{"items": [{"author": "Michael", "source_link": "https://www.facebook.com/jefftk/posts/157041844381587?comment_id=157047654381006", "anchor": "fb-157047654381006", "service": "fb", "text": "Seems to me that the expected value for the first envelope isn't actually \"X\" until it's opened and X is found in it. Before that, it's (50%)*X + (50%)*2*X, or 1.5*X, the same as the expected value of the other envelope -- thus no reason to switch. I was surprised not to see this answer in the Wikipedia article on this problem, so perhaps I have made a mistake -- but unlike the other answers, mine agrees with \"common sense\".", "timestamp": "1315192794"}, {"author": "Mark", "source_link": "https://www.facebook.com/jefftk/posts/157041844381587?comment_id=157065811045857", "anchor": "fb-157065811045857", "service": "fb", "text": "The problem with the 2Y vs. 0.5Y approach is that Y is not constant, so you can't just weight them evenly for determining expected value. Unlike the Monty Hall three door problem, nothing changes each time you choose an envelope, so the expected value is the same.", "timestamp": "1315196726"}, {"author": "Ben", "source_link": "https://www.facebook.com/jefftk/posts/157041844381587?comment_id=157094484376323", "anchor": "fb-157094484376323", "service": "fb", "text": "Ah, I hadn't heard the resolution on Wikipedia before. But I think I like it. It's that the statement \"the envelope has, with even odds, $2Y and $0.5Y\" must be false. Y is a random variable, which must fit some probability distribution. And no distribution exists such that the quoted statement is true.", "timestamp": "1315204691"}, {"author": "Jeff Kaufman", "source_link": "https://www.facebook.com/jefftk/posts/157041844381587?comment_id=157177457701359", "anchor": "fb-157177457701359", "service": "fb", "text": "@Michael: \"(50%)*X + (50%)*2*X, or 1.5*X\"
The average of 0.5 and 2.0 is 1.25, but this is minor.
\"Seems to me that the expected value for the first envelope isn't actually \"X\" until it's opened and X is found in it.\"
In the first framing, X was defined as the unknown amount of money in the first envelope. Then the other one has either half X or twice X, and an expected value of 1.25X. The first envelope has expected value X by definition: X is what we're calling whatever it happens to contain.", "timestamp": "1315225779"}, {"author": "Jeff Kaufman", "source_link": "https://www.facebook.com/jefftk/posts/157041844381587?comment_id=157179004367871", "anchor": "fb-157179004367871", "service": "fb", "text": "@Ben: the odds could be very close to even, no? Close enough that the expected value of switching is still positive, because we have something like (0.49%)2Y and (0.51%)0.5Y?", "timestamp": "1315226061"}, {"author": "Jeff Kaufman", "source_link": "https://www.facebook.com/jefftk/posts/157041844381587?comment_id=157180314367740", "anchor": "fb-157180314367740", "service": "fb", "text": "@Mark: say I'm sitting here, in this problem, with $Y in an envelope. I consider switching, and with even or close to even odds I could get $2Y or $0.5Y. How is $Y not constant? It's whatever happened to be in the envelope I was handed.", "timestamp": "1315226281"}, {"author": "Jeff Kaufman", "source_link": "https://www.facebook.com/jefftk/posts/157041844381587?comment_id=157180901034348", "anchor": "fb-157180901034348", "service": "fb", "text": "Does the problem change if the initial formulation becomes: someone puts $X in an envelope, and then flips a coin to decide between putting $2X and $0.5X in the other envelope?", "timestamp": "1315226403"}, {"author": "Josh", "source_link": "https://plus.google.com/118273920476267337216", "anchor": "gp-1315236195052", "service": "gp", "text": "I haven't read the whole wikipedia page about this, but my intuition is that when you say \"The other envelope has, with even odds, $2Y and $0.5Y\", this is incorrect, because Y is a different amount depending on whether you have the $1X envelope or the $2X envelope.", "timestamp": 1315236195}, {"author": "Ben", "source_link": "https://www.facebook.com/jefftk/posts/157041844381587?comment_id=157251857693919", "anchor": "fb-157251857693919", "service": "fb", "text": "@Jeff: The problem does change in your coin-flip scenario. Then, it becomes advantageous to switch; it's more like the Monty Hall problem, where after you make a choice, the \"host\" takes an action that reveals information.
