Posers and Puzzles

Posers and Puzzles

  1. Joined
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    21 Mar '09 23:22
    THE DIFFERENCE
    The difference in picking the second envelop to begin with, and switching after we open the first is this.

    With the latter we have information we did not have before, specifically a dollar amount which limits our possibilities to two equally likely cases.

    PARADOX #1 - Common Sense =VS= The Math
    It is counter to our common sense that this would make a difference, but the math works out differently than we naturally expect from a common sense perspective. This is the paradox.

    It shouldn't matter but it does, similar in vein to (albeit in a distinctly different manner from) the Monty Hall problem.

    PARADOX #2 - Is the amount really important?
    Or perhaps the paradox is in the fact that regardless of what we actually find, it is better to switch. In this respect, the fact we know the amount is important, but the actual amount is not.

    TESTING THE ANSWER
    The difficulty of testing the actual returns via a series of random trials is that the original terms of the trial is unbounded, making it impossible to select the random numbers required to test it.
  2. Standard memberPalynka
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    24 Mar '09 11:43
    I'm having trouble replicating my results. 😕

    Ok, here's what I'm getting now and you can tell me how to proceed or how to correct what I made wrong.

    We know that the amounts x cannot come from a uniform for all possible values of x (no such distribution exists) so let's set assume that the draws come a uniform distribution with an unknown bound of X.

    The question is then if we can infer anything about X from x. Using a similar reasoning as in the doomsday argument, we set a prior for X: P(X) = k/X (with k a normalizing parameter). The unconditional distribution of x is then such that P(x) = k/X.

    By Bayes' theorem, we get the posterior distribution (note that the k's cancel out as they should):
    P(X|x) = P(x|X)*P(X)/P(x) = x/X^2

    We know that for any x > X/2 then we must hold the larger amount an we can approximate the probability by a continuous one (X large) and integrate P(X|x) between x and 2x. This gives us a value of exactly one half. P(X<2x) = 1/2. Conversely, P(X>2x) = 1/2

    But in the first case (X<2x), we necessarily hold the larger envelope and swapping leads to a loss of -1/2x. In the second case (X>2x), then we might hold the larger or the smaller envelope. A naive interpretation of the last sentence would assume that we could be holding any of the two envelopes with equal probability and therefore the expected value of switching would be:
    1/2 * -x/2 + 1/2* (1/2*-x/2 + 1/2*x) = -1/8 x

    The expected value of switching would then be negative! The problem is that in the second case the probabilities may depend on X and x. I was able to find the correct expression to the expected value for switching when x<N/2 and it gave me a zero expected value of switching. I can't seem to be able to find it now, so it's probably wrong.

    The key intuition here is that if your amount in the original envelope is in the upper half of the distribution, then you cannot double. But if the distribution is uniform over a given interval, then you have 50% chance of being in that upper half!
  3. Joined
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    05 Apr '09 03:57
    I've now been thinking about this one for a while, and I think I have a little insight...

    In terms of running a test with millions of trials and an infinite ceiling, the reason why one would not be able to see a measured positive difference in the outcome from switching envelopes is the very lack of a "highest" value. since the envelopes themselves could be infinitely large in value, any data from a particular envelope choice is outweighed by an infinite number of infinitely larger cases. and this is true no matter how large the envelope is... there are an infinite number of cases that outweigh the current choice to the point where by magnitude, an expectation of 5/4x vs x is meaningless and immeasurable. and furthermore, since there is no known "highest value" envelope, there is no single exception case where the player would be making a bad play by switching envelopes... as you say it SEEMS paradoxical.

    it has also been said that the apparent paradox is that if one took the strategy of "picking" an envelope, and switching before it is opened, and THEN opening the envelope, that the person should get the same results as the person switching after seeing the envelope. yet the finite answer says that no matter what value you see you should switch.

    i contend that it is the ACT of observation of the envelope that defines it at a finite value. akin to the box containing "schroedinger's cat", the envelope you open has undefined value until you actually open it to see how much money is in it. until opening it has equal probability of being any value in the list of possible envelope values. but then AFTER opening the envelope, it is defined as being valued "x" and we are given a choice of choosing between the possibility of the other envelope being valued "2x" or "1/2x." and it turns out that for any FINITE value x, and with no promise of there being a "highest" valued envelope, there is more to gain from switching then there is to lose [with an expected value of (5/4)x].

    thoughts?
  4. Joined
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    05 Apr '09 04:12
    Originally posted by Palynka
    I'm having trouble replicating my results. 😕

    Ok, here's what I'm getting now and you can tell me how to proceed or how to correct what I made wrong.

