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Jul-16-08
 | | rinus: Envelope problem: two strategies
A) always change: Exp = 0.5*X + 0.5*2X = 1.5*X
B) never change: Exp = 0.5*X + 0.5*2X = 1.5*X
Result: the decision to change has no effect. |
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| Jul-16-08 | | YouRang: <rinus> Interesting article -- thanks for that. :-) |
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| Jul-16-08 | | hms123: <rinus> I am impressed--I never heard of it and I actually took a logic course in college based on Lewis Carroll's work. Further, it shows the difficulty of getting all of the assumptions right in these kinds of problems. I was on a doctoral committee 30+ years ago in the Math department where one of the questions on the written qulaifying exam was a probability problem. I don't remember the (irrelvevant) details, but basically the problem started with " You have a key that has a 0.50 proability of opening a bathrrom door"--and then went on with a myriad of irrelevant detail about the number of bottles in the medicine cabinet, the color of the pills, etc. etc. Well, when the smoke cleared the answer was 1/2--the probability of the key opening the door in the first place. None of the math doctoral students got it right. A lesson for us all about these sorts of things.--thanks--hms |
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| Jul-16-08 | | YouRang: <rinus: Envelope problem: two strategies A) always change: Exp = 0.5*X + 0.5*2X = 1.5*X
B) never change: Exp = 0.5*X + 0.5*2X = 1.5*X
Result: the decision to change has no effect.> Yes, that (I believe) is the correct answer for the 2-envelope case as originally given (where X is defined to be the lesser value of the two envelopes). But what happens if we open an envelope and look at it's contents? Let's say it's $5. Clearly, we know that the EV for keeping our envelope is $5. But how do you compute the EV for switching envelopes? Is it valid to say that the other envelope has a 50% chance of being $2.50 and a 50% chance of being $10? In that case, can we compute the EV at .5($2.50) + .5($10) = $6.25, meaning that we should swap? Or, do we somehow stick to our claim that the EV is the same -- and hence the EV for swapping must also be $5? If so, what is the mathematical justification? This, to me, is where this paradox gets interesting. :-) |
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| Jul-16-08 | | ganstaman: <YouRang: Or, do we somehow stick to our claim that the EV is the same -- and hence the EV for swapping must also be $5? If so, what is the mathematical justification?> At this moment, I'm not fully sure of the mathematical justification, but I'm 100% sure that the EV is still the same. Here's what I said before: <Let's say that we play this game together. We pick an envelope together. You look inside and I don't. So would it make sense for it to be neutral EV for me to keep or switch, and at the same time +EV for you to switch?> If we played many times together, would you (always looking and then switching) expect to end up with more money than me (always not looking but switching anyway since there is no change in EV)? Especially since the value you see doesn't affect anything (you are suggesting switching regardless of what value you see), then you don't even need to look -- you will use the same basic equations and get the same results. But you already said that when you don't look, the EV of switching is the same as not. |
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| Jul-17-08 | | hms123: <YouRang>
<Is it valid to say that the other envelope has a 50% chance of being $2.50 and a 50% chance of being $10? In that case, can we compute the EV at .5($2.50) + .5($10) = $6.25, meaning that we should swap?> No. You are assuming three envelopes. There are only two envelopes at a time. You have to list the cases the way I did in my original post. You either have the smaller (or the larger), so the other envelope can only be the larger (or the smaller). (It can't be both unless Schroedinger's cat filled the envelopes.) |
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| Jul-17-08 | | Waitaka: <If I give you an envelop and you inspect the contents to be $40. Would you swap it for an envelop that had a 50/50 chance for $40,000> I would prefer to swap it for the 50/50 $40,000 envelope, no doubt. Even if you told me that the second envelope had a 50/50 chance for $1,000 |
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Jul-17-08
