Envelope Dilemma?

OK, now we're getting somewhere, Baker. You're absolutely right (after you read the paper) that there is a difference in strategy whether or not the person opened the envelope or not. If they don't open the envelope, we can all agree that there is no advantage to switching. But once it is opened, the math would seem to indicate that there is an advantage to switching.

The paradox is that the math leads you to an illogical answer because of the quirkiness of infinity. IF you can assime that there is unlimited upside to the amount that can be in the other envelope, it would be the correct strategy to switch. But the moment you place an upper bound on the upside, the math is destroyed and the answer is functionally the same as if you never looked. That's why it isn't like a coin flip. The key to understanding this is recognizing that the upside isn't infinity, so the open-envelope strategy can't apply. It's 50/50 that the envelope you chose has the greater amount of money, not 50/50 that no matter how much money was in the envelope the other envelope might contain twice as much.

By the same logic and math path, let's say that one person opened one envelope and another person opened the other envelope, but neither knew more than what was in their own envelope. The math would say that they both should switch for the unknown envelope. But that can't be right -- it is a zero-sum game.

Yeah, it's tricky -- just don't let the math lead you off a logical cliff.

PS, Baker: Aren't you smart enough to know that fries always come with the value meal?

No one can argue with this:

The envelope you pick has no money in it.

Therefore, so does the other one.

Go have a beer.

OK, let's find a way to connect this problem with football...

Suppose you have two quarterbacks on your roster, Phillip Rivers and Drew Brees. You are told that one of the quarterbacks is much better than the other. The original plan is to start Rivers, but because of a holdout in training camp, Brees is named the starter.

Brees starts every game for the next two seasons, leading the team to two playoff appearances and ranking in the top 5 in NFL passer rating each of the two seasons. Now the question is: do you stay with Brees, and figure that he's mostly likely the better quarterback of the two? Or do you switch, and take the chance that Rivers is the next NFL superstar, in the mold of Manning/Brady? And does seeing Brees play for the past two seasons make a difference in whether to stay or to switch?

Think about this one for a while.

You have two doors to choose from.

Behind one is $25,000

Behind the other is $100,000

Bob Barker says you can choose Door A or Door B, but he'll give you $50,000 to walk away.

What do you do?

What if he said I'll give you $25,100 to walk away? You'd obviously say NO.

What if he said I'll give you $99,000 to walk away? You'd obviously say YES.

So at what dollar value would you change your answer from NO to YES?

Appears to me the answer would be $62,500; a figure where you'd be "betting" $37,500 to win $37,500.

Another way to look at it would be in 10 decisions of keeping or swapping the original envelope in question, assuming you picked the bigger amount 5 times and the smaller amount 5 times, which is what should happen logically.

Keep it 10 times and you end up with $10x

Always swap and you end up with .5x 5 times and 2x five times or a total of $12.5x

IRS gets half and your wife will get half of the other half after divorcing you for picking the wrong envelope.

Rudy, I have probably written more in regards to this problem than you have ever read. Go read my first post in this thread and you will see that I had full understanding of the difference in strategy depending on whether the contents of the envelope were known or not prior to reading the paper you cited.

The correct action (switching) isn't illogical but it is counterintuitive, there is a distinct difference in the two. You contend that the correct action within the contraints of the problem as presented is not to switch. The only way this could be proven is if you could estimate the funds available to the person making the offer, which you can't do with the information procided. In leui of that information can you identify the threshhold amount (of the offerers available funds) that would be required so that switching envelopes wouldn't be beneficial? I would guess you cannot. So basically you are just making an unfounded assumption that the offerer has an amount available small enough that 50000 would most probably be greater than 50% likely to be the larger of the two amounts offered.

Involving two people where each one gets to open an envelope creates a entirely different problem with very little in common with the one presented. Again you are wrong about what the correct action would be, you just can't use kindergarten logic to solve complex problems.

The paradox lies in the fact that you do not have enough information to solve the problem. To find any kind of solution, you will need to make some assumptions - and you'll get different answers based on the assumptions you make.

The key point in this debate is whether opening the first envelope provides information on the likelihood of the second envelope containing more money. Given a nonuniform distribution, it does. The problem is that we have little information about the distribution of amounts contained in the envelopes. And though we do not know what that distribution is, we cannot unilaterally assume that the distribution is uniform.

I ran this simulation 10,000 times using envelopes with a total dollar value averaging $500. The results were:

running simulation with swapping
total dollars = 2491909.0

running simulation without swapping
total dollars = 2493272.0

In short, the dollar expectation is identical regardless of whether one swaps envelopes.

Here is the java source code:

import java.util.Random;
import java.util.Date;

public class EnvelopeSim {

public static void runSimulation(long trials, long seed, boolean doSwap, boolean verbose) {

Random generator = new Random(seed);

double total = 0;

for (long i = 0; i<trials; i++) {

// generate the total amount of money

int amount = (int) (1000 * generator.nextDouble());

int[] envelopes = new int[2];

// distribute it such that once envelope has twice as much as the other

envelopes[0] = (int) (amount * .66666666666666666);
envelopes[1] = (int) (amount * .33333333333333333);

// choose an envelope at random

double choice = generator.nextDouble();

int selectedEnvelope = 0;
if (choice>=.5)
selectedEnvelope = 1;

// get the amount

int selectedAmount = envelopes[selectedEnvelope];

if (verbose)
System.out.println("envelope selected had " + selectedAmount);

// if we are swapping, do the swap

if (doSwap) {
selectedEnvelope = 1-selectedEnvelope;
selectedAmount = envelopes[selectedEnvelope];
if (verbose)
System.out.println("swapped envelope had " + selectedAmount);
}

total+=selectedAmount;
if (verbose)
System.out.println("total amount is " + total);

}

System.out.println("total dollars = " + total);
}

public static void main(String args[]) {

long seed = new Date().getTime();
System.out.println("running simulation with swapping");
runSimulation(10000,seed,true,false);
System.out.println("running simulation without swapping");
runSimulation(10000,seed,false,false);
}

If the second envelope has an equal chance of having $25,000 or $100,000 in it, the logical conclusion is that it has an expected value of $62,500. But common sense tells us that either envelope has an equal chance of being the high one or the low one. Maybe the point of the puzzle is to demonstrate that logic and common sense aren't the same thing.

Any idea on the expected value of Philip Rivers?