    We know that the amounts x cannot come from a uniform for all possible values of x (no such distribution exists) so let's set assume that the draws come a uniform distribution with an unknown bound of X. ...[text shortened]... is uniform over a given interval, then you have 50% chance of being in that upper half!
    i haven't really gone through your calculations for accuracy, but i think i understand your basic reasoning... however, i think most of it is based on an assumption i don't necessarily agree we can make.

    essentially, you are assuming there is a highest possible value M that the envelope could be, but we do not know how large that is. in fact, this assumption is exactly against the problem as proposed, and if it were true, of course there would be an error in the (6/5)x expected value interpretation i've been discussing: that if the envelope in question is the "highest possible" value, the player would be a fool to switch envelopes.

    then you propose (and please correct me if i'm wrong) that based on the assumption there is a largest possible value, but we don't know how large it is, we should make a guess as to whether our value is in the top half of a normal distribution that has a ceiling of M. this would be an educated guess of course, a probabilistic calculation that is analogous to the question "how likely is it that the host of this game show has so many envelopes that the value i see is one of the 'smaller' envelopes rather than one of the 'larger' ones."

    i don't know the "doomsday argument" but it seems to me that these assumptions are largely flawed considering that the posed problem gives us no indication of the magnitude of the highest possible value in an envelope, and in fact explicitly states that the values are drawn from a list that tends towards infinity. even if there WERE a ceiling value, our guess at its magnitude would be like throwing darts in the wind, in my opinion.
  5. Standard memberPalynka
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    05 Apr '09 10:20
    Thanks for the comments Aethereal, I'll try to go by them when I have more time , I just managed to skim them for the moment.

    Just so I know how to frame this, are you familiar (very, loosely, not at all) with the use of priors in probability?
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    05 Apr '09 20:24
    Originally posted by Palynka
    I'm having trouble replicating my results. 😕

    Ok, here's what I'm getting now and you can tell me how to proceed or how to correct what I made wrong.

    We know that the amounts x cannot come from a uniform for all possible values of x (no such distribution exists) so let's set assume that the draws come a uniform distribution with an unknown bound of X. ...[text shortened]... is uniform over a given interval, then you have 50% chance of being in that upper half!
    i've been examining this post and the info in the "doomsday argument" wikipedia page to help me better understand the math behind what you're doing here...

    but i think there's a logical error that is somewhat important: you discuss two cases, one in which X (the upper bound on x) is > 2x and one in which X < 2x and say that they have equal probability.

    however, the envelopes must always be double one another. so if X < 2x, the only possible case is that we are holding the biggest envelope in the set... but every envelope that is NOT the biggest envelope in the set contains an x such that X > 2x. so how can these cases have equal probability? in fact the larger the set gets, the less likely it is that we are holding the largest envelope in the set (the case with a (-1/2)x expected value) and the more likely we are holding an envelope that is not the largest in the set (the case with a [1/2*(-1/2*x) + 1/2*(x)] = +1/4*(x) expected value).

    and i think that as the set of possible envelopes tends towards infinity, the probability that we are holding the very largest envelope in the set tends towards zero, making the probability that we are NOT holding the very largest envelope tend toward 1.
  7. Joined
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    06 Apr '09 00:42
    Originally posted by Palynka
    Thanks for the comments Aethereal, I'll try to go by them when I have more time , I just managed to skim them for the moment.

    Just so I know how to frame this, are you familiar (very, loosely, not at all) with the use of priors in probability?
    i would say i'm between very and loosely familiar with the use of priors... what i am very good well versed in is extrapolation and inductive reasoning. i.e. i pick things up usually the first time through an explanation, if it's clearly explained, and am very fluent in the mechanics of algebra and calculus.

    so if it seems like i don't get the way in which you're framing the problem, a brief rundown of intent or a brief mathematical step-by-step walkthrough is usually enough to get me on the same page (for example, the maths you used in a previous post defining a prior and analyzing posterior was fine, and i agreed with the method except for the key point i outlined in my previous post)

    cheers!
  8. ALG
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    06 Apr '09 10:27
    I don't really see the problem

    I read somewhere in this thread you have 50% chance of gaining 100% and 50% chance of loosing 50%.
    ( 200% + 50% ) / 2 = 125% per switch

    So after 1.000.000 switches, you have X * 1,25 ^ 1.000.000 dollar?
    I think it doesn't matter what you choose.
  9. Standard memberPalynka
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    06 Apr '09 11:12
    Originally posted by Aetherael
    i contend that it is the ACT of observation of the envelope that defines it at a finite value. akin to the box containing "schroedinger's cat", the envelope you open has undefined value until you actually open it to see how much money is in it. until opening it has equal probability of being any value in the list of possible envelope values. but then AF ...[text shortened]... more to gain from switching then there is to lose [with an expected value of (5/4)x].
    This is exactly why I think the use of priors must be the correct way to formulate this. The act of opening the envelope updates the prior and so such a way would be consistent with indifference before opening but not after opening. The opening of the envelope reveals information.