| | rinus: <YouRang> felt where the pain is in the whole problem setting. We have an expectation about X cq 1000X; we expect both to be in a FINITE domain. For easiness of arguing let's suppose 1000X <= 1.000.000 as constraint; now we can handle the problem. Suppose you find 100.000 dollar in one envelope, would you change? |
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| Jul-17-08 | | hms123: There's a silly game show on TV here in the US. I occasionally stop for a moment while channel surfing. Anyway, the premise of the game is that there are a bunch (20 or so) suitcases (each held by an attractive, scantily-clad, young woman). Each suitcase has an amount of money in it ranging from $1 to $1,000,000. The contestant picks a suitcase, and before opening the suitcase gets an offer of an amount of money--the contestant has the option of taking the money and leaving the show. Each time the money is turned down, the suitcase is opened and the amount inside is taken down from a list that is seen by all. Being the crazy person that I am, I always calculate the expectation for the sums remaining on the list. Not surprising, the amount offered is always LESS than the expectation. Under these circumstances, it seems always correct to switch. I guess that depends a little on whether it is the absolute amount of $$ that matters or the "economic utility" of the $$ to the contestant. |
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| Jul-17-08 | | Waitaka: <Suppose you find 100.000 dollar in one envelope, would you change?> No, I would not change. Even if in the second envelope was a 50/50 $1,000,000.00 chance. I would keep the $100,000.00. |
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| Jul-17-08 | | Ziggurat: <hms123> I know that show - one of my friends was on it in Sweden. He ended up taking home "only" the equivalent of 5.000 US dollars - he could've stopped at ~50.000 but he chose the "optimal" switching strategy ... |
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| Jul-17-08 | | YouRang: <ganstaman><hms123> I agree that there are flaws in the mathematical arguments. I even said earlier that it's a mistake to assign a nonzero probability to an event that can't happen (e.g. having a 50% chance that an envelope contains 10 when no such envelope exists). But it's interesting (to me anyway) that when you don't know the value in either envelope, it's easy to see that the EV for either envelope is the same, but when shown the contents of one envelope, it can seem tricky. :-) Suppose the problem is presented this way:
There are two cards, A and B lying on a table. Each has a number on the side facing down, and all we know is that one of the numbers is double that of the other. The cards have been shuffled so that there's a 50-50 chance that card A has the lower number. I flip over card A, and it says 10. Now, one might reason that card B must have either 5 or 20, right? It *seems* reasonable to assign probabilities to these two values, and intuition might tell us that it's 50% chance of being 5, and 50% chance of being 20. If one got that far, one would compute the EV of card B using (50%)(5) + (50%)(20) = 12.5. This is higher than 10, so we should (wanting to maximize the value) exchange card A for card B. On the other hand, we might argue that we *know* that the EV for card A is the same as the EV for card B, and so once we know that card A is 10, we assume that the EV for card B must also be 10. With this, one might try to compute the probabilities for the values on card B: If we let Z represent the probability that card B is the lower value, then:
EV for card B = 10 = (Z)(5) + (1-Z)(20)
And solving for Z, we get Z=2/3! Does this mean that 2/3 of the time, card B will be less than card A? Of course there's a problem with these mathematical constructions. The problem is that the idea of assigning probabilities to the possible values for card B is flawed, which means that computing an EV for card B is flawed. |
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| Jul-17-08 | | YouRang: <rinus> <...we expect both to be in a FINITE domain. For easiness of arguing let's suppose 1000X <= 1.000.000 as constraint; now we can handle the problem.>> It would be nice to have a finite domain, but can we really? We want to select a random number X from some finite domain, and then have 2X also be in that same finite domain, with the same probability distribution for each value. It can't be done.