And you're right: there's an example on Wikipedia where the expected value of switching is 11Y/10. But they resolve that paradox by saying that the expected value turns out to be infinite, so Y and 11Y/10 are in fact the same.
Really cool problem, though. I'd be curious to see what people actually do if you ran this as a behavioral econ experiment.", "timestamp": "1315237294"}, {"author": "Jim", "source_link": "https://plus.google.com/112932743371759723915", "anchor": "gp-1315247622257", "service": "gp", "text": "Or to put it another way, you oughtn't use the same variable: the other envelope has, with even odds, $2Y or $0.5Z.", "timestamp": 1315247622}, {"author": "Jeff Kaufman", "source_link": "https://plus.google.com/103013777355236494008", "anchor": "gp-1315249801671", "service": "gp", "text": "Why isn't it fair to label whatever amount is in the envelope you hold as $X, in which case the other one has either $2X or $0.5X?", "timestamp": 1315249801}, {"author": "Josh", "source_link": "https://plus.google.com/118273920476267337216", "anchor": "gp-1315250165125", "service": "gp", "text": "I'm not sure how best to articulate this. Here's one way to think about it: If the other envelope has $2X or $0.5X, that implies that in one scenario it has four times as much money in it as in the other scenario. And that definitely isn't possible, right?", "timestamp": 1315250165}, {"author": "Jeff Kaufman", "source_link": "https://plus.google.com/103013777355236494008", "anchor": "gp-1315250496867", "service": "gp", "text": "I'm not sure how to think about it. I know that the EV=1.25X scenario has to be wrong, so something has to be impossible, but I'm not understanding your objection.", "timestamp": 1315250496}, {"author": "Jeff Kaufman", "source_link": "https://www.facebook.com/jefftk/posts/157041844381587?comment_id=157362404349531", "anchor": "fb-157362404349531", "service": "fb", "text": "@Ben: I think I misstated my revised scenario. Someone puts $X in an envelope, and then flips a coin to decide between putting $2X and $0.5X in the other envelope. Then the mix the two envelopes up behind their back and give you one at random.
I agree that as stated before you should switch once.
How do you get infinite expected value?", "timestamp": "1315252265"}, {"author": "Ben", "source_link": "https://www.facebook.com/jefftk/posts/157041844381587?comment_id=157426261009812", "anchor": "fb-157426261009812", "service": "fb", "text": "Okay, that makes sense. In the revised scenario, no need to switch, because the expected value in each envelope is (2x + 0.5X)/2 = 1.25X, as before.
You can get infinite expected value by having, for example, a distribution where with probability 1/2^n, the amount of money in the envelope is 4^n. (Let n > 0.) The probabilities sum to 1, so this is a legitimate distribution. And if you look at the expected value:
There's probability 1/2 that you have $4 (so that's +$2 to the expected value.)
There's probability 1/4 that you have $16 (so that's +$4 to the expected value.)
There's probability 1/8 that you have $64 (so that's +$8 to the expected value.)
And so on. There's no upper bound on expected value - hence, expected value infinity.", "timestamp": "1315260972"}, {"author": "Jim", "source_link": "https://plus.google.com/112932743371759723915", "anchor": "gp-1315275584219", "service": "gp", "text": "Wait a second. Simplify this to two envelopes with $X and $2X, and the option to keep your envelope or switch to a third envelope containing $1.5X. Whatever envelope you have has some amount of money $Y in it. The other envelope has, with even odds, 1.5Y or .75Y. The expected value of switching, then, is the average of those two, or $1.125Y, minus the $Y in the envelope. So you should always switch in this case too!\n
\nThere's your problem: you must not be able to calculate the expected value by averaging the numbers you think you do.", "timestamp": 1315275584}, {"author": "Jeff Kaufman", "source_link": "https://www.facebook.com/jefftk/posts/157041844381587?comment_id=157685340983904", "anchor": "fb-157685340983904", "service": "fb", "text": "\"In the revised scenario, no need to switch, because the expected value in each envelope is (2x + 0.5X)/2 = 1.25X, as before.\"
Really? X is defined as the contents of the envelope you have in front of you. So while I see that the expected value of the other one comes to 1.25X, I don't see how the expected value of yours can be anything other than X. If yours had an expected value of 1.25X as well, that would imply that while you know it contains X you should expect to gain 1.25X upon opening it.