    The problem I have with your final remarks in this post is it uses information from an impossible posterior (unbounded uniform). Although it is possible for that ex-post the probabilities are 1/2, this cannot be because the values came from an unbounded uniform!

    Another beauty of using priors is that we can use an unbounded uniform as a prior (reflecting our complete lack of information beyond the fact that the amounts are positive), which will imply a well-defined posterior after the envelope is opened.
  10. Standard memberPalynka
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    06 Apr '09 11:22
    Originally posted by Aetherael
    i haven't really gone through your calculations for accuracy, but i think i understand your basic reasoning... however, i think most of it is based on an assumption i don't necessarily agree we can make.

    essentially, you are assuming there is a highest possible value [b]M
    that the envelope could be, but we do not know how large that is. in fact, th ...[text shortened]... guess at its magnitude would be like throwing darts in the wind, in my opinion.[/b]
    Well, you can call it an assumption, but for me it is a logical necessity. The fact that unbounded uniform distributions do not exist leaves us with only two choices, if we don't want simply to answer: "the problem is incorrectly posed". These are:

    1) The distribution is not uniform.
    2) The distribution is not unbounded.

    Personally I think that weakening 2) is more true to the spirit of the question. Of course, arbitrarily fixing the limit would also not be true to the spirit of the question. So what I try to do is to impose as little as possible by not excluding ANY possible value ex-ante by using an unbounded prior.

    Like this, all values are possible ex-ante and the "true" distribution is still uniform.
  11. Standard memberPalynka
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    06 Apr '09 11:322 edits
    Originally posted by Aetherael
    i've been examining this post and the info in the "doomsday argument" wikipedia page to help me better understand the math behind what you're doing here...

    but i think there's a logical error that is somewhat important: you discuss two cases, one in which X (the upper bound on x) is > 2x and one in which X < 2x and say that they have equal probabilit ...[text shortened]... ro, making the probability that we are NOT holding the very largest envelope tend toward 1.
    however, the envelopes must always be double one another. so if X < 2x, the only possible case is that we are holding the biggest envelope in the set... but every envelope that is NOT the biggest envelope in the set contains an x such that X > 2x. so how can these cases have equal probability? in fact the larger the set gets, the less likely it is that we are holding the largest envelope in the set (the case with a (-1/2)x expected value) and the more likely we are holding an envelope that is not the largest in the set (the case with a [1/2*(-1/2*x) + 1/2*(x)] = +1/4*(x) expected value).

    I'm not sure I understand... I got the value of 1/2 through integrating: [;\int_n^{2n} P(N|n)dN ;].

    where n is the value in the envelope and N is the upper bound. I integrate over the upper bound.

    In fact, getting a "low" value of n is more likely when N is "low", so I don't think one can make an unconditionally claim. This is why I integrate over P(N|n).
  12. Standard memberPalynka
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    06 Apr '09 11:54
    But now you're making me think that there might be a way to apply the prior directly to n...

    I think my mistake is that when I'm in the P(N>2n), I assume equal likelihood of holding each envelope. Conditionally, this might not be true (and my intuition is that it isn't).
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    06 Apr '09 14:081 edit
    Originally posted by Palynka
    But now you're making me think that there might be a way to apply the prior directly to n...

    I think my mistake is that when I'm in the P(N>2n), I assume equal likelihood of holding each envelope. Conditionally, this might not be true (and my intuition is that it isn't).
    yeah this is the part where i was straying from your argument - the integration was fine, and the splitting into two cases to examine P(N>2n) and P(N<2n)

    but my point is that as the set of possible values in the envelope grows, P(N<2n) shrinks. Think of what the terms N and n mean in the setup of the problem we've been dealing with... we are trying to determine what the probability that the upper bound is less than double the value in our envelope. but this is ONLY true if we have chosen the largest possible envelope of all allowed in the set; that is, that we have chosen the envelope that sets the upper bound.

    just to use a concrete example... imagine the bounded set contains envelopes {1,2,4,8,16,32,64}. the player doesn't know of course that 64 is the ceiling value, but you can see that when N=64, the only time N<2n is when n=N, or rather when we are looking at the largest envelope. in this concrete example, we see that the probability we chose n=N is 1/7... as the number of envelopes in the set grows this probability will tend toward 0, and the probability that we did NOT choose the biggest envelope will tend toward 1.

