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Two envelopes problem

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Re: Two envelopes problem

#41

Post by Afforess »

sillysaurus2 wrote: post #38

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 o…

No. If I understand correctly, your last two outcomes are wrong. There are 6 possible outcomes, not four. This is because when you choose the 2X correctly, you may receive 2(2X) or X.

The 6 outcomes are:

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 choose to swap and wind up with 1/2X

You select X, them choose to swap and wind up with 2x

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

You select 2X, then choose to swap and wind up with 4X

Re: Two envelopes problem

#42

Post by Osmium »

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 other is 2x). I think it's misleading to introduce this "third" quantity of 0.5x... but perhaps I'm missing something subtle, in which case I'd be happy to be corrected!

Re: Two envelopes problem

#43

Post by cookingrobot »

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 randomly between 1-N, and they're not neccessarily going to tell you what N is. But in the long run you'll find that values closest to N/2 are more common than values further from N/2.

Re: Two envelopes problem

#44

Post by sillysaurus2 »

Afforess wrote: post #41
sillysaurus2 wrote: post #38

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 o…

No. If I understand correctly, your last two outcomes are wrong. There are 6 possible outcomes, not four. This is because when you choose the 2X correctly, you may receive 2(2X) or X. The 6 outcomes are: 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 choose to swap and wind up with 1/2X You select X, them choose to swap and wi…

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.

Re: Two envelopes problem

#45

Post by [deleted] »

Afforess wrote: post #41
sillysaurus2 wrote: post #38

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 o…

No. If I understand correctly, your last two outcomes are wrong. There are 6 possible outcomes, not four. This is because when you choose the 2X correctly, you may receive 2(2X) or X. The 6 outcomes are: 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 choose to swap and wind up with 1/2X You select X, them choose to swap and wi…

[deleted]

Re: Two envelopes problem

#46

Post by dnautics »

DoctorZeus wrote: post #40
dnautics wrote: post #25

Earlier quoted context omitted.

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.

Exactly, that's why your model is broken.

Re: Two envelopes problem

#47

Post by egreif1 »

This is a really cool problem. I don't claim to understand it, but here's one way of thinking about it. The paradox assumes that when we see the value of money in the envelope we open, we get no information about whether that's the smaller or larger amount of money. Now, in practice, is that possible?

Let X,Y be i.i.d. draws from some distribution with CDF F, and say we see Y=y when we open our envelope. If the distribution satisfies the constraints of the problem, then we have to have that, for all possible values of y: P(X > Y | Y = y) = 1/2 = P(X In practice, this is impossible. The reason why this is impossible is because P(X>Y | Y = y) = (1-F(y)). Since F(y) is a CDF, if we take the limit as y goes to infinity we have that (1-F(y)) = 1. So (1-F(y)) can't equal 1/2 regardless of y. That is, the probability that the other envelope contains the larger or smaller value is not independent of the value we observe in the envelope we open. And, more importantly, it can't be, such a distribution doesn't exist. You have to consider the sampling distribution to get a meaningful calculation.

Re: Two envelopes problem

#48

Post by jere »

olalonde wrote: post #35
jere wrote: post #19

Earlier quoted context omitted.

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.

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.

Re: Two envelopes problem

#49

Post by robotresearcher »

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.

Thanks, that is a wonderful way to put it. The swaps are not independent in their effects. That satisfies me.

Re: Two envelopes problem

#50

Post by iamshs »

sillysaurus2 wrote: post #44
Afforess wrote: post #41

Earlier quoted context omitted.

No. If I understand correctly, your last two outcomes are wrong. There are 6 possible outcomes, not four. This is because when you choose the 2X correctly, you may receive 2(2X) or X. The 6 outcomes are: 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 choose to swap and wind up with 1/2X You select X, them choose to swap and wi…

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.
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