Two Envelopes problem at MathProblems.info

Quote: gordonm888

This problem has two unknowns. let's restate it.

An unknown amount of money, X, has been placed into two envelopes. One envelope has 1/3X and the other has 2/3X.

The first unknown is: How big is X?
The 2nd unknown is : Which envelope has the larger amount of money?

Your desire is to maximize the amount of money that you get.

I think we're still missing THE key unknown: we don't know what is in the envelope we've chosen and we can't open it, or otherwise identify its contents, before we have to make a decision whether to switch.

As long as the unopened envelopes are exactly alike in weight, size, contour and so on -- so that there's no telling the difference between them before we open one -- there is exactly a 50/50 chance, for each envelope, that it contains the larger amount of money.

No matter where the envelopes are, across a table or in our hands, they both have to be treated as having an unknown amount of money until at least one envelope's contents can be identified.

The original poster said that the solution is that "there is no point in switching", but I don't think that this is entirely correct: Fact is, we should be ambivalent about switching. We ought not to care whether we switch or whether we don't, because, again, if we can't tell the difference between them before we open them, then no matter where the envelopes physically are, the chance of each envelope having the larger amount of money is exactly 50/50.

Everyone with the least big of math and logic skills will agree the player should be ambivalent about switching, assuming he doesn't open the first envelope. If you open an envelope, the questions gets more complicated because questions get asked about the wealth of the one stuffing the envelopes.

The point of the problem is to find the flaw in the argument that the other envelope has an expected amount 25% greater than the first envelope.

Let me break down the argument and you tell me which step is in error. Assuming one envelope has twice as much as the other and the player doesn't open the first one.

1. The player has a 50/50 chance of picking the smaller or larger envelope.
   
2. Therefore, the other envelope also has a 50/50 chance of being* the smaller or larger envelope.
   
3. Let x equal the amount in the chosen envelope.
   
4. If the player chose the smaller envelope, then the other one must have 2x.
   
5. If the player chose the larger envelope, then other other one must have x/2.
   
6. The expected amount of the other envelope is (2x + x/2)/2 = 1.25x.
   
7. Since 1.25x>x, switching is a good value.
   

Again, the question is at which specific step is my logic in error?

*amended

The problem is the x/2 or 2x

Instead let the envelopes hold x and x/2

Now, the EV of each envelope is 0.75 x, no matter which one you pick, or even if you open one first.

Knowing the amount of money in one envelope is useless info, since you don't know if you are looking at X or x/2

Once you start comparing x/2 vs x vs 2x, though, you are now comparing three values when there are only 2. If you are looking at X, then either one of 2x or x/2 are not valid possibilities.

Either you are looking at X and the other envelope is x/2, or you are looking at X/2 and the other envelope is x.

Quote: Wizard

Everyone with the least big of math and logic skills will agree the player should be ambivalent about switching, assuming he doesn't open the first envelope. If you open an envelope, the questions gets more complicated because questions get asked about the wealth of the one stuffing the envelopes.

The point of the problem is to find the flaw in the argument that the other envelope has an expected amount 25% greater than the first envelope.

Let me break down the argument and you tell me which step is in error. Assuming one envelope has twice as much as the other and the player doesn't open the first one.

1. The player has a 50/50 chance of picking the smaller or larger envelope.
   
2. Therefore, the other envelope also has a 50/50 chance of picking the smaller or larger envelope.
   
3. Let x equal the amount in the chosen envelope.
   
4. If the player chose the smaller envelope, then the other one must have 2x.
   
5. If the player chose the larger envelope, then other other one must have x/2.
   
6. The expected amount of the other envelope is (2x + x/2)/2 = 1.25x.
   
7. Since 1.25x>x, switching is a good value.
   

Again, the question is at which specific step is my logic in error?

I think the parentheses are misplaced. ((2x+x)/2)/2=.75x and represents what he replied above mine. 2 envelopes, 2 values, regardless of order.

Quote: Wizard

Everyone with the least big of math and logic skills will agree the player should be ambivalent about switching, assuming he doesn't open the first envelope. If you open an envelope, the questions gets more complicated because questions get asked about the wealth of the one stuffing the envelopes.

The point of the problem is to find the flaw in the argument that the other envelope has an expected amount 25% greater than the first envelope.

Let me break down the argument and you tell me which step is in error. Assuming one envelope has twice as much as the other and the player doesn't open the first one.

1. The player has a 50/50 chance of picking the smaller or larger envelope.
   
2. Therefore, the other envelope also has a 50/50 chance of picking being the smaller or larger envelope.
   
3. Let x equal the amount in the chosen envelope.
   
4. If the player chose the smaller envelope, then the other one must have 2x.
   
5. If the player chose the larger envelope, then other other one must have x/2.
   
6. The expected amount of the other envelope is (2x + x/2)/2 = 1.25x.
   
7. Since 1.25x>x, switching is a good value.
   

Again, the question is at which specific step is my logic in error?

