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Two envelopes problem

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Re: Two envelopes problem

#91

Post by delinka »

iamshs wrote: post #50
sillysaurus2 wrote: post #44

Earlier quoted context omitted.

Actually, you always wind up with X or 2X, never 0.5X and never 4X. Remember, the envelopes contain either X or 2X, and you always end up with either one or the other.

I myself understood the problem with Afforess's reply. You are explicitly pinning the value in the envelopes, but remember the other envelope has either half or double of the value than the one in your hand, and this is the crux of this paradox. Hopefully, I am thinking in correct terms.

But he has assigned the identifiers "X" and "2X" to the envelopes. He could have called them "A" and "B" and his logic still holds. The problem I see with the wiki article explanation is: while using actual dollar amounts as examples, it gives the impression that you get to open the envelope that you picked first to see the amount, but that you don't know if it's the smaller value or larger value.

So I choose "A" and I don't know if it contains $X or $2X. I don't even get to open it to know what amount is in the envelope. Whatever it contains, if I choose to switch, I get envelope "B" - I always have a 50% chance of choosing the larger amount or switching to the envelope with the larger amount because no information is revealed after the first choice.

for A->$X, B->$2X:

    Choose A, stay with A, receive $X
    Choose A, switch to B, receive $2X
    Choose B, switch to A, receive $X
    Choose A, stay with B, receive $2X
for A->$2X, B->$X:

    Choose A, stay with A, receive $2X
    Choose A, switch to B, receive $X
    Choose B, switch to A, receive $2X
    Choose A, stay with B, receive $X
50% chance of getting either amount, regardless.

Re: Two envelopes problem

#92

Post by RockyMcNuts »

olalonde wrote: post #52
RockyMcNuts wrote: post #27

Earlier quoted context omitted.

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 ex…

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 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.

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.

Re: Two envelopes problem

#93

Post by dools »

Osmium wrote: post #42
dools wrote: post #31

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 ei…

Unless you picked 2x to start with, in which case the other envelope does contain half of what you chose... i.e. there are two envelopes, x and 2x, and you choose one. There is a 50% chance that the envelope you didn't choose is half the value of the one you picked (you pick 2x, and the other is x), or there's a 50% chance the envelope you didn't choose is twice the value of the one you picked (you pick x, and the ot…

I think it's misleading to introduce this "third" quantity of 0.5x

Yes, it is misleading, that's my point :)

Put it another way: you have two quantities in two envelopes, A and B. If you choose B, what is the probability that the other envelope contains a 3rd quantity, C? The probably of that is zero, because C was not present in the initial two envelopes.

Still another way of putting it: I will place $10 in an envelope, and $20 in another envelope, then ask you to choose one. What are the chances that the other envelope contains $40? The chance of that is zero, because the only two amounts present at the start of the trick were $10 and $20.

Re: Two envelopes problem

#94

Post by dools »

jader201 wrote: post #11

The paradox depends on the possibility of always being able to double the amount. This isn't true. Swapping once will either double or half the amount; swapping back will just do the opposite.

Even better: if I put $10 into one envelope, and $20 in another envelope and get you to choose one, what is the probability that the other envelope contains $40? Zero.

Re: Two envelopes problem

#95

Post by thatthatis »

jere wrote: post #48
olalonde wrote: post #35

Earlier quoted context omitted.

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.

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…

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.

Re: Two envelopes problem

#96

Post by bluecalm »

thatthatis wrote: post #87
bluecalm wrote: post #66

Earlier quoted context omitted.

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…

If the new information is relevant, how is this simulation code wrong? https://gist.github.com/tedtieken/6567112

The code is wrong because there is an assumption that the distribution is 50% for 10$ and 50% for 20$. There no basis for this (how do you know you won't got 40$ if you see 20$ ?). See my other posts, I think this point is very well worth thinking about.
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