Posers and Puzzles

Posers and Puzzles

  1. Standard memberPalynka
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    06 Apr '09 19:48
    Originally posted by Aetherael
    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 distrib ...[text shortened]... there is double that or half that in the other envelope?" 🙂 just being cheeky.
    🙂

    I find it a very interesting puzzle to think about, not because I really want to find a solution, but because an apparently simple problem is actually not simple at all. I particularly like that it challenges some intuitive, but imprecise, interpretations about probabilities, which I feel helps me understand them at deeper level.

    If you're happy to remain with 3), I'm sorry to see you go. But I hope you are willing to continue to help me think about this and my intuition is that you are also intrigued by it. 😉
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    06 Apr '09 20:46
    Originally posted by Palynka
    🙂

    I find it a very interesting puzzle to think about, not because I really want to find a solution, but because an apparently simple problem is actually not simple at all. I particularly like that it challenges some intuitive, but imprecise, interpretations about probabilities, which I feel helps me understand them at deeper level.

    If you're ha ...[text shortened]... continue to help me think about this and my intuition is that you are also intrigued by it. 😉
    certainly 🙂 it's a really interesting problem
  3. Standard memberPBE6
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    07 Apr '09 17:411 edit
    Do you guys think the simple solution posted on Wikipedia makes sense?

    (from http://en.wikipedia.org/wiki/Two_envelopes_problem#Proposed_solution)

    *****

    Proposed solution

    The most common way to explain the paradox is to observe that A is used inconsistently in the expected value calculation, step 7 above. In the first term A is the smaller amount while in the second term A is the larger amount. To mix different instances of a variable or parameter in the same formula like this shouldn't be legitimate, so step 7 is thus the proposed cause of the paradox.

    For example, if the lower of the two amounts is denoted by C, the expected value calculation may be written as

    (1/2)C + (1/2)*(2C) = (3/2)C

    Here C is a constant throughout the calculation—as it should be—and we learn that 1.5C is the average expected value in either of the envelopes. So according to this new calculation there is no contradiction between the decisions to keep or to swap, and hence no need to swap indefinitely.

    *****
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    07 Apr '09 19:20
    Assuming the money in both envelopes are paper notes, the problem has a very obvious solution:

    If the envelope you open has an odd number of $ in it, switch and you will always find the bigger amount in the other envelope.

    If the envelope you open has an even number of $ in it, it doesn't matter whether you switch or not.

    This is the financier's solution, not the mathematician's solution 🙂
  5. Standard memberPalynka
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    07 Apr '09 21:13
    Originally posted by PBE6
    Do you guys think the simple solution posted on Wikipedia makes sense?

    (from http://en.wikipedia.org/wiki/Two_envelopes_problem#Proposed_solution)

    *****

    [b]Proposed solution


    The most common way to explain the paradox is to observe that A is used inconsistently in the expected value calculation, step 7 above. In the first term A is the smaller am ...[text shortened]... tion between the decisions to keep or to swap, and hence no need to swap indefinitely.

    *****[/b]
    In short...no. It seems written by someone that didn't really understand why there is a paradox.
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    07 Apr '09 23:011 edit
    Originally posted by Palynka
    In short...no. It seems written by someone that didn't really understand why there is a paradox.
    i was going to post something along these lines... the writer seems to attribute the paradox to a misuse of variables, where there is none. and he shows his misunderstanding of the expected value calculation through the math he presents as [(1/2)c + (1/2)2c]/2 = (3/2)c. this is like calculating the expected value of "keeping it half the time and switching half the time, given that i am holding the smaller envelope" which does not remotely resemble the real problem...

    *edit* the writer of the wikipedia article did not seem to really get what the writers of the math article cited were really saying... in fact it's almost reasonable, but in my estimation is only applicable to the "closed envelope" version of the problem where the value of the envelope is unknown before expected value is calculated. this, in my opinion, is clearly fallacious, and the paper does a very decent job of explaining why... however it's not relevant to the "open envelope" version we've been talking about and so is attacking a "different" paradox, so to speak. i don't know if that makes sense but check out the article!

    http://www.sorites.org/Issue_20/dever.htm
  7. Standard memberPBE6
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    08 Apr '09 01:12
    Don't be fooled by the length of this post!! It is quite possibly utter nonsense. But if you don't seek help in editing your thoughts, how will you ever learn? 🙂

    I wonder if Bayes theorem is even applicable here, maybe that's the source of the apparent paradox? To be fair I don't know enough math to propose anything definitive, but here's my chain of thought, just for teh lulz.

