A gambling game that creates money!

[efiusa]

If you did start with the premise that your envelope contained X to begin with, then you have a 50% chance of switching to 2X and 50% chance switching to .5X

Therefore, when you started with X and switched to 2X, the calc is .5X + .5(2X) = 1.5X

When you started with X and switched to .5X, the calc is
.5X + .5(.5X) = .75X

Since each of these have a 50% chance of happening,
.5 (1.5X + .75X) = 1.125X

I think you might want to get a new math teacher

[SheetWise]

[ QUOTE ]
If you did start with the premise that your envelope contained X to begin with, then you have a 50% chance of switching to 2X and 50% chance switching to .5X

Therefore, when you started with X and switched to 2X, the calc is .5X + .5(2X) = 1.5X

When you started with X and switched to .5X, the calc is
.5X + .5(.5X) = .75X

Since each of these have a 50% chance of happening,
.5 (1.5X + .75X) = 1.125X

I think you might want to get a new math teacher

[/ QUOTE ]

First of all -- puting all of the other absurd assumptions aside -- if the values are .5X and 2X the range is 1:4 -- which violates the original premise.

[betafemale]

This is the way I see it:

First of all, we know that one envelope contains twise as much money as the second one. So one envelope is X and the other 2X, the sum is always 3X (X + 2X). 50% of the time we would get X and 50% of the time we would get 2X. That makes our EV 1.5X (3X / 2) no matter if we swap or not. Also it does not matter if we watch what is in the envelope or not.

Hope that makes sense.

[mosta]

The ev is 1.5, not 1.25.

The mistake is that the problem is misstated when you make that argument. You have two quantities A and B, such that one is double the other. When you make the confusing argument, you make it in terms of three quantities, X, A, & B (an amount, half the amount and double the amount). This is FALSE. Stop, crumple and discard your sheet, and start over. There are not three quantities, only TWO. No variables. Money doesn't fluctuate. The envelops are sealed.

As soon as you list 3 quantities you are talking nonsense and need to start over.

The correct statement of the problem: there are TWO quantities A & B, with B being double A. You will either switch A for double A, or B for half B. (Don't say I have X and will switch for double X or half X. Say Either I have A and will switch for double A, or I have B and will switch for half B. Substituting to say you hvae "X" rather than "A or B" is the error.)

Add this up and EV is 1.5, switch or no switch.


edit: since I think there are a couple confused responses above, I'd better do the formal computation:

B = 2 * A

2 cases

i. I am given A and switch. Outcome is 2A.

ii. I am given B and switch. Outcome is A.

cases are equally likely.

therefore EV = .5*2A + .5*A = 1.5A.
voila.

--------------------------------------------------------------
equivalent different statement:

i. I am given A and switch. Outcome is B.

ii. I am given B and switch. Outcome is .5B.

EV = .5*B + .5*.5B = .75B.
B = 2A.
Therefore EV = .75*(2A) = 1.5A

again!
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lets do one more:

i. I am given A and switch. Outcome is 2A.

ii. I am given B and switch. Outcome is .5B

EV = .5*2A + .5*.5B = A + .25B.
B = 2A.
Therefore EV = A + .25*(2A) = 1.5A

yet again!!
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I don't deny (or belittle) the fact that this "paradox" is confusing, but really it is very, very, very trivial. Ie, there is nothing mathematically, or theoretically complicated about it at all. It is just a beguiling but simple misstatement of a problem.

If you get to peek into the first envelop, don't let that beguile you into saying there are three possibilities. Talk about two quantities only, A & B.

[pzhon]

[ QUOTE ]
Substituting to say you hvae "X" rather than "A or B" is the error.

[/ QUOTE ]
A common version of the envelope paradox lets you see that there is $100 in the first envelope. Are you saying that when you see the $100, it is an error to say that you have $100, and you must instead say that you have A or B? That is ridiculous.

[ QUOTE ]
Add this up and EV is 1.5, switch or no switch.