Note that I have amended item#2 to correct it to what I presume the Wizard meant to say.

I believe that the item in error is "The expected amount of the other envelope is (2x + x/2)/2 = 1.25x."

First, let's address the state of the envelopes and define that one envelope was filled with an amount A (that is unknown to the player) and the other envelope was filled with amount 2A.

In the scenario in which the player initially chooses the smaller envelope, the player will gain A by swapping envelopes, and the observed quantity of x =A.

In the other, equally probable scenario in which the player initially chooses the larger envelope, the player will lose A by swapping envelopes, because the observed quantity of x =2A.

Thus, swapping envelopes has an EV =0 because you either gain A or lose A.

Even though the player opens an envelope and observes a quantity of money, x, he does not know whether the observed quantity x is 1/3 of the total or 2/3 of the total. Within the context of this problem statement, the identity (or significance) of x is conditional because it depends upon whether the chosen envelope is the smaller or the larger envelope. Therefore, the expected value cannot be expressed in terms of x because x has different meanings in the two scenarios that you are summing over.
***
I have an unusual talent for explaining things, but I found the explanation of this logical error to be very difficult to explain clearly. I hope the above is satisfactory.

Quote: Wizard

1. The expected amount of the other envelope is (2x + x/2)/2 = 1.25x.
   

Again, the question is at which specific step is my logic in error?

I think it is here. This is where you assume an underlying distribution stating that it was equally likely, before stuffing the envelopes, that the max value was your observed value or double your value.

Using real dollar values, let's say your envelope has $4. You are stating that you know that there is equal chance that the large value is $8 or $4. If, indeed, there is an equal chance that the large envelope contains $8 or $4, then you would switch envelopes. But, what if there is a 0% chance of the large envelope being $8? This might be the case, but you don't know.

Absent the underlying distribution, all you know is there is an envelope x and x/2. If you have x and switch, you get x/2. If you have x/2 and switch, you get x.

Quote: beachbumbabs

I think the parentheses are misplaced. ((2x+x)/2)/2=.75x and represents what he replied above mine. 2 envelopes, 2 values, regardless of order.

I stand by the way I had them. For further clarity, I could have said (2x + (x/2))/2 = 1.25x.

Quote: gordonm888

Within the context of this problem statement, the identity (or significance) of x is conditional because it depends upon whether the chosen envelope is the smaller or the larger envelope. Therefore, the expected value cannot be expressed in terms of x because x has different meanings in the two scenarios that you are summing over.
***
I have an unusual talent for explaining things, but I found the explanation of this logical error to be very difficult to explain clearly. I hope the above is satisfactory.

I agree. I couldn't have explained it any better. The way I think of it to myself is there is a correlation to the amount in the original envelope and the ratio of the other envelope to the original one. Given that, it doesn't seem kosher to use the expected value formula. Still, I can't find a satisfactory "ah ha!" way of explaining it.

Quote: CrystalMath

But, what if there is a 0% chance of the large envelope being $8? This might be the case, but you don't know.

If you're saying the other envelope isn't random because the outcome is already determined, then how would you answer the question of the probability of tails of a fair coin flipped and then placed under a cup?

Quote: Wizard

If you're saying the other envelope isn't random because the outcome is already determined, then how would you answer the question of the probability of tails of a fair coin flipped and then placed under a cup?

You know the distribution of a coin toss. You don't know the distribution of the possible envelope values.

Quote: CrystalMath

You know the distribution of a coin toss. You don't know the distribution of the possible envelope values.

I agree. The problem statement does not address this aspect. It does not say "an amount of money, picked at random from a distribution in which all values are equally probable out to infinity, is placed into an envelope and 2x that amount is placed into another envelope." It simply says that an amount of money is placed into the two envelopes at a ratio of 2:1.

If you were to see $100 in the 1st envelope, you can deduce the total amount in the two envelopes is either $150 or $300, but have no basis for knowing whether those have equal or highly assymetric probabilities. In real life, you could take many factors into account and make an educated guess, but in a mathematical problem you do not have that latitude.

The weakness in the statement of the problem is that it provides no direct, bounding or contextual information about the total amount of money being placed in the envelopes, and then proceeds to impose an Expected Value methodology for utilitarian player decisions. By definition, of course, it is not possible to calculate absolute EVs. If relative EVs are to be used as a figure of merit, then they must be expressed as fractions of the total money that was placed in the envelopes in a way that makes no implicit or explicit judgements about the probability of the total money being any particular value. Because the problem statement provides no basis for making such judgements.
****
A fun thought exercise is to genericize this problem by saying that the ratio of money in the two envelopes is n:1, where n can be any positive non-zero number other than 1.

So, say n = 1,000,000:1. You open up an envelope and see $10,000 in it. Do you swap? Hell NO, because you impose real life factors and assume that it is unlikely that the other envelope has $10 billion and far more likely that it has one penny. You take the $10K and run.

