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Two envelopes problem

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Re: Two envelopes problem

#61

Post by bluecalm »

jamii wrote: post #23
bluecalm wrote: post #8

Earlier quoted context omitted.

Yes, it's not a paradox it's just seductive flawed reasoning. Yes, at any point EV of picking an envelope at random is 3/4n (n being higher amount of money out of the two). It is all there is to it. The "paradox" is introduced by silent assumption that distribution of amounts put in envelopes is uniform which is impossible (because you can't pick numbers from infite set uniformly even if there was infinite amount of…

You can explicitly state the distribution and still run into the same problem: https://news.ycombinator.com/item?id=6387344 . The underlying problem is basically that probability theory in non-finite spaces has some gotchas - one of which is that the expectation of a random variable does not always exist.

Interesting point and nice read. Still the problem is in assumption about underlying distribution of amounts in envelopes (in original case impossible uniform distribution). The reasoning is based on this assumption and leads to nonsense. What you are saying (I think) is that assuming some other distribution (possible one, instead of impossible one) could still lead to nonsense or doesn't lead anywhere at all.

Re: Two envelopes problem

#62

Post by tmoertel »

Arbitrarily label the two envelopes A and B. Assume that no envelope is empty. Let «X» denote the amount of money in envelope X.

We are told that either «A» = 2«B» or «B» = 2«A». Because these propositions are exhaustive and mutually exclusive, we know that

    P(«A» = 2«B») + P(«B» = 2«A») = 1.      (1)
Since we have no knowledge that would make either proposition more plausible than the other, we also have (by symmetry) that

    P(«A» = 2«B») = P(«B» = 2«A»).          (2)
From (1) and (2) we can solve for the individual probabilities of the propositions:

    P(«A» = 2«B») = P(«B» = 2«A») = 1/2.    (3)
Therefore, from the initial conditions, we have no reason to prefer either envelope.

Now we are given the information that we have picked one of the envelopes at random (let's say it's A). We are further given the information that A contains $20, that is «A» = $20. How does this new knowledge affect the probabilities?

It has no effect because we can't say anything more about either proposition without also knowing «B» as well, and we don't know it. That is,

    P(«A» = 2«B» | «A» = $20) = P(«A» = 2«B»)
and

    P(«B» = 2«A» | «A» = $20) = P(«B» = 2«A»).
Therefore, our probability assignments from (3) remain unchanged, and we have no reason to prefer one envelope to the other, let alone swap A for B.

Re: Two envelopes problem

#63

Post by thatthatis »

Here's my contribution:

-----------------

Approach 1: Absent new information, we cannot improve our outcomes.

In the montey hall problem there is either obscure new information, or a obscure change in the rules between firs choice and second choice.

Montey hall collapses to an initial choice of the prize behind door A or the prizes behind both door B and C. When the true, collapsed, choice is revealed the common sense reasoning is correct.

In this problem there is no new usable information.

In the two envelopes problem the new information appears relevant but is actually not on it's own any more useful to reasoning about expected value than knowing that there is a red or green piece of paper in either envelope.

-----------------

Approach 2: Keeping the quantities symbolic.

There are two envelops with quantities x and y inside. We are told that 2x = y. After choosing it is revealed that our envelope has quantity z. It is not revealed if z=x or z=y.

Let's consider the universe of possibilities.

Possibility one (50% chance): z = y. Value of switching: -x

Possibility two (50% chance): z = x. Value of switching: +x

Therefore the value or switching is: .5 * -x + .5* +x = +/-0

The red herring here is that knowing value z feels like it is information about values x and y, but it isn't.

The slight of hand is in trying to say that the other envelope is worth either .5z or 2z. This is false because there is an unknown but fixed universal constant variable x. We don't know from the available information if we are in universe z=x or universe z=y.

In short: The two envelopes paradox mistakes an unknown constant for an unknown variable. Knowing that z = 20 doesn't change the universal constants x and y.

Re: Two envelopes problem

#64

Post by bluecalm »

jamii wrote: post #22

You can even explicitly define the distribution as P($2^i) = 1/(2^i) and the 'paradox' remains. The problem is that you are reasoning about the expected gain of swapping but that expectation is a sum over a series which is not absolutely convergent ( http://en.wikipedia.org/wiki/Absolute_convergence ) so it's value depends on the order in which you sum over the different cases. In fact, if you repeatedly simulate the…

Well, yeah :) The problem is still in assuming something about the distribution without having any good reason to. In original "paradox" reasoning it was impossible uniform distribution. What you show is that even assuming possible distribution could still lead nowhere. It still doesn't mean there is any reason to assume it's P($2^i) = 1/(2^i). I may just as well assume it's exactly 100$-200$ and that switching from 100$ gives me guaranteed payoff. It would be as baseless as assuming uniform distribution or the one given by you.

I think the main lesson from this puzzle is that you can't assume stuff without good reason to.

Re: Two envelopes problem

#65

Post by jamii »

scythe wrote: post #55

This is a fantastic and subtle paradox, and not at all taken to quick resolution. A resolution follows, first to spot a problem (if there is one) gets a cookie. Consider the amount in the lower to be f(x), the higher to be x, given f(z) such that f(z) all distributions, as the problem clearly applies to all distributions. You open the first envelope, which contains A. The second envelope contains B. The challenge is…

The problem here is that you are adding together two values for a specific case of x and then adding together the result for all values of x (is that clear? probably not).

