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Two envelopes problem

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Re: Two envelopes problem

#71

Post by jamii »

bluecalm wrote: post #61
jamii wrote: post #23

Earlier quoted context omitted.

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.

Not so much that it leads to nonsense as that naively applying expectations doesn't always work. This is a contrived example, but it's not uncommon in eg random walk theory to hit upon cases like this where the expectation does not exist at all.

People commonly think of mathematics as being purely about formal proof but the reality is an interplay between proof and intuition. Usually when a mathematician encounters a problem in a familiar area they immediately know the answer by intuition which then guides the production of a correct proof. When you first enter a new area of mathematics your intuitions are all completely wrong and you have no idea where to start with a proof. Good teachers will introduce edge cases like this problem to refine your intuition until it is useful enough to be a guide.

http://terrytao.wordpress.com/career-advice/there%E2%80%99s-...

Re: Two envelopes problem

#72

Post by russellsprouts »

cookingrobot wrote: post #43

Here's how I see it - the important unstated variable is how big a budget does the host of the game have? If you know that you're playing with a billionaire who loves this party trick, and is probably willing to give away a million dollars on this, then the $20 you found in you envelope is probably the smaller value. If your envelope contains $800,000 it's probably the bigger one. The host has chosen a number randoml…

However, it says in the problem that you are offered the switch before you open the envelope. So, you can't know if your original choice was $20 or $80,000.

Re: Two envelopes problem

#73

Post by bluecalm »

jamii wrote: post #67
bluecalm wrote: post #64

Earlier quoted context omitted.

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.

Well, there are distributions which don't have well defined expected value and I agree that it's valuable lesson. I would argue though that this is not the best example to show it. The point of two envelope problem comes much earlier and specifically giving distribution defined by you would make it completely different problem and assuming it in original form would already be a mistake.

I think there are better examples to show how infinite sums and relying on expected value based on those might leads to problems. Like this one for example:

http://plato.stanford.edu/entries/paradox-stpetersburg/

Re: Two envelopes problem

#74

Post by [deleted] »

bluecalm wrote: post #66
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 p…

[deleted]

Re: Two envelopes problem

#75

Post by jamii »

harryh wrote: post #24
jheriko wrote: post #21

the flaw here is assuming that the value being higher on average in the envelope is a sufficient criteria to make swapping the best choice. this is not how probability works, if i have a million empty envelopes and one with a billion dollars in it, the average value of each envelope being £1000 doesn't mean anything in the face of the 1 in a million chance of actually picking the right one.

Actually, that's exactly how expected value works.

I think the point he is getting at is that expected value (or even expected utility) is not actually a good measure of human preferences.

Additionally, in this problem there are certain distributions over the amounts in the envelope for which the expected value is not defined, making it an interesting problem to study.

http://en.wikipedia.org/wiki/Thinking,_Fast_and_Slow is a good book exploring mathematical models for decision making.

Re: Two envelopes problem

#76

Post by jamii »

bluecalm wrote: post #73
jamii wrote: post #67

Earlier quoted context omitted.

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.

Well, there are distributions which don't have well defined expected value and I agree that it's valuable lesson. I would argue though that this is not the best example to show it. The point of two envelope problem comes much earlier and specifically giving distribution defined by you would make it completely different problem and assuming it in original form would already be a mistake. I think there are better examp…

The expected value in the St Petersburg game is actually well-defined - it converges to positive infinity. That is a subtly different result than not converging at all. Both problems are useful.

Re: Two envelopes problem

#77

Post by pierrebai »

The paradox is interesting to discover and discuss errors in reasoning, but we should agree from the start that any proof that one should always switch (or that any behaviour could increase the expected gain) is flawed.

This can be easily proven by:

1. Taking at face value that choosing an envelope is random. 2. Adding a second player that is given the other envelope.

In that scenario, any reasoning that leads one player to always switch would apply to the other player too. Both players would swap their envelopes. Clearly, the expected value for both cannot go up. Thus any reasoning that mandates switching /has/ to be wrong.

(If you have issue with adding another player, instead assume a parallel universe where the player chose the other envelope.)

The error in reasoning is usually to assume that the envelope contains X and then reasoning about X/2 and 2X. In reality, the envelopes always contain X and 2X. When you actually use the parallel universe version, clearly one has X and the otehr has 2X. Nobody has X/2 (nor 4X, nor any other expected value your theory may come up with).

Re: Two envelopes problem

#78

Post by scythe »

jamii wrote: post #65
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 w…

In my original version of the post I had intended to argue for a geometric mean, which was sort of a joke, as it only works where f(x) is a constant k*x. So the statement "A false resolution follows, first to spot the problem..." appears in the post. Upon seeing this an enterprising and intelligent person may have clicked "reply"

While writing this I realized that inverting the definition of A gave a different resolution based loosely on operator theory, and posted that instead. A couple minutes later I realized the mistake and changed the beginning to refer to the first to spot "a" problem.

The argument you've presented is good, but you have attacked the well-foundedness of the problem itself rather than the logic I've applied. I'm not sure if you get the cookie.

Re: Two envelopes problem

#79

Post by jamii »

staccatomeasure wrote: post #68
jamii wrote: post #51

Earlier quoted context omitted.

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.

Well, with the distribution I gave it could be $2 (with probability 1/2), or $4 (with probability 1/4), or $8 (with probability 1/8) etc. That distribution doesn't have a maximum.

For a simpler example, imagine tossing a coin and counting the number of tosses it takes you to get a head. There is no limit on the number of tosses it might take, it just gets less and less likely as the numbers get bigger. We could get any positive integer and there are an infinite number of integers.

Similarly, the envelope problem has an infinite number of possibilities because there is no maximum amount that could be in the envelope. The article doesn't specify the exact odds of any given amount showing up in the envelope so I gave some specific odds that demonstrate the problem.

Probability theory still works when you have an infinite number of possibilities but it has a few subtleties that aren't commonly taught until undergrad. One of those is that the expected/average value of a random variable doesn't always exist so you have to be careful when reasoning informally.

Re: Two envelopes problem

#80

Post by thatthatis »

jamii wrote: post #58
thatthatis wrote: post #54

If it is possible to switch and gain on the average we could write a 10,000 or so iteration monte-carlo simulation of this switching algorithm to demonstrate the gains. The envelope has either x or 2x. One envelope has amount z, but the agent doesn't know if z = x or z = 2x. If it is true that not switching leads to an EV of 3x/2 but switching leads to a higher EV: then a single switch each round should come back wit…

Go ahead and run a few million simulations of the problem and graph the running mean over time. You might be surprised :) Be sure to use exact arithmetic (eg http://docs.python.org/2/library/fractions.html )

https://gist.github.com/tedtieken/6567112

Run the simulator and you'll see, the ev is 3x/2, regardless of switching behavior.

Feel free to modify if you think I've innacurrately conceptualizer the problem.

Also, please excuse some non pythonic names, I'm writing the code on an iPhone.

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