12 ms·

I love this problem because it is so simple, and the false line of reasoning is so compelling that it would hardly raise an eyebrow if you saw it in an academic

by sjburt 4y ago

I love this problem because it is so simple, and the false line of reasoning is so compelling that it would hardly raise an eyebrow if you saw it in an academic paper and yet the conclusion is so obviously wrong. Decision problems are tricky and in non intuitive ways.

Schroedingersat 4y ago
Step 3 (and thus also 1 and 2 on reflection) stands out to me immediately as describing two different situations for the entire game. The case where switching gives you 2A and the case where switching gives you A/2 describe 2 completely different universes, not two different actions in one.
- NamTaf 4y ago
  Agreed. Step 6 is where it breaks down, because it's redefined A as being one of the two envelopes to being the halfway point between those two envelopes. Until step 6, it's describing two parallel views of the universe (depending on which envelope you first get - hence steps 4 and 5), and then mashes them together in a way that doesn't work. Take the practical example of $100 and $200: A is either $100 or $200 depending on which envelope you receive first, so the equation is either 0.5*A + 0.5*2A, or it is 0.5*A + 0.5*(A/2). It can never be 0.5*2A + 0.5*(A/2)
  - RHSeeger 4y ago
    Even after looking at those numbers, it still felt wrong to me. It eventually occurred to me that this was because my intuition was telling me the total should add up to A. And it doesn't; it adds up to 9/8 A. It took me a while to realize that my intuition was wrong, and that there's no reason the total should be A. The thing that helped me reconcile this was realizing that the A in the two equations are different values; then replacing them. (0.5⋅(0.5⋅A + 0.5⋅2.A)) + (0.5⋅(0.5⋅A + 0.5⋅0.5⋅A)) = 9/8 A ^ but the A in the left grouping (where it's 100) is different than the A in the right grouping (where it's 200). Replacing the As with their actual values 0.5⋅(0.5⋅100+0.5⋅2⋅100)+0.5⋅(0.5⋅200+0.5⋅0.5⋅200) = 150 And, since one envelope has 100 and the other has 200, an expected outcome across both envelopes is, as calculated, 150.
- xmprt 4y ago
  The breakdown for me was in step 4-5. > If A is the smaller amount, then the other envelope contains 2A. > If A is the larger amount, then the other envelope contains A/2. These are conditional probabilities so you can't simply add them up and compute an expected value of switching like they show in the example.
joe_the_user 4y ago
Well, the simple way to put is that "A" isn't fixed. The "expected value" argument in steps 6-7 is acting like "A" is a single value when "A" will be larger or smaller depending on what envelope you picked.
- dllthomas 4y ago
  I think that subtly misses the point. The problem is that you're implicitly using a distribution that... isn't a distribution. And with this particular not-a-distribution, whether you should switch or not does not depend on the value of A. But with any actual distribution (... I think?) it does, at which point... no paradox. It's true that it's not clear (at least to me, but perhaps more generally) what distribution we should assume, but as long as we avoid treating something that isn't a distribution as if it were one we avoid the worst of it.
  - eru 4y ago
    Yes. The big problem is that you can't have a uniform distribution on the natural numbers. (And by extension, you can't have a uniform distribution on the rounded-to-integer version of a distribution on the real numbers.)
    - xigoi 4y ago
      Who said there is a uniform distribution on the natural numbers?
      - dllthomas 4y ago
        I think the assumption that "for all x, p(x) = p(2x)" implies a uniform distribution on the natural numbers. Imagine we have a distribution that satisfies that assumption, and then someone tells you they've sampled from that distribution and found that the result is of the form (say) 7*2^k, for some k > 0. That conditional distribution for k would seem to have to be uniform, right?
    - dllthomas 4y ago
thekiptxt 4y ago
This reminds me of how chess tactics feel so obvious when I’m on a tactics trainer, yet I can’t identify them in games where they actually happen. Or code that “obviously” has a bug only after it caused an issue in production. Would be nice if someone warned me before I do a PR: “ok, fyi there is a subtle but devastating bug in the new_feature.cpp”
- maest 4y ago
  This is a class of "hard to find, easy to verify" solutions.
- na85 4y ago
  Well if debugging is the process of removing bugs from code, then logically the act of writing the code must be called "bugging" and the probability of having a bug in any new piece of code asymptotically approaches 1 with increasing SLOC ;^)
  - dbt00 4y ago
    Every program has bugs, every program has inefficiencies. Therefore every program can be eventually reduced down to a single instruction, which will be incorrect.
    - Archelaos 4y ago
      > Every program has bugs Where is the bug in this programm? END
      - stephencanon 4y ago
        What’s the specification?
        the_sleaze9 4y ago
        It was supposed to return a generated image of Reverend Thomas Bayes in the style of stained glass.
deleted 4y ago
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by sjburt 4y ago

Schroedingersat 4y ago

Step 3 (and thus also 1 and 2 on reflection) stands out to me immediately as describing two different situations for the entire game. The case where switching gives you 2A and the case where switching gives you A/2 describe 2 completely different universes, not two different actions in one.