Quote: CrystalMath

You know the distribution of a coin toss. You don't know the distribution of the possible envelope values.

How about x and 2x?

Quote: Wizard

How about x and 2x?

Nope.

Is 2x more likely to be $4 or is it more likely to be $8?

Here is hopefully a simple English way to see the flaw in the 1.25x argument.

Suppose, instead, one envelope has $1 and the other has $1,000,000. You pick one but before opening it the host says to you, "I'll make you a deal. I will multiply whatever is in your envelope by 100, but only if you choose the $1 to begin with. However, if you chose the larger amount, I will multiply it by 0.99.

The average ratio of the new envelop to the old one is (100 + 0.99)/2=50.495. Over 50 times as much -- sounds good -- or does it? Consider the dollar amounts, not the rate of change, it obviously is a terrible offer.

Without switching our EV is $500,000.50. With switching is is (100+990000)/2 = $495,050.

The problem statement does not say that this is a game show.

When the ratio is 1,000,000 to 1 it is easy to notice when we impose our real-life biases and say "if I find a penny in the 1st envelope I am definitely swapping, because the 2nd envelope is more likely to have $10,000 than 10^-6 cents." But the problem statement provided no basis for assuming that 10^-6 cents is unlikely to be in the envelope.

Instead, the problem uses a 2:1 ratio which sets a trap -it makes it easy for us to overlook the fact that we are imposing our sensible real life values and not recognizing the absence of any information about the total amount in the envelopes:

"Oh, there's $100 in the first envelope. So the total amount of cash in the two envelopes is either $150 or $300, and if this were a gameshow those amounts would be equally probable because 'who cares about the difference between $150 and $300?' and so I'll pick the scenario where the total amount of money is $300 because that's a scenario I prefer because I get more money."

But the problem statement never said this was a gameshow and provides no basis for assuming that $150 and $300 are equally probable values for the total cash within the envelopes. When we assume that, we have brought external factors into the problem.

EVs must be expressed as fractions of the total amount in the two envelopes, because there is no basis for doing otherwise. So , the total amount in the 2 envelopes is A, then one envelope has 1/3A and the other has 2/3A. If you picked 1/3A and swap, you will gain 1/3A. If you picked the envelope with 2/3A and swap then you lose 1/3A. It is symmetrical.

But, yes, in real life if you are on a gameshow you may gain an edge by opening the 1st envelope, looking at the amount and applying game theory and making a guess as to whether it is likely that the gameshow put 2X into the second envelope.

Interestingly, or not, if this were a game show and the host said he would flip a fair coin to either multiply by 2 or divide by 2 whatever was in your envelope, you should definitely take it, unless you're extremely risk averse.

Quote: Wizard

Interestingly, or not, if this were a game show and the host said he would flip a fair coin to either multiply by 2 or divide by 2 whatever was in your envelope, you should definitely take it, unless you're extremely risk averse.

Or if life changing amounts are in play... say $1500?

Quote: Ayecarumba

Or if life changing amounts are in play... say $1500?

Ouch! That is what makes TPiR hard to watch sometimes.

Quote: Dalex64

The problem is the x/2 or 2x

Instead let the envelopes hold x and x/2

Now, the EV of each envelope is 0.75 x, no matter which one you pick, or even if you open one first.

Knowing the amount of money in one envelope is useless info, since you don't know if you are looking at X or x/2

Once you start comparing x/2 vs x vs 2x, though, you are now comparing three values when there are only 2. If you are looking at X, then either one of 2x or x/2 are not valid possibilities.

Either you are looking at X and the other envelope is x/2, or you are looking at X/2 and the other envelope is x.

Dalex has a good point.

If we say the envelope that we choose has x and the other x/2, then the EV of each envelope is (0.5)[x + x/2] = 0.75x.

If we say the envelope that we choose has y and the other 2y, then the EV of each envelope is (0.5)[y + 2y] = 1.5y.

The flaw is if we assume x=y. The other envelope that we didn't choose isn't necessarily 50/50 to have either x/2 or 2y. We have to consider the amounts in the envelopes in the one way or the other -- as x and x/2, or y and 2y.

And so whichever way we pick, the EV of the two envelopes is the same, and so we should be ambivalent about switching.

Hello forum members,

I did a simulation of the problem in Excel and as weird as it might seem: The expected value for switching is 1.25.

