A real poser this time

Originally posted by DoctorScribbles
I can't tell if you're serious or joking. If you're really serious that you're not convinced that switching gives you no benefit, I'll provide both a theoretical proof and a reproducible Excel simulation to back it up. But I do hope that you're joking - the 50/50 result should be completely obvious.

First, for the theoretical proof.

Let X and 2X be the amounts in the envelopes.
Let us call the two envelopes A and B.
I deduce from the problem statement that

1. A contains X and B contains 2X with probability .5, and
2. A contains 2X and B contains X with probability .5

Condsider two cases based on which envelope is initially chosen.

Case 1: The player initially chooses A.
His expected value for staying is: .5*X + .5*2X = 1.5X
His expected value for switching is: .5*2X + .5X = 1.5X
In this case, both strategies have an equal expected value.

Case 2: The player initially chooses B.
His expected value for staying is: .5*X + .5*2X = 1.5X
His expected value for switching is: .5*2X + .5X = 1.5X
In this case, both strategies have an equal expected value.

In both cases, the switching and staying strategies have equal expected values.
Therefore, in no case does employing the switching stragey give the player a larger expected value. The player must conclude that switching gives him no
extra benefit that he doesn't receive by staying.

QED

Here's some more food for thought. This is the solution provided on "Shack's Math Problems" where this problem was taken from. Note that the solution is "doesn't matter" as evereyone has been saying, and as I originally thought (according to my first post). However, note the 2nd last paragraph:

Question:
http://mathproblems.info//group1.html

Solution:
http://mathproblems.info//prob6s.htm

Problem 6 Solution

The expected value of this gamble can not be taken. The assumption that there is a 50% chance of doubling or cutting in half your money is not valid. To take an expected value you must know absolutely what the possible outcomes are. In this case the money in the other envelope is not random. There is either a 100% chance of it being twice your envelope or a 100% chance of it being half your envelope.
Here is another way of puttin it: Suppose the amount in envelope one is x and the amount in envelope two is y. Assume you chose envelope x. Then does E(y)=1.25*E(x)? If so then E(x) must equal 1.25*E(y) if your first choice were envelope y. If E(x)=1.25*E(y) and E(y)=1.25*E(x) then the only values that satisfy both equations are x=0, y=0, which does not meet the condition that both envelopes contain a non-zero amount of money. So the argument for switching has been disproven by contradiction.

Jeff Spirko sent in a more appropriate way of looking at the choice of switching. Suppose the envolopes contained $1 and $2. By switching you would either gain $1 or lose $1. Regardless of the actual amounts in the envelopes the point remains the same, that the average gain of the two possibilities is zero.

Adam Atkinson adds another interesting observation he credits to a graduate student at Cambridge University, whose name he forgot. Using an increasing function, mapping values of x from 0 to infinity to 0 to 1, the player can improve his probability of ultimately choosing the correct envelope to above 50%. For example if the amount in the initial envelope is x and the player keeps his original envelope with probability x/(c+x), where c>0, then he will be more likely to not switch with the larger envelope. Note that the function must be decided upon before the first envelope is chosen and a method must be available to choose a precise random number.

Finally, here is an exhaustive analysis of this paradox: www.u.arizona.edu/~chalmers/papers/envelope.html.

Originally posted by DoctorScribbles
First, for the theoretical proof.

Let X and 2X be the amounts in the envelopes.
Let us call the two envelopes A and B.
I deduce from the problem statement that

1. A contains X and B contains 2X with probability .5, and
2. A contains 2X and B contains X with with probability .5

Condsider two cases based on which envelope is initially ch ...[text shortened]... onclude that switching gives him no
extra benefit that he doesn't receive by staying.

QED

Originally posted by PBE6
However, note the 2nd last paragraph:

Originally posted by PBE6
Close. The expected value for staying is X, and the expected value for switching is 1.25X.

Originally posted by DoctorScribbles
That is irrelevant to the problem you posed. In that paragraph, the player gets to open his first choice and examine the contents of the envelope. In the problem you posed, he does not.

Originally posted by DoctorScribbles
Nonsense. The expected value for staying is 1.5X as I claimed.

Originally posted by PBE6
The way you stated it, you're looking at half of both cases. One envelope contains X, the other contains either 0.5X or 2X, so the "expected value" is 1.25X for switching. But as Mr. Shackleford points out, expected value does not apply here.

Originally posted by PBE6
The expected value of this gamble can not be taken. The assumption that there is a 50% chance of doubling or cutting in half your money is not valid. To take an expected value you must know absolutely what the possible outcomes are. In this case the money in the other envelope is not random. There is either a 100% chance of it being twice your envelope or a 100% chance of it being half your envelope.

Originally posted by DoctorScribbles
Nonsense. Expected value applies and it is as I stated it. My evaluation is in accordance with the definition of expected value. Yours is not; 1.25*X doesn't come from any process or equation that is well-defined in accordance with the notion of expected value.

If you are saying that 1.25*X = .5(2X + .5X), then I don't dispute that, but I do ...[text shortened]... e definition of expected value, and also say exactly what you think it is the expected value of.

Originally posted by PBE6
Here's some more food for thought. This is the solution provided on "Shack's Math Problems" where this problem was taken from. Note that the solution is "doesn't matter" as evereyone has been saying, and as I originally thought (according to my first post). However, note the 2nd last paragraph:

Question:
http://mathproblems.info//group1.html

So ...[text shortened]... e is an exhaustive analysis of this paradox: www.u.arizona.edu/~chalmers/papers/envelope.html.

A simulation, in C++, using GCC 3.3.1 (by ways of Dev-Cpp: www.bloodshed.net)

#include <iostream>
#include <stdlib.h>
#include <time.h>

using namespace std;

int main()
{
const int TRIALS = 50;
const int DRAWS = 5000;
double env1, env2, stay, sw, result;

//set random seed based on current time
srand(time(NULL));

for (int i=0; i<TRIALS; i++)
{
stay = 0;
sw = 0;
result = 0;
for (int j=0; j<DRAWS; j++)
{
if (double(rand())/double(RAND_MAX) < 0.5)
{
env1 = rand()/1000;
env2 = 2*env1;
}
else
{
env2 = rand()/1000;
env1 = 2*env2;
}
stay += env1;
sw += env2;
result += (env1 < env2 ? 1 : 0);
}
cout << i << " " << stay << " " << sw << " " << result << endl;
}
//uncomment if using Dev-Cpp
//system( "PAUSE" );
}
As you will see, the chances are equal.

BTW, The forum won't display the indentation spaces 🙁

Originally posted by piderman
A simulation, in C++, using GCC 3.3.1 (by ways of Dev-Cpp: www.bloodshed.net)

#include <iostream>
#include <stdlib.h>
#include <time.h>

using namespace std;

int main()
{
const int TRIALS = 50;
const int DRAWS = 5000;
double env1, env2, stay, sw, result;

//set random seed based on current time
srand(time(NULL));

...[text shortened]... you will see, the chances are equal.

BTW, The forum won't display the indentation spaces 🙁

What about what I wrote?

If you see the choice of swapping as betting 0.5x to have a 50% chance of gaining an extra x.


Edit: To prove it doesn't matter if you swap.

You should see the problem differently because you already have X, so you have 0.5x to lose and x to gain.