Two envelopes problem

[Andrey_Dmitriev]

Colleagues, I would like to discuss with you "Two Envelopes Problem"

In few words the problem is quite simple:

"...The player is given two indistinguishable envelopes, each of which contains a positive sum of money. One envelope contains twice as much as the other. The player may select one envelope and keep whatever amount it contains, but upon selection, is offered the possibility to take the other envelope instead..."

For example, €10 is the amount in the selected envelope.
The other envelope may contain either €20 or €5
If €10 is the smaller amount, the other envelope contains €20 with probability 0.5
If €10 is the larger amount, the other envelope contains €5 also with probability 0.5
So the "expected" value of the money in the other envelope is: 0.5*€20 + 0.5*€5 = €12.5
This is greater than €10, so swapping is always favored...

More detailed description in Wikipedia (click).

My quick and simple LabVIEW simulation shows that its not true (my human logic say this too as well)...

Probably someone else will be able to provide small LabVIEW snipped which will show this problem simulated in action?

with best regards, 

Andrey.

Message Edited by Andrey Dmitriev on 08-24-2009 03:37 PM

[Neil.Pate]

This is similar to the three doors game, same thing but there are three choices.

In that instance it is always better to swap your guess, but this can be proved conclusively (unlike your example!)

As the wiki article suggests, this is a bit of a paradox!

My take on this is that the probability is 50% as even if you change your mind 100 times you still only have two possibilities (unlike the 3 doors problem where one of the unknowns is removed before you make your guess).

[Hornless.Rhino]

I stumbled across this article a few minutes ago, and thought I'd post it here.

http://www.physorg.com/news169811689.html

[Cory_K]

That is a pretty funny trick.

Reminds me of this one:

Let A = B

Add A to both sides
A + A = A + B

2A = A + B

Subtract 2B from both sides

2A - 2B = A + B - 2B

2A - 2B = A - B

Factor out a 2 from the left

2(A - B) = A - B

Divide both sides by (A-B)

2 = 1

2 = 1 ?????? Hmmm

[Hornless.Rhino]

Another version of Corey's is...

A=B

A^2=A*B

A^2 - B^2=A*B - B^2

(A+B)(A-B)=A*B - B^2

(A+B)(A-B)=B(A-B)

A+B=B

B+B=B

2B=B

2=1

[Neil.Pate]

Divide by zero error. Universe halted...

[Hornless.Rhino]

How about adding a machine epsilon block and calling it round off error then?