Here is how I did the simulation. Please notify me of any errors you might detect:

Column A for envelope #1
set to a constant value of 1

Column B for the random factor of envelope #2


Column C for envelope #2
calculated by multiplying the amount of envelope #1 and the randomly chosen factor in column B
in approx. 50% of the cases holds half the amount of envelope #1
in approx. 50% of the cases holds twice the amount of envelope #1

I ran the simulation multiple times over a set of 1,000,000 rows each with the following results:

total amount of envelope #1 (sum of column A)
1,000,000

total amount of envelope #2 (sum of column C)
always close to 1,250,000

number of envelopes #2 having amount 0.5
always close to 500,000

number of envelopes #2 having amount of 2
always close to 500,000



As an additional test I varied the amount of envelope #1:

=INT(RAND()*10)+1 for random integers [1;10]
=INT(RAND()*100)+1 for random integers [1;100]

I calculated the ratio

(sum of amounts of envelope #2) / (sum of amounts of envelope #1)

which during all simulations stayed close to 1.25.



Please try for yourself . . .

Quote: ThomasK

Hello forum members,

I did a simulation of the problem in Excel and as weird as it might seem: The expected value for switching is 1.25.

Here is how I did the simulation. Please notify me of any errors you might detect:

Column A for envelope #1
set to a constant value of 1

Column B for the random factor of envelope #2

=(INT(RAND()*2)*3+1)/2

step by step explanation

INT(RAND()*2) generates a random approximation of 50% zeros and 50% ones

( . . .)*3+1 maps to the values to zero and four respectively

(. . .)/2 eventually creates the needed factors of 0.5 and 2 to calculate the (random) amount of the second envelope

Column C for envelope #2
calculated by multiplying the amount of envelope #1 and the randomly chosen factor in column B
in approx. 50% of the cases holds half the amount of envelope #1
in approx. 50% of the cases holds twice the amount of envelope #1

I ran the simulation multiple times over a set of 1,000,000 rows each with the following results:

total amount of envelope #1 (sum of column A)
1,000,000

total amount of envelope #2 (sum of column C)
always close to 1,250,000

number of envelopes #2 having amount 0.5
always close to 500,000

number of envelopes #2 having amount of 2
always close to 500,000



As an additional test I varied the amount of envelope #1:

=INT(RAND()*10)+1 for random integers [1;10]
=INT(RAND()*100)+1 for random integers [1;100]

I calculated the ratio

(sum of amounts of envelope #2) / (sum of amounts of envelope #1)

which during all simulations stayed close to 1.25.



Please try for yourself . . .

It's because of your false assumption that x/2 and 2x have the same probability.

Try another simulation: put $1 in one envelope and $2 in another. Randomly choose one envelope and calculate the average. Then, switch envelopes every time and calculate that average.

Thanks, Chrystal Math.

If I define envelope #1 as the small amount and envelope #2 as the large amount and then randomly choose an envelope it is completely irrelevant whether I stick to my first choice or switch to the other envelope. The sums are always close to another. So the expected value is 1.0.

From doing these experiments I suspect that the original 1.25 equation might belong to the following experiment:

You are offered an amount x as instant cash or you may choose to select one of two envelopes.
One envelope having half the amount of x and the other having twice the amount of x.
Should you take the instant cash or should you choose one of the envelopes?
The instant cash has an expected value of 1.0.
Choosing one of the envelopes has the 1.25 expectation.
Because now the x/2 and 2x do have the same probability . . .

Quote: ThomasK

From doing these experiments I suspect that the original 1.25 equation might belong to the following experiment:

You are offered an amount x as instant cash or you may choose to select one of two envelopes.
One envelope having half the amount of x and the other having twice the amount of x.
Should you take the instant cash or should you choose one of the envelopes?
The instant cash has an expected value of 1.0.
Choosing one of the envelopes has the 1.25 expectation.
Because now the x/2 and 2x do have the same probability . . .

Yes, I posted that point yesterday I think it was. If the other envelope is truly 50/50 being half or double, then it will have 25% more, on average. The paradox, is that you don't gain more by switching, on average in the two-envelope situation.

Regarding the Excel experiment, consider if you do the following for many rows:
Column A: 1 or 2, each having 50% chance. This can be done with int(rand()*2)+1
Column B: 3 minus amount in column A.
Column C: column A/column B (or the inverse of that).

Take an average of column C and you'll get close to 1.25 the more rows you have. Still, the average of columns A and B will approach 1.5 the more rows you have.

Let's say someone hands you $100 and says you can keep it ($100 value) or flip a coin - heads and you get an extra $100 (doubling your money) or tails and you lose $50 (cutting it in half).

Surely, we should all agree flipping the coin is the proper, +EV, play.

How is this any different than the envelope problem? You open the envelope and see $100 in it. You don't know if the other has $50 in it or $200 in it. Given it's a mathematical problem....and not this stupid BS about "well herp de herp, it's not 50/50 because, well you just don't know" nonsense. Forget the whole "utility of money" stuff and "you weren't specifically told it's 50/50".

I'm not saying switching is better or worse, just that I haven't read a good explanation as to why switching is not better. If I first pick a $100 envelope, why aren't the $50 and $200 envelopes each 50/50? If they aren't 50/50, what are their respective probabilities?