    The simple solution proposes the following:

    There are two sums, A and B. Let A be the smaller sum, and B (=2A) be the larger sum. You are presented with an envelope, and the sum inside is revealed. The sum is either the smaller sum A or the larger sum B, but you don't know which one you have. Assuming that you switch based on the flip of a coin*, if you just so happened to have revealed the smaller sum then your expected value is (1/2)A + (1/2)2A = (3/2)A. If you just so happened to have revealed the larger sum then your expected value is (1/2)B + (1/2)B/2 = (3/4)B = (3/4)2A = (3/2)A.

    *Now, I realize that the entire point of the question is to determine the proper strategy when presented with either sum. In that case, the "flip of the coin" strategy begs the question. That's a fair objection.

    Having looked up Bayes theorem on Wikipedia just to try and reassure myself of the details, I came across a typical example problem. Say there is a private school with 60% boys and 40% girls, and all the boys plus half the girls wear trousers, while half the girls wear skirts. You see one of the students from a distance, and realize that they're wearing trousers. What is the probability that the student is a girl?

    Bayes theorem provides a clear answer. If A = "the student is a girl", and B = "the student wears trousers", then P(A|B) = P(B|A)*P(A)/P(B) = 0.5*0.4/0.8 = 0.25. But what really happened here? You knew some information about the students initially, including how what percentage were girls and boys, and how many of those wear trousers. If you draw a Venn diagram of the situation, say with labels "Bt" for boys wearing trousers, and "Gt" and "Gs" for girls wearing trousers and skirts, respectively, it's clear to see that you've really sorted the population into subgroups along two dividing lines, gender and clothing. In one sense, Bayes theorem formalizes the process of comparing the probabilities of selecting a member from any of these subgroups by telling you which subgroups are included and which ones aren't. The basis for this discrimination is additional information provided by the observation of one particular attribute.

    Now, what if we were to arrange an unknown number of tiddleywinks in a row, and pick one randomly while blindfolded. The tiddleywinks were indeed in a definite order, but it's not possible to distinguish one tiddleywink from another on the basis of appearance alone. However, we are asked "given that you picked your particular tiddleywink, what is the probability that the next tiddleywink you pick was laying to the right of your original tiddleywink?". The fact that you have selected a particular tiddleywink, but one which you can't identify, gives you no help in determining either the number of tiddleywinks nor the relative left/right distribution of tiddleywinkdom relative to your tiddleywink. Is the expected value of the left/rightedness of the next tiddleywink a paradox? Possibly, but it really seems to be a case of non-information masquerading as information.

    What does this have to do with the problem at hand? Well, I think that the "information" provided by opening the envelope isn't really useful information at all (unless you assume a finite limit on the possible amount of money in the envelope, which is really a practical consideration and not really relevant to a purely conceptual arrangement). It really just reveals that indeed you have selected a sum somewhere on the continuum of sums from the lowest possible sum to the highest possible sum (if there even is a highest possible sum). If that's the case, then what relevance does it have to the problem at hand? I think the answer is "none at all". If that's the case, on what basis do we make our Bayesian discrimination?
  8. Standard memberPBE6
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    08 Apr '09 02:00
    Further to my previous post, I don't think I addressed the problem of distribution.

    Even for a finite case, the distribution matters. For instance, in the tiddleywink example, let's say there are 2 tiddleywinks, and I select the left tidleywink with probability P1 and the right tiddleywink with probability P2. Obviously, P1 + P2 = 1. Now, the probability that the next tiddleywink I pick lies to the right of the original is P1(1) + P2(0) = P1. This only works out to 50%/50% if P1 = P2 = 50%. If not, the probability is skewed and the assumption that all selections are equal is wrong.

    But I suppose this brings me back to the problem of "information". If I knew that the tiddleywinks would be selected with a particular known distribution before I picked, or even just that the distribution were skewed, I wouldn't be able to assume that all choices were equal. However, what if I don't know this? Would my calculation be wrong if I didn't consider a piece of information I didn't have in the first place? Which distribution should I pick in the absence of any information?

