Hacker News Board Hacker News, browsed like a classic message board
Live data from Hacker News

Two envelopes problem

en.wikipedia.org

Post reply on HN
Print view View original

31–40 of 96 posts

Re: Two envelopes problem

#31

Post by dools »

I read through the common resolution but I just can't get why you need all this. The problem seems to be at the start: you select one of two amounts, then you say the other amount is either double or half what you chose.

That means you have 3 amounts in the equation: 0.5x, x and 2x. But in reality there are only 2 amounts: x and 2x.

So you have to state the problem like: you choose an envelope. The other envelope either contains x or 2x. It can't contain 0.5x because that amount never existed in the first place.

Re: Two envelopes problem

#33

Post by jamii »

millstone wrote: post #30

This is one of my favorite problems. Here is one variation can help clarify thinking about it: "There's two envelopes. One contains twice as much money as the other. No envelope contains more than $N." This changes the problem dramatically. If your envelope contains more than $N/2, of course you should not switch: the other envelope necessarily contains less. If it contains $N/2 or less, perhaps you should switch: th…

Exchanging limits and expectations is not always valid (eg http://www.stanford.edu/class/msande322/docs/app_B.pdf‎). In this case the expected gain is not convergent in the limit. See https://news.ycombinator.com/item?id=6387344

This is one of my favourite problems too :)

Re: Two envelopes problem

#34

Post by colomon »

andrewflnr wrote: post #12
oneeyedpigeon wrote: post #6

Earlier quoted context omitted.

I don't have a BS in maths, but the important bit, as I read it, is that you don't know the values involved. I.e. if you have 20, there is either 10 in the other envelope, or 40. You have no way of knowing which is the case, so it's in your interest to swap since the benefits outweigh the risks. This is totally counterintuitive, though, so I'm fully willing to accept I'm missing something! And I don't buy the 'indefi…

Once you have the second envelope in hand, since you don't actually know what's in it, the exact same expected-value argument applies to switching back: there might be $10 or $40 in the other (first) envelope, "so it's in your interest to swap since the benefits outweigh the risk". While the second swap would reverse any advantage possibly gained, it would also reverse any possible harm sustained You dont know. The s…

Env A has $20 (we have it and know). Env B has $10 or $40 (equal chance).

We swap. Now we hold Env B, which has $10 or $40 in it. We know the other envelope (Env A) is $20. Why would we switch again?

Re: Two envelopes problem

#35

Post by olalonde »

jere wrote: post #19
EllaMentry wrote: post #4

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 b…

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

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.

Re: Two envelopes problem

#37

Post by ataggart »

jere wrote: post #19
EllaMentry wrote: post #4

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 b…

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

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.

Re: Two envelopes problem

#38

Post by sillysaurus2 »

I'd like to offer my own humble solution. Maybe it's flawed. Maybe you can embarass me. :)

Here it is: the goal is to choose a strategy which statistically maximizes our return. I.e. strategy A is superior to strategy B if it yields higher returns after, say, 1,000,000 iterations.

So there are two envelopes, X and 2X. You select one, then you're offered a chance to change your selection. What do you do?

Let's write out all the possibilities:

You select X, then you choose to stay, and wind up with X.

You select 2X, then you choose to stay, and wind up with 2X.

You select X, then you choose to swap, and wind up with 2X.

You select 2X, then you choose to swap, and wind up with X.

Those are the only four possibilities. You're forced to choose one of these possibilities randomly, because you have no information to guide your choice. Since two of them yield 2X and two yield X, and since your choice is necessarily random, then therefore all strategies will converge on the same expected value. In short, it doesn't matter what you do. You always have a 50% chance of X or 2X, regardless of your sequence of choices.

At first glance this is similar to the Monty hall problem, but the critical difference is that information is revealed during the Monty hall problem. No extra info is revealed here.

I assert that the envelopes could contain X and 1000X and it still doesn't matter what you do.

Ok, go, embarass me!

Re: Two envelopes problem

#39

Post by [deleted] »

bluecalm wrote: post #8
EllaMentry wrote: post #4

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 b…

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…

[deleted]

Re: Two envelopes problem

#40

Post by DoctorZeus »

dnautics wrote: post #25
DoctorZeus wrote: post #18

Earlier quoted context omitted.

Taking out the specific dollar amount doesn't change the math at all. If X designates the amount in the envelope I've selected, than 50% chance the other envelope contains .5 * X and a 50% chance it contains 2 * X, so the expected value of the other envelope is .5 * .5 * X + .5 * 2 * X = 1.25 * X which is greater than X.

but you are implicitly turning the scenario into one with three values, 1/2X, X, and 2X, so something went wrong with what you're doing. There was only ever a universe of two values.

Sure, these are variables, so they correspond to different possibilities. X can be anything, and given X, the amount in the other envelope is one of two different possible values - 1/2X OR 2X.

So there are far more than three possible values, but only two actual values.

Post reply on HN