Quote: RS

Let's say someone hands you $100 and says you can keep it ($100 value) or flip a coin - heads and you get an extra $100 (doubling your money) or tails and you lose $50 (cutting it in half).

Surely, we should all agree flipping the coin is the proper, +EV, play.

How is this any different than the envelope problem? You open the envelope and see $100 in it. You don't know if the other has $50 in it or $200 in it. Given it's a mathematical problem....and not this stupid BS about "well herp de herp, it's not 50/50 because, well you just don't know" nonsense. Forget the whole "utility of money" stuff and "you weren't specifically told it's 50/50".

I'm not saying switching is better or worse, just that I haven't read a good explanation as to why switching is not better. If I first pick a $100 envelope, why aren't the $50 and $200 envelopes each 50/50? If they aren't 50/50, what are their respective probabilities?

I don't think you're allowed to open the envelope and see the $100 before you can choose to switch though, right?

If you are, then yea, it'd seem like the (fair) 'coin flip' scenario - switching would basically be betting $100 for a 50/50 shot at either $50 or $200, which you'd definitely want to take (return = (0.5)[50+200] = $125)....

There are 3 x. 2 of them are put in one envelope and 1 of them is put in the other.

You open an envelope. You don't know if you are looking at 1x or 2x.

If you are looking at 2x, you assume there is an equal chance of getting 1x or 4x. But, there were only 3x to start with, so you can't get 4x.

Likewise, if you are looking at 1x, you assume there is an equal chance of getting 0.5x or 2x. Again, 0.5x is impossible.

If you use the 50/50 argument, you are guaranteed to make one of the above mistakes.

Quote: docbrock

I don't think you're allowed to open the envelope and see the $100 before you can choose to switch though, right?

The way this problem is often stated is that the first envelope is opened and some nominal amount like $100 is seen. However, I prefer to not phrase it that way, lest the conversation veer off into the direction of who is stuffing the envelopes, how much money does he have, and why is he stuffing them? It only confuses the issue. However, it can help to make the issue more understandable to those who have an algebra phobia.

After you open the 1st envelope, re-frame the question as "How much total money do I think was placed in the 2 envelopes?" rather than as "Is the other envelope larger?" If you think that it is likely that 3X the money in the first envelope was placed in both envelopes then yes, go ahead and switch.

The Two Envelopes problem is a semantics issue - the problem statement says nothing about who and why there is "money in two envelopes" and offers no limits on what might be in there. So, we each impose our own imaginary context on the problem -its a game show and if I find $100 in the 1st envelope, its easy to believe the 2nd has $200 or $50 with equal probability.

But how about this problem statement:

Money is placed in two envelopes, with one envelope having 1,000,000 X more money than the 2nd envelope. You choose one envelope and open it up and find $10,000. Do you swap with the 2nd envelope?

I think most people will NOT swap -because they believe it is more likely that the 2nd envelope has 1 cent rather than $10 billion.

Your expectation - your EV - is based on your judgement about whether the two possibilities for the 2nd envelope are equally probable -or have very assymetric probabilities.

If the second envelope, after the first is opened, has either $50 or $200, does there have to be a 50% chance that it has $50 and a 50% chance that it has $200? Can it be 2/3 that it has $50 and 1/3 that it has $200?

Quote: docbrock

If the second envelope, after the first is opened, has either $50 or $200, does there have to be a 50% chance that it has $50 and a 50% chance that it has $200? Can it be 2/3 that it has $50 and 1/3 that it has $200?

You can phrase the original question any way you want.

Quote: docbrock

If the second envelope, after the first is opened, has either $50 or $200, does there have to be a 50% chance that it has $50 and a 50% chance that it has $200? Can it be 2/3 that it has $50 and 1/3 that it has $200?

It is undefined.

When you open one envelope, you know that the envelopes either contain {100, 200} or {50, 100}, two possibilities for the contents of the envelopes, but there is no information given to determine how probable either of those two possibilities are.

So if there's a 2/3 chance that the envelope contains $50 and a 1/3 chance that it contains $200, the EV of the envelope will be $100, which is the same amount of money that's in the envelope we've already opened:

p(50) + (1-p)(200) = 100
50p + 200 - 200p = 100
-150p = -100
p = 2/3

I don't know how to do it, but if we can prove somehow that these are the actual probabilities, we'll have shown that we'd have to be ambivalent about switching, correct? Because the actual value of our envelope and the expected value of the other envelope would be the same?

Quote: docbrock

p = 2/3

I think it would be difficult to prove that a randomly selected envelope had a 2/3 chance of being the larger amount.

We know the odds are 50% for each, because there are only 2 that we're picking from (right?).

In a math problem, if we're told one is 1,000,000 times the other and we open the $10k envelope, then we should switch or stay based on the highest EV and not falling back on some, "But in the real world...the other likely doesn't have $10B in it..." nonsense.

I think the best way to prove it, sorta, is by negation or whatever you call it in logic.