    On a quantum level, it is apparent that there are events for which we cannot know the outcome beforehand on a fundamental level, and any skewness in the outcome must be determined by repeated experiment. On a macroscopic level, it seems that all events are deterministic in any relevant sense provided we had enough information about the processes going on above the quantum level, which we frequently do not (by "relevant", I mean we would be able to determine the difference by measurement...considering that most people have difficulty counting the number of objects in a group above 50, and that not everyone has access to atomic-scale measurement devices, not to mention our ability to model atomic interactions in real-time is extremely limited, quite a bit of information escapes our everyday grasp). In the case of the tiddleywinks, the event organizer would likely know the distribution of the tiddleywinks (or at least the tiddleywinks would have had an impact on the ground when they were laid, hence making a record of their position and distribution), so it might be that the word "probability" is being employed loosely at best by many parties. But for me, the blindfolded selector with no information, "probability" is all I've got.

    In the mathematical sense, none of the above has much relevance. Observer effects and additional information can be included in probability calculations (and in fact form the basis of Bayesian probabilities), while infinite domains coupled with the mathematical impossibility of assigning uniform distributions to them, throws off the whole equation. But this is because the mathematical representation of the problem is idealized, unconstrained by the physical limitations of reality. Our intuitive sense of probability was honed to deal with practical problems (like roughly calculating the relative risks and rewards of hunting a large dangerous animal for food), and even our trained sense of probability relies on manipulating our intuitive sense by breaking down complex problems into easily digestible components (which is why writing down equations step by step helps so often when correcting a mistake in a probability question, or when addressing a complex problem outside our normal expertise). I think this may be part of the reason why the results of this envelope problem seem so paradoxical, because the conditions of the problem are so far removed from practical experience.
  9. Standard memberPBE6
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    08 Apr '09 02:08
    NOTE: I'm going to be much briefer here.

    With all the above nonsense out of the way (my apologies to those who spent nonrefundable minutes of their lives reading it 😕), I think the problem with the paradox lies either in the question itself, or with the transcription of the question into mathematical terms. I also think that the intuitive answer for the expected value of (5/4)X, while "wrong" in the classical sense since it lights the way to the paradox, is misinterpreted as a finite answer, when in reality it seems to be an inarticulate indicator that the infinite and infinitessimal values in the question have not been reigned in sufficiently.

    And with that vague statement out of the way, I welcome your comments. Fire away! Time for me to learn a little more about math, the hard way (again!!)... 😕
  10. Standard memberPalynka
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    08 Apr '09 09:28
    Originally posted by PBE6
    [b]Don't be fooled by the length of this post!! It is quite possibly utter nonsense. But if you don't seek help in editing your thoughts, how will you ever learn? 🙂

    I wonder if Bayes theorem is even applicable here, maybe that's the source of the apparent paradox? To be fair I don't know enough math to propose anything definitive, but here's my chain ...[text shortened]... he case, on what basis do we make our Bayesian discrimination?[/b]
    Ok, apologies if I sound brusk, no arrogance intended, just that there's a lot to go through.

    There are two sums, A and B. Let A be the smaller sum, and B (=2A) be the larger sum. You are presented with an envelope, and the sum inside is revealed. The sum is either the smaller sum A or the larger sum B, but you don't know which one you have. Assuming that you switch based on the flip of a coin, if you just so happened to have revealed the smaller sum then your expected value is (1/2)A + (1/2)2A = (3/2)A. If you just so happened to have revealed the larger sum then your expected value is (1/2)B + (1/2)B/2 = (3/4)B = (3/4)2A = (3/2)A.
    This doesn't seem appropriate to think about this problem. To see this clearly, imagine the values were fixed, known and equal to, say, 10 and 20. Then by opening the envelope, you would have full information about which envelope contains what. Yet nothing would change in your thought experiment and your expected values would remain the same.

    Is the expected value of the left/rightedness of the next tiddleywink a paradox? Possibly, but it really seems to be a case of non-information masquerading as information.
    Mmm... That isn't not a paradox. I think the probability would be exactly one half for any distribution (as long as you cannot "pile" them in one spot and there's more than one).

    Well, I think that the "information" provided by opening the envelope isn't really useful information at all (unless you assume a finite limit on the possible amount of money in the envelope, which is really a practical consideration and not really relevant to a purely conceptual arrangement).
    How is it not relevant? You are calculating expected values based on a probability distribution for the sums that doesn't exist! That it seems intuitive to do so, doesn't mean it's right. Moreover, that sleight of hand leads to a paradox!