Suppose our domain is integers from a low value of L to a high value of H. If we pick X=H (or any value above [(L+H)/2]), then 2X > H. Also, for any value of X we pick, the value 2X excludes all odd numbers. If the user opens an envelope and sees the number 23 (say), they know for sure that the other envelope must have 46. |
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| Jul-17-08 | | hms123: <YouRang>
<On the other hand, we might argue that we *know* that the EV for card A is the same as the EV for card B, and so once we know that card A is 10, we assume that the EV for card B must also be 10.>We might argue that--but I wouldn't. I will agree that the EV for the two cards is the same BEFORE you know the value of either. For the most part, the EV is the mean of the underlying distribution. So without knowing the underlying distribution we can't know the EV. SO, after we know that card A=10, the EV for card B is still the mean of the original distribution. Unless by some miracle that mean happens to be = 10, then what we really know is that the value of A (10) is no longer what we expected. If we knew the original distribution, then we could make judgements about whether A was an outlier, or above or below the mean, etc. |
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| Jul-17-08 | | hms123: <YouRang> <Ganstaman> Depending on what got included in an infinite domain, the odd/even issue might not be relevant. For example, suppose you included irrational numbers as possibilities (so the card says "Congratulations! You have won sqrt(2) dollars!" I don't think that 2*sqrt(2) is clearly an even number. Is it? perhaps by definition? (like 0! =1 ) Interesting. |
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| Jul-17-08 | | YouRang: <hms123: <YouRang>
<On the other hand, we might argue that we *know* that the EV for card A is the same as the EV for card B, and so once we know that card A is 10, we assume that the EV for card B must also be 10.>
We might argue that--but I wouldn't. >
Again, so that you don't misunderstand me:
I'm not proposing that this is a valid argument. Rather, I'm presenting this as a hypothetical line of thought that one *might* reasonably (or at least intuitively) think was valid. But as I stated at the end, the idea of computing an EV for card B is flawed -- so that this line of thought is, in fact, not valid. |
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| Jul-17-08 | | YouRang: <hms123: <YouRang> <Ganstaman>
Depending on what got included in an infinite domain, the odd/even issue might not be relevant. For example, suppose you included irrational numbers as possibilities (so the card says "Congratulations! You have won sqrt(2) dollars!" I don't think that 2*sqrt(2) is clearly an even number. Is it? perhaps by definition? (like 0! =1 ) Interesting.> Yes, but I think the odd/even issue problem goes away by simply using rational numbers (you don't need to include the irrational numbers). In fact, in the case where the two values are X and 2X, it is sufficient to limit the range to a subset of rational numbers -- that is, all numbers of the form: P / (2^Q), where P and Q are integers.
But in any case, we still have an infinite value space. This whole concept of "selecting a number at random" from an infinite value space is flawed (assuming every number has an equal chance of being selected). |
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| Jul-17-08 | | hms123: <YouRang>
<This whole concept of "selecting a number at random" from an infinite value space is flawed (assuming every number has an equal chance of being selected).> Good point--it sparked a thought, namely, that without an upper limit on the infinite space there can't be an EV. (I think the mean of a rectangular distribution is just the midpoint.) |
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Jul-17-08
 | | Sneaky: There's so much to say here I don't know where to begin, but let me start with a small point:
<amadeus:
<Sneaky: Either (1) the other envelope contains $20 and you picked the bigger one, or (2) the other envelope contains $80 and you picked the smaller one. Both (1) and (2) are equally likely possibilities with a 50% chance of each.> I don't think so. One is 100%, and the other is 0%, the choice of values was already set.> Let me ask you a hypothetical. Suppose I pull all four aces of a deck and shuffle them under the table, then grab one of them at random and place it face down in front of you. I ask you to guess whether it's a black ace (clubs or spades) or a red ace (hearts or diamonds). Would you not agree that there is a 50% chance that it's a red ace, and a 50% chance that it's a black ace? Or do you feel that there is a 100% chance that it "is what it is" and a zero percent chance of the other? Just like we cannot peer into the future to know what result some dice will show, we cannot peer through the cards to see what randomly selected value lies underneath. In statistics, lack of knowledge can lead to a probabilistic computation. True, if we had x-ray vision then we'd have more information and the probability would vanish, but we don't. So we treat face-down cards (and unopened envelopes!) just like we treat future dice-rolls. I don't see what distinction it is that you are really trying to make. |
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Jul-17-08