If switching is always better, and we change the problem to involve two people. Person A picks an envelope and has the ability to switch if he wants. Person B is left with whichever envelope remains. If A stays the same, then B stays the same. If A switches, then B also switches. You can see that when A and B switch, there is never an increase in overall value.


But now I've confused myself because I have another thought but it doesn't make sense in my head (it's still morning for me, so maybe my brain just being wonky). If you're given $100 and the option of a coin-flip, and we determined that's +EV so you either end with $50 or $200....is it really +EV? On average you expect 1 heads for every 1 tails. If you get 1 of each, you end up with $100. If you get 2 tails then 2 heads, you go 100-50-25-50-100. 2 heads then 2 tails is 100-200-400-200-100. heads/tails/heads/tails is 100-200-100-200-100.

Although I suppose it'd be +EV because if you run good (2H, 0T) you'll end up with $400 compared to if you run bad (0H, 2T) and end up with $25. Up $300 vs down $75. But it's weird, since if you run at expectation, coin wise, you'll be right back where you started....but if it's +EV and you ran at expectation, how are you back where you started?

#TheWorldMayNeverKnow

Since it's always better to switch, just pick the envelope you don't want and be done with it.

Quote: CrystalMath

Since it's always better to switch, just pick the envelope you don't want and be done with it.

I like that!

Several posts suggested to use the total amount T of both envelopes and express the individual envelopes as fractions of this sum. This idea seems to be rather helpful:

First I try to show that the expected value of randomly choosing one of two envelopes (i.e. probability 50%) always is T/2.
In a second step further below I try to show why the „instant cash“ variant has the advantage of 1.25. From this I try to apply the same type of calculation to the „two envelope problem with peeking in the first envelope“


1) Randomly selecting one out of two envelopes.

The two envelopes of the original problem (without peeking in first envelope):
E(T) = 50% * (1/3)*T + 50% * (2/3)*T
= (1/6)*T + (2/6)*T
= (3/6)*T
= (1/2)*T

The two envelopes of the instant cash variant (comparison with instant cash see further below):
E(T) = 50% * (1/5)*T + 50% * (4/5)*T
= (1/10)*T + (4/10)*T
= (5/10)*T
= (1/2)*T

Two envelopes in general, one having a fraction f of the total amount:
E(T) = 50% * f*T + 50% * (1-f)*T
= (f/2)*T + (1/2)*T - (f/2)*T
= (1/2)*T

An alternative equivalent transformation
E(T) = 50% * f*T + 50% * (1-f)*T
= 50% * T * (f + 1 - f)
= 50% * T * 1
= (1/2)*T

Special cases:
If f was 0.0 or 1.0 respectively only one of the envelopes would hold the total amount T and the other envelope would be empty.
If you now randomly chose one of the envelopes over an inifinite number of trials the average of the drawn amounts should be T/2.

Side: Generalization with probabilities other than 50%.

E(T) = p * f*T + (1-p) * (1-f)*T
= p*f*T + 1*T - f*T - p*T + p*f*T
= (2*p*f - f - p + 1)*T

I can’t seem to think of some intuitive interpretation for now …

A test with p=50%:

E(T) = (2*50%*f - f - 50% + 1)*T
= (1*f - f +50%)*T
= 50%*T
= (1/2)*T


2) Comparing the „instant cash“ to the random draw of an envelope.

E(„instant cash“) = 100% * (2/5)*T
= (2/5)*T
= (4/10)*T
From above:
E(„envelope“) = (5/10)*T

Advantage = E(„envelope“) / E(„instant cash“)
= (5/10)*T / (4/10)*T
= (5/10) * (10/4)
= (5/4)
= 1.25


3) Comparing the peeked amount to the second envelope.

Let peeked amount = (1/3)*T - the small envelope
Advantage = E(„peeked“) / E(„other“)
= (100% * (1/3)*T) / (100% * (2/3)*T)
= (1/3) * (3/2)
= (3/6)
= (1/2)

Let peeked amount = (2/3)*T - the large envelope
Advantage = E(„peeked“) / E(„other“)
= (100% * (2/3)*T) / (100% * (1/3)*T)
= (2/3) * (3/1)
= (6/3)
= 2

Since we don’t know whether we saw the small or the large envelope both situations have the same probability of 50%.
Both advantages add up to a total of 2.5 and we can apply the equation given in 1):
E(advantage) = 50% * (1/5) + 50% * (4/5) = (1/2).
This value is the same for either sticking to the chosen envelope or switching to the other envelope and therefore there is no advantage in switching.

Nice. I took a shot at a solution too... it's not quite so formal but it may work. Check my work though.

We're presented with two envelopes and we're told one envelope has double the amount of the other. We're told to choose an envelope, open it, and then we're given the choice to switch or to stay with the envelope we originally chose.

