. Since no information about the contents was available before selecting the first envelope switching envelopes is a neutral act. While this sounds like just a lazy intuitive explanation, it calls out the key component - a lack of information about the outcome - that can help avoid the paradox in more complex si…
The actual solution is explained right there on the page, the expectation for both envelopes (one has x, one has 2x) is x*3/2, so switching changes nothing.
The "solution" demanded is supposed to be pointing out the specific logical error in the erroneous calculation that says switching should win. Citing the correct logic is not considered sufficient.
A simple non-mathematical resolution is that no information has been gained by selecting the first envelope. Since no information about the contents was available before selecting the first envelope switching envelopes is a neutral act. While this sounds like just a lazy intuitive explanation, it calls out the key component - a lack of information about the outcome - that can help avoid the paradox in more complex si…
Yeah I was looking for someone saying this to double down. I think the "paradox" comes down to information loss. The tricky bit is this framing: "when you're holding an envelope, the other envelope contains >= $MONEY, so there's no reason not to switch". But this omits (loses) the information of " both envelopes contain >= $MONEY". That is, as soon as you switch to the other envelope, the situation is still true, the…
The way that they fool you is by saying that you either double your money or you halve it, leading you to think that your expected ratio is 2.5/2. But that isn't correct. That ratio doesn't mean anything. That's the wrong step.
There is a related puzzle which I've also heard called the Two Envelopes Problem. The premise is that you are handed two envelopes. Inside each envelope is a piece of paper with a number written on it. I have written down these numbers by sampling from some probability distribution, but you don't know what it is. All you know is that the two numbers are different. You get to pick one envelope and then look at the pap…
1. If the number you pick is positive, then there are technically (although arbitrarily small) more negative numbers that can be chosen for the other envelope. So you have a slightly larger than 50% chance of winning. 2. The inverse is true if you chose negative, however.
That said - I think this would generally imply you should always switch, which is why I'm not convinced I'm right. Consider a purely positive set, for example. If my number shows up as "12" - then I know there are infinitely more numbers larger than it, but only 11 numbers lower than it. But this starts to feel like the original two envelopes paradox.
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tldr; I'm not sure of the answer, although I think if I saw the answer I would have an "Aha!" moment.
You cannot calculate the expected value (of the envelope you get, when you pick one randomly and then swap) like that, as A is not a constant and actually depends on if you took the less or more valuable envelope initially.
If you use A1=X in case you took the less valuable envelope initially and A2=2X in the other case, the expected value is (1/2)*(2*A1) + (1/2)*(A2/2) = 1.5X
and if you don't swap, you get: (1/2)*A1 + (1/2)*A2 = 1.5X
... which makes sense.
If you change the experiment so that the value of the second envelope does not depend on the first one (if you first chose and only then the value of the other envelope is randomly either double or half), A is constant and you actually should swap.
The Two Envelopes Problem: the most boring paradox in the world - https://news.ycombinator.com/item?id=6388645 - Sept 2013 (12 comments)
Two envelopes problem - https://news.ycombinator.com/item?id=6387044 - Sept 2013 (88 comments)
Two Envelopes Problem - https://news.ycombinator.com/item?id=1582219 - Aug 2010 (88 comments)