It's analogous to another age old problem: what is the value of 1 + (-1) + 1 + (-1) + 1 + (-1) ...

You could argue that (1 + -1) + (1 + -1) ... = 0 + 0 ... = 0.

You could also argue that 1 + (-1 + 1) + (-1 + 1) = 1 + 0 ... = 1.

There are actually way to group the numbers in that series to get any integer answer you want :D

Since we haven't actually defined a distribution over x the problem is not well-defined anyway. But lets pick, say, P(x=2^i) = 1/(2^i) for all i>1.

Then the expected value of A is

    E(A) = (1/2 * $2) + (1/4 * $4) + (1/8 * $8) ...
         = +infinity.
Similary E(B) = +infinity.

Now for E(B-A)

    E(B-A) = (1/2 * $1) + (1/2 * -$1) + (1/4 * $2) + (1/4 * -$2) ...
Like the example above, we can add this up in different ways:

    E(B-A) = ((1/2 * $1) + (1/2 * -$1)) + ((1/4 * $2) + (1/4 * -$2)) ...
           = $0 + $0 ...
           = $0
Or

    E(B-A) = (1/2 * $1) + ((1/2 * -$1) + (1/4 * $2)) + ((1/4 * -$2) ...
           = $0.5 + $0 + $0 ...
           = $0.5
When adding up infinite numbers of things you have to be very careful. I go into more detail in this thread - https://news.ycombinator.com/item?id=6387344

Cookie please :D

Re: Two envelopes problem

#66

Post by bluecalm »

thatthatis wrote: post #63

Here's my contribution: ----------------- Approach 1: Absent new information, we cannot improve our outcomes. In the montey hall problem there is either obscure new information, or a obscure change in the rules between firs choice and second choice. Montey hall collapses to an initial choice of the prize behind door A or the prizes behind both door B and C. When the true, collapsed, choice is revealed the common sens…

Approach 1: There is in fact new information when you look into the envelope it's also very valuable because it allows you make judgement taking into account your knowledge about the world and specific situation (who puts money in the envelopes, what are general preferences of people in such situations etc.).

Approach 2: You made the same mistake. Seeing the money is actually valuable and very real information. The problem is that original reasoning makes wrong use of it. It doesn't mean there is no information or that we can/should ignore it.

Re: Two envelopes problem

#67

Post by jamii »

bluecalm wrote: post #64
jamii wrote: post #22

You can even explicitly define the distribution as P($2^i) = 1/(2^i) and the 'paradox' remains. The problem is that you are reasoning about the expected gain of swapping but that expectation is a sum over a series which is not absolutely convergent ( http://en.wikipedia.org/wiki/Absolute_convergence ) so it's value depends on the order in which you sum over the different cases. In fact, if you repeatedly simulate the…

Well, yeah :) The problem is still in assuming something about the distribution without having any good reason to. In original "paradox" reasoning it was impossible uniform distribution. What you show is that even assuming possible distribution could still lead nowhere. It still doesn't mean there is any reason to assume it's P($2^i) = 1/(2^i). I may just as well assume it's exactly 100$-200$ and that switching from…

I'm not assuming that that is the distribution. I'm saying that the problem as presented is underspecified, but even if you give this specific distribution in the problem you can still cause confusion. Assuming stuff is bad but the original 'paradox' still doesn't go away if you nail everything down. It's a useful problem for education people about the subtleties of infinite sums.

Re: Two envelopes problem

#68

Post by staccatomeasure »

jamii wrote: post #51
staccatomeasure wrote: post #36

Earlier quoted context omitted.

No, quite the opposite unfortunately :(

I guess I skipped a lot of background knowledge :) The quantity everyone is arguing about is 'the expected gain from swapping envelopes' ( http://en.wikipedia.org/wiki/Expected_value ). Informally, you might say 'the average gain from swapping envelopes'. The way you work this out is you take all the possible things that could happen and for each one multiply the probability of it happening by the amount you gain if…

Sorry but right after you finish the basic probability coverage it stops making sense.

-> In the case of the two envelopes the problem comes from the fact that there are an infinite number of possible amounts in the envelope.

Right there.

Re: Two envelopes problem

#69

Post by russellsprouts »

Correct me if I'm wrong, but here's my analysis.

There are two envelopes. One contains a value x, and the other has 2x. You pick an envelope at random. Let's call the envelope you choose z. The variable z is a random variable that has an equal chance of being x or 2x, so the expected value of z is 1.5x.

If you switch, there are two possibilities. In one case, you picked x, so switching gains you .5x over the expected value of 1.5x.

In the other case, you picked 2x, so switching loses you .5x from the expected value of 1.5x. So, the expected gain from switching should be zero.

Re: Two envelopes problem

#70

Post by bluecalm »

Many explanations here contain the same mistake: Some reasoning -->> seeing the money doesn't change anything and/or doesn't count as information.

If you arrive at this point you've made a mistake. Seeing the money is valuable information. The trick is not in ignoring it but making good use of it like this:

-use your best knowledge about person who puts money in the envelope, general human tendencies and how the world is to approximate distribution of money in the envelopes

-pick according to your personal utility of money assuming the distribution

What the the wrong reasoning in original puzzle suggests is to assume uniform distribution which is impossible but more importantly baseless. As other poster shows there are baseless distributions you could assume which leads you nowhere as well.

Just don't let it detract from the main point: there is new information but making use of it is not that easy.

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