    It really just reveals that indeed you have selected a sum somewhere on the continuum of sums from the lowest possible sum to the highest possible sum (if there even is a highest possible sum).
    Again, if all values are equally likely, then there must be a highest possible sum. There is no escape from that. You either drop one or the other. Using priors and Bayesian definitions of probability is the only possible way to say:
    1) There is no pair of positive reals (x,y) for which P(x) > P(y) (i.e. unconditionally, there is no value which is more likely than any other)
    2) The lowest sum is drawn from a probability distribution
  11. Standard memberPalynka
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    08 Apr '09 09:39
    Originally posted by PBE6
    Even for a finite case, the distribution matters. For instance, in the tiddleywink example, let's say there are 2 tiddleywinks, and I select the left tidleywink with probability P1 and the right tiddleywink with probability P2. Obviously, P1 + P2 = 1. Now, the probability that the next tiddleywink I pick lies to the right of the original is P1(1) + P2(0) = P ...[text shortened]... ecause the conditions of the problem are so far removed from practical experience.
    This only works out to 50%/50% if P1 = P2 = 50%. If not, the probability is skewed and the assumption that all selections are equal is wrong.

    But I suppose this brings me back to the problem of "information". If I knew that the tiddleywinks would be selected with a particular known distribution before I picked, or even just that the distribution were skewed, I wouldn't be able to assume that all choices were equal.


    As long as there is anonymity in your pick (guaranteed by the blindfold), this is wrong. You're drawing from a distribution at random and asking what is the likelyhood of me being to the left or the right of the distribution. Obviously, it's 50%. If the distribution is skewed, then anonymity ensures your pick will follow the exact same skewness and you will be half of the times below and half of the times above the median. Remember, your example is about position relative to the median, not the mean!

    But this is because the mathematical representation of the problem is idealized, unconstrained by the physical limitations of reality. [...] I think this may be part of the reason why the results of this envelope problem seem so paradoxical, because the conditions of the problem are so far removed from practical experience.

    Most people won't even understand there is a paradox, so I don't see how it is intuition which leads to the paradox. Quite the contrary. When it comes to conditional probability, wrong intuitive interpretations are rampant, from the Monty Hall problem to the Prosecutor's fallacy.
  12. Standard memberPBE6
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    08 Apr '09 20:571 edit
    Thanks for the reply! I don't want to claim this as a defense of the bad ideas, just the length - but I was stoned last night and feeling very verbose. 😀

    Another question:

    In the original problem, you don't know anything about the amount of money in either envelope or the probability distribution from which the amounts were drawn (including whether they were drawn from a finite or an infinite domain). If that's the case, what information do you really obtain by opening one of the envelopes, except that now you know precisely how much money is in envelope A?

    In the other experiment you mentioned, having a definite $10 in one envelope and a definite $20 in the other, you already have quite a bit of information before you choose. After choosing, you have even more information because you know precisely which envelope has the larger amount. In this example I'm clear on what information was revealed, but what information was revealed in the original problem?
  13. Standard memberPBE6
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    08 Apr '09 21:00
    The intuitive use of probability I was referring to in my other post was that of treating all probability "objects" the same. That is, imagining infinite and infinitessimal quantities to be merely "large" and having the same properties as finite quantities. Careful analysis can help us refine our instincts. I agree that it may not play a part in this paradox, but it's always a risk when dealing with those quantities.
  14. Standard memberPalynka
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    08 Apr '09 22:011 edit
    Originally posted by PBE6
    In this example I'm clear on what information was revealed, but what information was revealed in the original problem?
    We know for a fact that there is some information being revealed:

    - About the distribution: we now know that the support doesn't have an upper bound lower than x.
    - About the possible values in the other envelope (now restricted to x/2 and 2x)

    And possibly more. What exactly is revealed, and how to incorporate it, is the crux of the question, of course.
  15. Standard memberPalynka
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    08 Apr '09 22:05
    Originally posted by PBE6
    The intuitive use of probability I was referring to in my other post was that of treating all probability "objects" the same. That is, imagining infinite and infinitessimal quantities to be merely "large" and having the same properties as finite quantities. Careful analysis can help us refine our instincts. I agree that it may not play a part in this paradox, but it's always a risk when dealing with those quantities.
    Definitely. If you look at Aethereal's example of the values being powers of 2, the paradox could be resolved neatly due to infinity. (Note that you still need a Bayesian and not a frequentist interpretation of probabilities or the problem is still mis-specified.
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