| | Sneaky: <BUT THIS RAISES A QUESTION: Is it really possible to pick any positive rational number at random?> Because the rational numbers can be placed in a one-to-one correspondence with the positive integers (aka the "counting numbers" 1, 2, 3...) then this problem can be simplified slightly be asking instead, "Is it really possible to pick any positive integer at random?" The answer is: YES it is possible, however it's impossible to do it in such a way that all positive integers have an equal probability of being chosen. A simple demonstration that it's possible to make an algorithm to select any counting number from the infinite set of numbers: Write the number 1 on a chalkboard. Throw two dice. If the dice come up snake-eyes then you keep the number on the chalkboard (#1) but if they come out anything else then you erase the number on the chalkboard and increment it (now it says #2). Then throw the dice again, with the same rules: snake-eyes means you keep that number, anything else means you increment and keep going. Under that system, any integer, no matter how large, could be selected. But the distribution is not equal. It's far more likely to end up with the number 3 then the number 1000. I believe that it's impossible to select "any number, so that all numbers are equally likely" for the simple reason that because there are infinitely many numbers, if they all had an equal probability of being selected, they would all have a zero chance of being selected. |
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Jul-17-08
| | Sneaky: Incidentally, when I wrote: <Because the rational numbers can be placed in a one-to-one correspondence with the positive integers> This is by no means an "obvious assertion" but the proof is both beautiful and well known. If anybody demands the proof I'm sure somebody can dig it up online somewhere. Conversely, the real numbers CANNOT be placed on a one-to-one correspondence with the integers and there is also a beautiful proof of that. We're getting a little off the beaten path here, but I felt that had to be mentioned in case somebody was scratching their heads over my claim. |
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| Jul-17-08 | | YouRang: <<BUT THIS RAISES A QUESTION: Is it really possible to pick any positive rational number at random?>
Because the rational numbers can be placed in a one-to-one correspondence with the positive integers (aka the "counting numbers" 1, 2, 3...) then this problem can be simplified slightly be asking instead, "Is it really possible to pick any positive integer at random?"> Yes. I mentioned 'positive rational' numbers in that post because that was the case in the situation I was discussing. I realize that the folly occurs with any infinite set where we expect each number to have equal probability of being selected. BTW, another interesting proof is that all the real numbers may be mapped 1-to-1 with just the real numbers between 0 and 1. It's actually not too hard to prove that one. In fact, the tangent function pretty much does the trick. :-) |
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| Jul-17-08 | | hms123: <Or do you feel that there is a 100% chance that it "is what it is" and a zero percent chance of the other?> There's an interesting (to me at least) issue in even defining what probability means. One school takes the "long run" view to justify saying that (e.g.) "the probability of heads is 1/2"--what they mean is something like "if we flip a fair coin a sufficiently large number of times the empirical result will differ from 1/2 by less than any amount we care to name." Another school takes the view that one coin flip is just that--one coin flip--so that saying the probability is 1/2 is based more on our understanding of the physical nature of coins--they have two sides etc. What does it mean to ask the probability of a particular team's winning the Super Bowl? One answer is "it is either 0 or 1--we just don't know yet but will find out when the games is over." There is no long run, so statements like "in the long run the Colts will beat the Giants 9 times out of 10" are meaningless. Of course, we could take the (subjective) betting odds to reflect the probability of one team's beating the other, but what does that have to do with reality? (I am glad to stop these posts any time they get tedious--just let me know--nicely--thanks--hms) |
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| Jul-17-08 | | YouRang: <Would you not agree that there is a 50% chance that it's a red ace, and a 50% chance that it's a black ace? Or do you feel that there is a 100% chance that it "is what it is" and a zero percent chance of the other?> Perhaps, rather than saying: <there is a 50% chance that it's a red ace>, it would be more precise to say: <there is a 50% chance that MY GUESS that it's a red ace is correct>. After all, the uncertainty really lies with our knowledge of which card was picked, but not with the card itself. This doesn't settle anything of course. :-)
However, the example using the four aces is different since you're selecting at random from a finite set. The case in the 2-envelope problem involves "a number" picked at random, which is a flawed concept and may well be why the probability calculations don't make sense for that problem. |
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Jul-17-08
| | rinus: <YouRang> Is <Cantor> a CG member? |
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