So we choose an envelope and it has its amount of money in it. Then we're offered to switch. The other envelope could have amount A, amount B, or either A or B. There are four possible scenarios:

Our envelope	A	B
x	2x	x/2
x	x/2	2x
2x	x	N/A
x/2	x	N/A

Each scenario has its own particular EV of switching, and each scenario has its own amount of money in the envelope which we can choose to keep:

Our envelope	A	B	EV of Switching
x	2x	x/2	1.25x -- from (0.5)(2x + x/2)
x	x/2	2x	1.25x -- from (0.5)(x/2 + 2x)
2x	x	N/A	x
x/2	x	N/A	x

The EV of our original choice is down the left-hand column:
= (x + x + 2x + x/2) / 4
= 4.5x/4
= 1.125x

The EV of switching is down the right-hand column:
= (1.25x + 1.25x + x + x) / 4
= 4.5x/4
=1.125x

So the EV of our original choice = the EV of switching, so we shouldn't care whether we switch or not.

Does that check out?

Thanks for the arguments above. I'm always happy to see others challenged by this problem. No arguments from me with the last two posts.

However, I still can't get away from the core simplicity of the question. The other envelope has half or it has double. Nobody can argue that there are any other possibilities. Half or double. Certainly I had a 50/50 chance at picking the lower/higher envelope. So, the other envelope should be 50/50 it is larger or smaller. Or 50/50 it is half or double. Take the average and you're at 1.25. Where can that obvious chain of thought possibly be wrong?

Let me ask another question. Suppose you flip a fair coin in the dark and cover it with a cup. The bottom of the coin is 50/50 to be heads or tails. The top also is 50/50 to be heads or tails. However, once you assign a value x to either side, let's say the top, the other side is no longer random. If the top of the coin is x, then the bottom can't also be x. So, each individually can be heads or tails but they both can't be the same thing.

What is my point? I'm not sure. Even the original question says the two envelopes can't be the same. I thought I was onto something but I'm back to square one -- again.

Quote: Wizard

Thanks for the arguments above. I'm always happy to see others challenged by this problem. No arguments from me with the last two posts.

However, I still can't get away from the core simplicity of the question. The other envelope has half or it has double. Nobody can argue that there are any other possibilities. Half or double. Certainly I had a 50/50 chance at picking the lower/higher envelope. So, the other envelope should be 50/50 it is larger or smaller. Or 50/50 it is half or double. Take the average and you're at 1.25. Where can that obvious chain of thought possibly be wrong?

Let me ask another question. Suppose you flip a fair coin in the dark and cover it with a cup. The bottom of the coin is 50/50 to be heads or tails. The top also is 50/50 to be heads or tails. However, once you assign a value x to either side, let's say the top, the other side is no longer random. If the top of the coin is x, then the bottom can't also be x. So, each individually can be heads or tails but they both can't be the same thing.

What is my point? I'm not sure. Even the original question says the two envelopes can't be the same. I thought I was onto something but I'm back to square one -- again.

Your fault is X.

Your expected value never falters away from 1.25x, in either envelope. No matter which envelope you pick, you will either take .5x or 2x. Of course the mean value is 1.25x. The difference between one envelope and the other is 1.5x.

You might as well restate the problem. In one envelope is x dollars and in another envelope is 4x dollars. Choose an envelope. I don't see the problem here.

Quote: boymimbo

Quote: Wizard

Thanks for the arguments above. I'm always happy to see others challenged by this problem. No arguments from me with the last two posts.

However, I still can't get away from the core simplicity of the question. The other envelope has half or it has double. Nobody can argue that there are any other possibilities. Half or double. Certainly I had a 50/50 chance at picking the lower/higher envelope. So, the other envelope should be 50/50 it is larger or smaller. Or 50/50 it is half or double. Take the average and you're at 1.25. Where can that obvious chain of thought possibly be wrong?

Let me ask another question. Suppose you flip a fair coin in the dark and cover it with a cup. The bottom of the coin is 50/50 to be heads or tails. The top also is 50/50 to be heads or tails. However, once you assign a value x to either side, let's say the top, the other side is no longer random. If the top of the coin is x, then the bottom can't also be x. So, each individually can be heads or tails but they both can't be the same thing.

What is my point? I'm not sure. Even the original question says the two envelopes can't be the same. I thought I was onto something but I'm back to square one -- again.

Your fault is X.

Your expected value never falters away from 1.25x, in either envelope. No matter which envelope you pick, you will either take .5x or 2x. Of course the mean value is 1.25x. The difference between one envelope and the other is 1.5x.

You might as well restate the problem. In one envelope is x dollars and in another envelope is 4x dollars. Choose an envelope. I don't see the problem here.

But that's a different problem.

The way you've setup the problem, the EV of swapping is 1.25X!

Based on how you've defined X, it follows that:

1) If the smaller envelope was picked, then the total amount of money in both envelopes is 3X.

2) If the larger envelope was picked, the total amount in both envelopes is 1.5X.

