Yes, I had an algorithms class in college, although my recollection of it is pretty foggy (to say the least) now.

I gather that this is related to coming up with some sort of computer simulation of this problem? I think it might be very tricky to avoid inadvertantly programming some distortion into it.

Again, no longer trying to add confusion. I still believe this "switch or keep" problem is unsolvable because of the flawed, yet intuitively reasonable premise.

I think the following example might give you (us) some light (sorry my poor English).

Let's say we put X dollars in the Envelop A. Then we flip a coin, in order to determine if the envelope B will have 2X or X/2 dollars.

Now our outcome is:

KEEP THE ENVELOPE
50% X
25% 2X
25% X/2
EV: 1.125X

SWITCH THE ENVELOPES
25% X => 2X
25% X => X/2
25% 2X => X
25% X/2 => X
EV: 1.125X

Keeping or switching, we have the same EV (1.125X), and you have no idea of what to do (because you don't know if your envelope contains X, 2X or X/2, whatever the value of X).

That, I believe, was your mistake: assuming that you have X, and the other one is 50%2X or 50%X/2. But that's not the case.

Actually, your envelope, whatever its value, represents 1.125X -- that is (50%X + 25%X/2 + 25% 2X). Just like the other one.

QED? Does that make any sense?

Only God knows. But I think for now that's all folks :)

Moreso than anything else I've read. It seems to take care of the issue wikipedia mentioned that Sneaky asked about:

<Sneaky:
<1. Denote by A the amount in the selected envelope

7. So the expected value of the money in the other envelope is

(1/2)*(2A) + (1/2)*(A/2) = (5/4)*A

The most common way to explain the paradox is to observe that A isn't a constant in the expected value calculation, step 7 above. In the first term A is the smaller amount while in the second term A is the larger amount. To mix different instances of a variable or parameter in the same formula like this shouldn't be legitimate, so step 7 is thus the proposed cause of the paradox.>

<You're now heading into the murky land of approximations, and I'm not sure I want to follow. But I'll give one example: if we see that one envelope has 0.03 in it, then the other envelope might have either 0.01, 0.02, 0.05, 0.06 or 0.07 (each with varying degrees of probability). Do you really want to deal with such possibilities?>

If you let 2X > 0 (no upper boundary on 2X cq X) , then your problem will statistically totally float away.

The problem is a supposed/felt 'upper boundary' on 2X cq X ; look at <sentriclecub>'s posting:

<I for one would still "switch" if I opened $5, regardless of the "factor of increase/decrease" but we all agree that if you opened $100 million, its a sure-keep.>

Monte Carlo Methods:

Suppose you want to calculate the integral of a function via MCM; example:

function ... X^3

INTEGRAL[0..1](X^3)dX = 0.25 as we all know.

We take the square formed by X.[0..1], Y.[0..1] , throw at random darts at that square ( or generate random reals for X cq Y) and check if the so found points are above or below the function.

But you've forgotten why you had the upper bound in the first place.

Without an upper bound, you are left with the task of "randomly selecting a number from an infinite set". It is impossible. :-)

Maybe part (but only part) of our communication problem is that you are thinking of 'real numbers' in a computer sense, while I am thinking of them as they are defined mathematically.

Now from the obtuse problem:

<A closer look at Carroll’s solution reveals that he indeed made use of uniformity. However, he applied that assumption only to the interior of the region bounded by Figure ABDCE. The computed value of about 0.64 provides the correct answer to a different question.

The correct problem for Carroll’s solution would be:
Given two arbitrary points A and B, a triangle is constructed by choosing a third point at random from the region for which AB is the triangle’s longest side. What is the probability that this triangle is obtuse-angled?>

Do you see the difference in how A is defined in each case?>

I was sort of hoping for a DIFFERENT example. I mean, Wikipedia claims "this EV formula of the envelope problem is illegitimate". I neither agree nor disagree, but I make a simple request: "Give me an example of an illegitimate EV formula". And you come back with "The envelope problem." Feel my frustration! It's as if this rule about "mixing different instances of a variable or parameter in the same formula" was invented just for the envelope problem, and nothing else?!

Anyhow, to answer your question, sure I see the difference.

In case #1, I snipped out a bunch of what you wrote, on purpose, because it's all irrelevant for the purposes of defining A. You defined A as the the amount of money in the envelope that I'm holding. Period, end of sentence--that's all I needed to know. Then you added < I have another envelope which contains ... bla bla bla yadda yadda yadda>. No problem with having another envelope, you can have a thousand other envelopes, but that doesn't change the definition of A. You also add <You have the option to ... yadda yadda yadda>. Those are the rules of the game; you could have also listed major league baseball rules, but none of that changes the definition that my envelope contains A dollars.

In case #2 you define A identically in the sense that "my envelope contains A". However you also define A in terms of (X,2X). I suppose thats where the value of A starts to take on different characteristics.

And to make matters stranger, if you peek into the envelope and see a specific amount of money, for example $20, then it's no longer a variable. It's reminds me a little bit of Schrödinger's cat: when the observer takes a peek, the wave function collapses.

I think it can be done if there's a known upper limit for 2X (let's say 100.000 dollars)

There are two ways to go then; unawareness of the odd problem, or just using the centwise rounding on generated random reals.