As for the infinite expected value, thanks for the example. Playing with a new problem: let's say we used that function with 4^n for one envelope and another one with 4^n+C (for some very large C) for the other envelope. Both would have infinite expected value. But someone who chose the second envelope would on average do better by C.", "timestamp": "1315316484"}, {"author": "Ben", "source_link": "https://www.facebook.com/jefftk/posts/157041844381587?comment_id=157972310955207", "anchor": "fb-157972310955207", "service": "fb", "text": "\"X is defined as the contents of the envelope you have in front of you.\"
But you already defined X as the amount the person puts in an envelope, before flipping a coin to determine whether 2X or 0.5X goes in the other. So let's use Y to refer to the amount in the envelope in front of you.
In that case, X is fixed, and Y is a random variable that takes on X with p=0.5, 2X with p=0.25, and 0.5X with p=0.25. And the other envelope has exactly the same distribution. So again, it makes no difference whether you switch.
That's a good point about how one variable with infinite expected value may not be as good as another with infinite expected value. It seems that expected value isn't a very useful measure in such cases.", "timestamp": "1315359554"}, {"author": "Jeff Kaufman", "source_link": "https://www.facebook.com/jefftk/posts/157041844381587?comment_id=157975904288181", "anchor": "fb-157975904288181", "service": "fb", "text": "@Ben: \"X is fixed, and Y is a random variable that takes on X with p=0.5, 2X with p=0.25, and 0.5X with p=0.25\".
So the envelopes have X and 2X, and Y is what's in the envelope in front of you? Then isn't Y just X with p=.5 and 2X with p=.5?", "timestamp": "1315360197"}, {"author": "Ben", "source_link": "https://www.facebook.com/jefftk/posts/157041844381587?comment_id=157981000954338", "anchor": "fb-157981000954338", "service": "fb", "text": "I was going off of this statement of the scenario:
\"Someone puts $X in an envelope, and then flips a coin to decide between putting $2X and $0.5X in the other envelope. Then the mix the two envelopes up behind their back and give you one at random.\"", "timestamp": "1315361070"}, {"author": "Jeff Kaufman", "source_link": "https://www.facebook.com/jefftk/posts/157041844381587?comment_id=157981650954273", "anchor": "fb-157981650954273", "service": "fb", "text": "OK; I was forgetting which one we were working with. You're right.", "timestamp": "1315361164"}, {"author": "Eric", "source_link": "https://plus.google.com/105086402673996622245", "anchor": "gp-1315367357279", "service": "gp", "text": "Jeff, I think I finally (FINALLY) see the problem with the logic in the proposed \"paradox\" viewpoint. The proper treatment of the problem is that the two envelopes are random variables, Y and Z, each with a 50% probability of containing either 2X dollars or X dollars.\n
\n
\nNow, let S be the value of switching: S = Z - Y. Consider its expected value: E[S] = E[Z - Y]. In your second analysis, you stop here, and note that 50% of the time, Z - Y = X, and 50% of the time, Z - Y = -X. Thus, E[Z - Y] = 0. This is completely correct.\n
\n
\nIn the first (and traditional, I will note) analysis, you go on to subtract the expected value of Y (conditioned on Y) from the expected value of Z (conditioned on Y). But Y and Z are not independent random variables, so the result is not the expected value of their difference! I believe there are some more sophisticated arguments to try to salvage this viewpoint, but I think they all share the same flaw: subtly treating two dependent random variables as if they were independent.", "timestamp": 1315367357}, {"author": "George", "source_link": "https://www.facebook.com/jefftk/posts/157041844381587?comment_id=158439854241786", "anchor": "fb-158439854241786", "service": "fb", "text": "I am completely satisfied with the explanation of why there is nothing funny going on here presented in Grinstead and Snell (free PDF at http://www.dartmouth.edu/.../probability_book/book.html) in example 4.28 on page 178.