NamTaf 4y ago
Agreed. Step 6 is where it breaks down, because it's redefined A as being one of the two envelopes to being the halfway point between those two envelopes. Until step 6, it's describing two parallel views of the universe (depending on which envelope you first get - hence steps 4 and 5), and then mashes them together in a way that doesn't work. Take the practical example of $100 and $200: A is either $100 or $200 depending on which envelope you receive first, so the equation is either 0.5*A + 0.5*2A, or it is 0.5*A + 0.5*(A/2). It can never be 0.5*2A + 0.5*(A/2)
- RHSeeger 4y ago
  Even after looking at those numbers, it still felt wrong to me. It eventually occurred to me that this was because my intuition was telling me the total should add up to A. And it doesn't; it adds up to 9/8 A. It took me a while to realize that my intuition was wrong, and that there's no reason the total should be A. The thing that helped me reconcile this was realizing that the A in the two equations are different values; then replacing them. (0.5⋅(0.5⋅A + 0.5⋅2.A)) + (0.5⋅(0.5⋅A + 0.5⋅0.5⋅A)) = 9/8 A ^ but the A in the left grouping (where it's 100) is different than the A in the right grouping (where it's 200). Replacing the As with their actual values 0.5⋅(0.5⋅100+0.5⋅2⋅100)+0.5⋅(0.5⋅200+0.5⋅0.5⋅200) = 150 And, since one envelope has 100 and the other has 200, an expected outcome across both envelopes is, as calculated, 150.
xmprt 4y ago
The breakdown for me was in step 4-5. > If A is the smaller amount, then the other envelope contains 2A. > If A is the larger amount, then the other envelope contains A/2. These are conditional probabilities so you can't simply add them up and compute an expected value of switching like they show in the example.

joe_the_user 4y ago

Well, the simple way to put is that "A" isn't fixed. The "expected value" argument in steps 6-7 is acting like "A" is a single value when "A" will be larger or smaller depending on what envelope you picked.

dllthomas 4y ago
I think that subtly misses the point. The problem is that you're implicitly using a distribution that... isn't a distribution. And with this particular not-a-distribution, whether you should switch or not does not depend on the value of A. But with any actual distribution (... I think?) it does, at which point... no paradox. It's true that it's not clear (at least to me, but perhaps more generally) what distribution we should assume, but as long as we avoid treating something that isn't a distribution as if it were one we avoid the worst of it.
- eru 4y ago
  Yes. The big problem is that you can't have a uniform distribution on the natural numbers. (And by extension, you can't have a uniform distribution on the rounded-to-integer version of a distribution on the real numbers.)
  - xigoi 4y ago
    Who said there is a uniform distribution on the natural numbers?
    - dllthomas 4y ago
      I think the assumption that "for all x, p(x) = p(2x)" implies a uniform distribution on the natural numbers. Imagine we have a distribution that satisfies that assumption, and then someone tells you they've sampled from that distribution and found that the result is of the form (say) 7*2^k, for some k > 0. That conditional distribution for k would seem to have to be uniform, right?
  - dllthomas 4y ago

thekiptxt 4y ago

This reminds me of how chess tactics feel so obvious when I’m on a tactics trainer, yet I can’t identify them in games where they actually happen. Or code that “obviously” has a bug only after it caused an issue in production. Would be nice if someone warned me before I do a PR: “ok, fyi there is a subtle but devastating bug in the new_feature.cpp”

maest 4y ago
This is a class of "hard to find, easy to verify" solutions.
na85 4y ago
Well if debugging is the process of removing bugs from code, then logically the act of writing the code must be called "bugging" and the probability of having a bug in any new piece of code asymptotically approaches 1 with increasing SLOC ;^)
- dbt00 4y ago
  Every program has bugs, every program has inefficiencies. Therefore every program can be eventually reduced down to a single instruction, which will be incorrect.
  - Archelaos 4y ago
    > Every program has bugs Where is the bug in this programm? END
    - stephencanon 4y ago
      What’s the specification?
      - the_sleaze9 4y ago
        It was supposed to return a generated image of Reverend Thomas Bayes in the style of stained glass.

deleted 4y ago

[deleted]