I think both ways will reveal the exact amount in the first envelope, which makes it wise to change, and that the amounts will differ at most 1 cent.

Logically must follow that if the meant point is at $38254.47 for a 100.000 dollar maximum for 2X, then it will be $3825.45 for a 10.000 dollar maximum for 2X.

From your post, I got the impression that you didn't see why wikipedia described the 'A' to be "non-constant" in the EV formula: (.5)(A/2) + (.5)(2A) = 1.25A (or what I called case 2).

I hoped to at least illumninate the difference between a valid constant (case 1) and the invalid constant (i.e. non-constant) in case 2.

2X(max) = 100.000, generate a random 2X in that domain, with that a random X, and do the rounding.

Put 2X and X random in envelope 1 and 2.

Choose 'change point' at 10.000 dollar, i.e. choose to change if found amount in envelope 1 is below 10,000.

For this change point let the computer run for 10^9 times.

See the EV.

Repeat the whole experiment for 20,000 dollars 'change point'.

The maximum amount that might be found in either envelope is 100000, right?

Note then, then when we find a value of 50001 or higher, that it is NEVER a good idea to switch envelopes. The other envelope will always be half (within a penny), because doubling it would exceed the maximum of 100000.

with maximum of 100.000

<change point> 10.000 means if you find more than 10.000 dollars: DON'T CHANGE

If less : CHANGE

**************

<change point> 34.000 means if you find more than 34.000 dollars: DON'T CHANGE

If less : CHANGE

**************

The question is: what is the optimal <change point> ?

I predict that it will optimal to switch when you have values close to 0, and optimal to not change when you have values close to 100000 (actually, above 50000).

But this won't shine any light on the orginal problem. The original problem had no boundaries. Presumably, no matter what value you saw in your envelope, it has no bearing on whether you should expect the other envelope to contain half that amount or double that amount.

As in the obtuse problem too (btw, I've written a program for the given bounded cases in the obtuse problem too, and my results were equivalent to those given in that document).

My prediction is around 35.000 dollars.

maximum 2X 100,000 dollar (uniformly distributed)

maximum X 50,000 dollar (uniformly distributed)

minimum for both 0

X and 2X were put randomly in the envelopes.

<change point> if we find an amount of money below this point in envelope 1, we'll switch to envelope 2.

1,000,000 runs per <change point>

Results:

<change point> <expect> <switches>

0.0E+0000 ~~ 37434.28 ~~ 0
1.0E+0004 ~~ 37825.48 ~~ 150453
2.0E+0004 ~~ 38983.36 ~~ 299473
3.0E+0004 ~~ 40843.37 ~~ 450342
4.0E+0004 ~~ 43393.54 ~~ 600422
5.0E+0004 ~~ 46833.16 ~~ 749991
6.0E+0004 ~~ 45484.06 ~~ 799426
7.0E+0004 ~~ 43837.64 ~~ 849979
8.0E+0004 ~~ 41993.62 ~~ 900125
9.0E+0004 ~~ 39834.74 ~~ 950021
1.0E+0005 ~~ 37454.49 ~~ 1000000

As expected 'never change' (<change point> = 0) same expectation as 'always change' (<change point> = 100,000).

Suspicion: optimal <change point> = 50,000

A closer look:

<change point> <expect> <switches>

4.5E+0004 ~~ 44999.72 ~~ 676025
4.6E+0004 ~~ 45353.90 ~~ 690285
4.7E+0004 ~~ 45765.48 ~~ 704507
4.8E+0004 ~~ 46100.14 ~~ 720271
4.9E+0004 ~~ 46462.66 ~~ 734785
5.0E+0004 ~~ 46829.87 ~~ 749939
5.1E+0004 ~~ 46690.40 ~~ 755434
5.2E+0004 ~~ 46622.25 ~~ 759764
5.3E+0004 ~~ 46460.17 ~~ 765786
5.4E+0004 ~~ 46345.87 ~~ 769450
5.5E+0004 ~~ 46168.17 ~~ 775623

Make NO switch if envelope 1 contains $50,000 or more.

<change point> <expect> <switches>

1 <sentricle> = 400 dollars

Let's say that the casino version of the game works like this. The player pays some sum of money for the right to play the game. Then the croupier rolls a 6 sided die in secret. The croupier then consults a chart which tells him what sums to put in the envelopes:

1 = (1,2)
2 = (2,4)
3 = (4,8)
4 = (8,16)
5 = (16,32)
6 = (32,64)

The envelopes are then loaded and the player gets to pick one of them. The player then gets to peek in his envelope and choose if he wants to switch or not.

Clearly, the player should always switch if he finds $1 and never switch if he finds $64.

But what if he finds some other value? Say the player finds 16. He now knows that the dealer either rolled a 4 or a 5, and has no reason to believe that one possibility is more likely than the other. There is a 50% chance that the other envelope contains $8, and a 50% chance that the other envelope contains $32.

So if he happens to find $16 then it seems to me that it would perfectly profitable to switch. It fact, it would be foolish not to. The same thing can be said for any figure other than $64.

<Would YOU switch? What do you want, an automatic $16 or a chance to win either $8 or $32?>