That said, here is a much more interesting candidate paradox that ALSO involves two envelopes. Suppose we have two indistinguishable envelopes and we fill them with two, distinct, positive integer numbers of dollars, X and Y. The particular specific values of X and Y are unknown to you, but you are given one envelope uniformly at random and you open it, letting you observe the money inside. Then you are asked if you wish to switch. Does it trouble you that there exists a strategy for when to switch that guarantees a probability strictly greater than 50% of getting the larger amount of money?", "timestamp": "1315445068"}, {"author": "Jeff Kaufman", "source_link": "https://www.facebook.com/jefftk/posts/157041844381587?comment_id=158454504240321", "anchor": "fb-158454504240321", "service": "fb", "text": "@George: If I'm reading Grinstead and Snell correctly, they conclude that there is at least one case where you should switch (and presumably keep switching forever). At the end of example 4.29 at the bottom of page 180 they show a probability distribution where the expected value of switching is greater than zero. Am I misreading them?", "timestamp": "1315448311"}, {"author": "George", "source_link": "https://www.facebook.com/jefftk/posts/157041844381587?comment_id=158460487573056", "anchor": "fb-158460487573056", "service": "fb", "text": "Yes you are reading the wrong example. 4.29 answers some subtly different but related questions.", "timestamp": "1315449475"}, {"author": "George", "source_link": "https://www.facebook.com/jefftk/posts/157041844381587?comment_id=158462607572844", "anchor": "fb-158462607572844", "service": "fb", "text": "The text in G&S for 4.29 is confusing, see the paper they cite: http://econ.as.nyu.edu/docs/IO/9393/RR90-26.pdf
4.28 is the example that addresses the situation I think you were talking about (unless *I* was confused!)", "timestamp": "1315449905"}, {"author": "Jeff Kaufman", "source_link": "https://www.facebook.com/jefftk/posts/157041844381587?comment_id=158576647561440", "anchor": "fb-158576647561440", "service": "fb", "text": "@George: I know you were trying to reference 4.28, but 4.29 seemed to be an equivalent setup. Rereading the leadup text to 4.29, now see the difference: you've opened your envelope and know what's in it. Further, you know the probability distribution that determined the envelopes. But I think they still come to the \"always switch\" conclusion.
In the referenced paper this is \"exchange condition always satisfied\", example 3, from page 5. If you should always switch, no matter what you see in the envelope, then it does't matter what's in the envelope. Switching without considering at all what's in your envelope can't be right.", "timestamp": "1315483553"}, {"author": "George", "source_link": "https://www.facebook.com/jefftk/posts/157041844381587?comment_id=158631500889288", "anchor": "fb-158631500889288", "service": "fb", "text": "However, you can't switch forever since you opened one and observed the money inside. Also, the distribution of money inside is a bit odd to make you always want to switch.", "timestamp": "1315493243"}, {"author": "George", "source_link": "https://www.facebook.com/jefftk/posts/157041844381587?comment_id=158632060889232", "anchor": "fb-158632060889232", "service": "fb", "text": "I encourage you to think about the other paradox I mentioned with the strategy that guarantees a >50% chance of getting the larger amount of money (in the setup I described) regardless of the distribution of the quantities in the envelopes.", "timestamp": "1315493338"}, {"author": "Jeff Kaufman", "source_link": "https://www.facebook.com/jefftk/posts/157041844381587?comment_id=158664980885940", "anchor": "fb-158664980885940", "service": "fb", "text": "@George: \"you can't switch forever since you opened one and observed the money inside. Also, the distribution of money inside is a bit odd to make you always want to switch.\"
Because the \"always switch\" is not dependent on 'x', the amount of money you saw inside, you don't have to open it before deciding to switch. Then you switch forever.
\"I encourage you to think about the other paradox I mentioned with the strategy that guarantees a >50% chance of getting the larger amount of money (in the setup I described) regardless of the distribution of the quantities in the envelopes.\"
There are two envelopes, with arbitrary integer quantities X and Y, where X<Y. You open one of them and observe z. You know z is either X or Y, but don't know whether it is the larger. Should you switch? That's the question, no? I think I remember talking about this with you in sharples a while ago. All you need is some source of random positive integers that assigns non-zero probability for every integer. Then you ask it for a number, let's say it gives you w. Now you compare z to w, and if z is greater than w you keep the envelope you already have, otherwise you switch.
In the cases where w<X or w>Y, then this has no effect on your expected value: whether z=X or z=Y has no effect on your switching. If w<X you will always switch, and if w>Y you will never switch. So we can ignore these.
If X>w>Y, then if z=X you will switch and if z=Y you won't switch. Which is the right choice. Because there is some chance that w is in this range, our odds of switching correctly are greater than 50%.", "timestamp": "1315498926"}]}