    all other envelopes contribute to the P(N>2n), which has a positive expectation of +(1/4)n from the math in your earlier post.

    i guess what i mean is, though we are using a uniform distribution for the possible values of n, that these possible values are discrete numbers, and are ONLY the powers of 2... so to think that half of them would be larger than N/2 and half of them smaller than N/2 is folly. in fact only one value in the distribution will be ever be larger than N/2 and that envelope will be the upper bound N.
  14. Standard memberPalynka
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    06 Apr '09 14:521 edit
    Originally posted by Aetherael
    yeah this is the part where i was straying from your argument - the integration was fine, and the splitting into two cases to examine P(N>2n) and P(N<2n)

    but my point is that as the set of possible values in the envelope grows, P(N<2n) shrinks. Think of what the terms N and n mean in the setup of the problem we've been dealing with... we are trying to the distribution will be ever be larger than N/2 and that envelope will be the upper bound N.
    but my point is that as the set of possible values in the envelope grows, P(N<2n) shrinks.

    The set of possible values is not bounded, so I don't know what you mean by that. My prior already includes all possible values of N up to infinity. And I'm integrating over that support (up to infinity). Think about it. If the integral of n up to 2n is 1/2, then the integral of 2n to 3n is smaller than 1/2. So your idea that P(N<2n) "shrinks" when N increases* is already included.

    *I'm interpreting your statement like that, but note that "possible values" are up to infinity. There is no value for which I impose a zero-probability of happening.

    {1,2,4,8,16,32,64}
    Ha! Ok, I may have misled you with my initial posts. In my replicating calculations I was thinking of the set of possible values being the reals, not just a sequence of the powers of 2. This was where we were miscommunicating (my bad, sorry...)

    i guess what i mean is, though we are using a uniform distribution for the possible values of n, that these possible values are discrete numbers, and are ONLY the powers of 2... so to think that half of them would be larger than N/2 and half of them smaller than N/2 is folly.
    Yes, absolutely correct. See above.

    Somewhere along the way I realized exactly what you're saying now and dropped the powers of 2 assumption. It is too restricting and I suspect solves the problem exactly in the way you also suspect. Before opening the envelope, switching is infinity-infinity (indeterminate) and after opening the envelope, switching is optimal. The paradox disappears because the indeterminacy solves the conundrum.

    It doesn't seem very satisfying, right? Well, to me it isn't. I think it imposes too much which isn't in the question (the powers of 2), along with the unknown interval size. So I'm now thinking how to work it out in a more general setting...
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    06 Apr '09 19:311 edit
    Originally posted by Palynka
    [b]but my point is that as the set of possible values in the envelope grows, P(N<2n) shrinks.

    The set of possible values is not bounded, so I don't know what you mean by that. My prior already includes all possible values of N up to infinity. And I'm integrating over that support (up to infinity). Think about it. If the integral of n up to 2n is 1/2, n interval size. So I'm now thinking how to work it out in a more general setting...[/b]
    ok now i see where we separated in terms of my understanding of your argument. now it seems ok, except i will ask you to consider this:

    1) wouldn't setting the problem over the reals contain an infinitude of irrational numbers that, in the spirit of the problem, wouldn't really work as a "dollar amount" in the envelope? since they are uniformly distributed throughout the real line, this wouldn't hurt our ability to uniformly distribute the RATIONALS, and so i can see an argument for using this as a setting for the problem. however,

    2) using the rationals requires an infinitesimally accurate system of "value," that is to say the number you see in the envelope could be as tiny as .0000....0001, which i think again is not within the spirit of the problem. also the rationals include things like 5/7 dollars and 41/3 dollars, which it seems to me goes against the spirit of the problem as a dollar amount. if it were posed as "you open an envelope and there is a number in it... and are told it is either exactly half or exactly double the number in the other envelope" etc. then i could see this as plausible. but i think these restrictions are not flippant, but really within the spirit of the problem, and as such

    3) i believe the most "natural" setting (har har) for the problem is over the natural numbers. but then since we get into the issue that any non-power of 2 would give you too much information about the other envelope, so we restrict our scope to {1, 2, 4, 8, 16, ...}

    perhaps these restrictions are too strong, but i personally think it is more in line with the spirit of the problem. that being said, infinitesimally accurate decimals and a placement over the rationals is entirely arguable, and i think that all the maths you've done so far would work over the rationals...i just think it doesn't seem right to tell the player "in your envelope is 147 and 93756/142019274 dollars... do you think there is double that or half that in the other envelope?" 🙂 just being cheeky.
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