[EllaMentry]

That is a very long article based on flawed argument.

Given no other information, assuming someone gave you 2 envelopes and told you one has $40 vs $20, common sense dictates choose 1 randomly and walk away - with no other information it is illogical to reason any other way.

The chance you choose the lower value is 1/2.

Now, if you are allowed to look inside the envelope (which gets introduced further down) then it becomes a different game.

Get $20...well by swapping you may get $10 or $40...you should probably swap.

Get $2000...well by swapping you may get $1000 or $4000...you should probably swap.

I think this works all the way up...someone with a bit more background on game theory may be able to formalise it, but the realisation that swapping forever leads to $0 nullifies this "paradox"

[bluecalm]

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 money in "adversary" disposal). The assumption is then used for conditional probability calculations: "if we see 10$ there is 50% chance the other envelope contains 20$" - BEEP, ERROR, THINK AGAIN.

Perhaps good exercise in clear thinking but not really a paradox. Good analogy is this: "If we pick random building and climb to the roof of it there is 50% chance first building we see is higher than the one we just climbed". This is obviously true, now following "paradoxical" reasoning we get: "If we climb a building randomly and see it's the Empire State Building there is still 50% chance first building we see will be higher".

This is exact analogy to reasoning about 2 envelopes problem which is supposed to lead to a paradox.

[jamii]

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.

[bluecalm]

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.

[jamii]

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-...

[thatthatis]

Code that shows ev stays at 3x/2 (if x is the lower amount) or 3n/4 (if n is the higher amount)

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

[jere]

Common sense in the Monty Hall problem says just pick a door and stick with it. The chance you chose the goat is 1/2, right?

No. Common sense is often wrong.

http://en.wikipedia.org/wiki/Monty_Hall_problem

[ataggart]

Except that in the Monty Hall problem something has changed after the initial choice: one of the previous options is proven to be a goat. No such change occurs with the two envelopes, no new information is made available, thus no basis for switching.

[olalonde]

There is additional information in the Monty Hall problem: the host never opens the door with a car. In the two envelope's problem, the "host" merely restates the question no matter which envelope was chosen.

[jere]

Folks, it's an analogy. An analogy compares two different things... trust me, I know they're different and I know in which ways.

All I was saying is common sense doesn't get you far in the Monty Hall problem. It really doesn't:

>[Vos Savant] received thousands of letters from her readers; 92% of the general public, 65% of universities, and many with PhDs, were against her answer.

Thus, it's kind of silly to say we're wasting time by going past a common sense analysis of something. It's also revisionist to say common sense does help you solve Monty Hall.

[olalonde]

Sorry for misunderstanding you. I guess I lacked a bit of common sense...

[thatthatis]

The montey hall problem can be restated in a way that makes it amenable to common sense. Door A or (Door B and Door C).

Or, restate the problem with 100 doors.

I'm not sure every problem like this one can be reduced to a common sense analog, but I'm suspect of problems that can't be.

[RockyMcNuts]

Suppose you restate it. You own a $100 stock. It has a 50% chance of doubling, and a 50% chance of going down 50%.

Should you sell it, or hold onto it? Expected value of holding is ($200 + $50)/2 = $125 . So it seems like you should hold on!

If you repeat that wager indefinitely, the standard deviation of the net wins (number of ups-downs) goes up according to a square root law.

And the value of your stock goes up exponentially as the number of net wins goes up.

So over time the EV goes to infinity like (2 ^ (n/2)). The EV of say the -1SD outcome goes to zero. The EV of the +1SD outcome goes to 1/that. So the average EV overall goes to infinity.

And yet the expected value of the growth rate is 0. For every 16x win there's a loss down to 1/16th. But the average of those 2 outcomes diverges to infinity.

All this to say, when you're looking at exponential returns (or other processes), you need to measure growth rates, not average outcomes. And it has nothing to do with log utility.

[olalonde]

Javascript simulation (might freeze your browser so would be a good idea to run it in Node):

    var t = 100; for (var i = 0; i < 1e6; i++) { if (Math.random() > 0.5) { t = t * 2; } else { t = t / 2; } console.log(t); }

It's interesting how quickly the return goes from extremely small to extremely big amounts.

[RockyMcNuts]

yeah. Now I wonder what happens when you have $10,000, and always put 1% of your portfolio in that stock? I think that turns it into a positive growth expectation. You do it 10 times and outcomes are 50/50, 5 times you made about $100, 5 times you lost about $50.

On the other hand, if you bet your whole stack each time, you doubled up 5 times and lost half your stack five times, you're even.

if you bet your whole portfolio every time, I think your long run growth rate is 0. if you bet a small amount each time, I think your growth rate is positive.

The Kelly Criterion or gambler's curse in action. In the first case you're taking a positive EV bet and turning into a long run no-growth situation by overbetting.