[/ QUOTE ]
How does this 1.5 (or 1.5 A) relate to the $100?

[ QUOTE ]

I don't deny (or belittle) the fact that this "paradox" is confusing, but really it is very, very, very trivial. Ie, there is nothing mathematically, or theoretically complicated about it at all.

[/ QUOTE ]
As a professional mathematician, I'm telling you that this is not trivial, and you have not resolved the paradox, although others have (see my past posts). You have made some simple, correct statements. (Actually, you were sloppy, and your statements could easily be interpreted as applying to a context in which they are wrong, as is seen by the above questions.) However, you haven't come close to describing the complexity of the situation, or the real but subtle mistake made in one side of the paradox.

[mosta]

I'll look for your posts later. I am interested. I'll just say that I've heard people try to "solve" this problem with arguments about the finitude of money or the nature of the continuum or something similarly esoteric, and I have found that less than compelling.

Let me just say that as far as my post went I knew when I was writing it that I could have written it better (like many, having done this before). Instead of A and B, I should have said X and 2X.

I think it's a mistake to say, this is X, that is 1/2 X or 2 X.

I think it's correct to say, this envelop (I peek) is X or 2 X.
If it is X and I switch..., if it is 2X... . Therefore EV = 1.5 X.

And it seems to me to be a simple contradiction when one argues thus:
1. there are two envelops A and B.
2. I see the first envelop, X, therefore the other envelop is 1/2 X or 2 X.

The first statement has two quantities and remains true. The second has 3 quantities and is inconsistent.

But that doesn't affect anything you said. No time to search, or read
Wiki, right now.

[mosta]

Well I read the wiki and I'm not going to press the issue. I pretty much agree with the paper in n. 20 and think I've been in line with it at least loosely. (though I hate to pick a philosopher to rest on.) I find it very hard to accept that you need more than 8th grade math to be able to avoid this paradox.

you could say you're changing the terms of the problem after you start, to something inconsistent with the starting point (we could have a game where you pick a number and I will double it or halve it--and this is *not* that game).

or you could say that you are expressing the value of Env 2 as a function of Env 1 and that is simply untrue. I wonder if you can make an argument examining a game of picking between $A and ($A + or - $100) and then apply that analysis by analogy.

when you know that the envelops contain X and 2X and then you look in the first envelop, that tells you nothing and changes nothing. any reasoning based on that information has to be misguided. it tells you nothing about whether what you saw is X or 2X. you can't blend the two situations into one analysis b/c they are inconsistent--the involve 3 values instead of 2.

[Reckless1der]

I saw this a couple of years ago on Mike Caro's webpage. He called it the Johnny Moss paradox. The flaw as I see it is that there are only 2 envelopes, not 3. You were 50/50 to pick the bigger envelope, the first time and 50/50 if you switch. It is EV neutral.

[pzhon]

[ QUOTE ]
I saw this a couple of years ago on Mike Caro's webpage.

[/ QUOTE ]
Caro's article is mostly correct, although not very clear.

It is wrong where he says the following: "Since you don't know what range Johnny Moss decided on for the checks (but that obviously he had to use some range), it does you no good (or no harm) to switch. If you had information letting you know when you were at the high or low end, then you could beat the system by always switching when low, never switching when high. But you don't have this information... look no further."

It is as though Caro said calling is the same as folding, so it doesn't matter, and you don't need to think about it. However, raising may be better than calling or folding.

What Caro argues is that always switching is worth just as much as always staying. However, he doesn't consider the possibility that you can profitably switch some of the time, and stay some of the time. In fact, you can set a threshold t, and switch if you see less than t, and stay if you see more than t. If both amounts are above t, or if both amounts ae below t, then you break even on average compared with either constant strategy. However, you improve when t is between the envelope amounts. You can make this strategy slightly more complicated (letting t vary randomly, or switching with probability p(amount seen) with p(x) decreasing) to do strictly better than always switching or always not switching whenever the expected value of either is finite.

This has been covered in past threads.

[greg nice]

ill rich is right