If you require the total amount in both envelopes to be the same in both scenarios (e.g. Y), you have constrained the problem such the only legal value of X is zero. At that point, the (apparent) paradox is removed. The EV of the other envelope is 1.25X = 0.

---

If I picked $100 in my envelope and asked whether I should switch, then I need to resort to evaluating probabilities of how much total money was placed into envelopes. I will factor any piece of information you give to help me refine the probabilities.

If you tell me I picked the smaller envelope, then the most important unknown is removed. I have perfect information whether to switch or not.

---

Let's consider the scenario where I picked $100. It follows that the total amount in the envelopes is either $150 or $300. If I ascribe equal weight to both scenarios, then I absolutely should switch given my EV of swapping is +$25. If the only two options for the total amount were $150 or $300, then the EV for my first pick was $112.50 (25% @ $50, 50% @ $100, and 25% @ $200).

If after picking $100, you (or Monte Hall) tell me l that either $150 or $300 was placed in the envelopes, then I absolutely will switch. The problem is that you can't reveal this information if l pick any amount other than $100 while keeping the problem interesting (and without changing the value of the other envelope after I pick the first one).

Going back to the original framing of the two envelope problem, it is completely valid to say that the EV of swapping is +0.25X if it is revealed that the total amount placed in the envelopes is either 1.5X or 3X, with equal probability. There's no problem...the EV of the total amount is 2.25X. I got X. EV of other envelope is 1.25X. EV of swap is +0.25X.

---

Coming full circle...if I placed Y (total) and was forced to present you a decision where the EV of swapping was +0.25X by telling you that Y was either 1.5X or 3X, I could not always do it truthfully unless Y was zero.

If you want X to be a variable and allow Y to be a probability distribution, then the EV of swapping is +0.25X.

If you want Y to be a variable and allow X to be a probability distribution, then the EV of swapping is zero (0.5*Y/3+0.5*(-Y/3)).

If X & Y are both variables (but actual values), the only scenario where the EV of swapping equals the scenario of when Y is a probability function is when X = Y = 0.

I know this point has been made before, but I can't think of a way to put the explanation of the paradox in much simpler language than the following.

The fallacy in the argument that the EV(other envelope)=1/2*[2*x + x/2) ] = 1.25x, where x is the chosen envelope is that there are two different x's. In other words, we're calling two different things the same variable, x.

Instead, let's define:

x₁ = Chosen envelope
x₂ = Other envelope
L = lower of the two envelopes


EV(x₂) = 1/2 * [ E(x₂ | x₁ is low) + E(x₂ | x₁ is high)
= 1/2 * [2*L + L]
=(3/2)*L

In other words, the value of the other envelope is half way between the lower and higher envelope, which is obviously what the first envelope is.

Any criticisms or simpler way to explain it?

Are you asking for the simplest explanation to the paradox? I wrote my long reply under the assumption that there was some confusion.

First of all thank you very much for your comments on my post. I’m glad I did the math right.
The discussions about x are still ongoing so I’d like to add an idea and am looking forward to your opinions:


Quite a while ago I’ve read Daniel Kahneman’s „Thinking Fast and Slow“ which has impressed me a lot. I recall one of the psychological phenomena called „priming“. For some reason I feel intrigued to think that both the envelope problem as well as its counterpart, the Monty Hall problem, are based on the phenomenon of "priming" and that this might be what connects both problems:

In the envelope problem the amount of the first envelope seems to be the priming effect. We try to find the answer relative to this (absolute) value.
In the Monty Hall problem we are asked for a decision when we are faced with two closed doors. An answer of „fifty-fifty“ almost instantly comes to mind.

I’d like to explain this in more detail.

Two Envelopes problem.

The priming effect.
In the envelope problem the first thing we hear about is an envelope. From that moment on we seem to be fixated to solve the problem relative to this first envelope’s amount. Even worse: A variant of the game allows us to see the amount in this first envelope which makes it even easier to expect to find a solution relative to this amount.
Furthermore it seems to be easier to work with the absolute amount of the first envelope instead of considering the amount as a faction of the sum of both envelopes. (Would this correspond to Kahneman’s „Cognitive Ease“?)

Mathematical support.
1) The goal is to calculate an expected value.
2) The general equation for the expected value of an experiment with two outcomes is
E(X) = 50% * a + 50% * b.
3) While the experiment is carried out infinite times the parameters a and b may not change but have to remain constant.
4) For the envelopes this is true if we define a = (1/3)*T and b = (2/3)*T with T being the total sum of both envelopes and a and b being fractions of this total.
On the other hand:
5) If a and b are defined as functions of the absolute value x of the first envelope then a and b will change during the repetition of the experiment because a and b will randomly and arbitrarily vary by the factors 0.5 and 2.
5a) If we tried to keep a at a fixed value by changing x according to the individual factor applied in each trial then b will vary even more because besides the randomly varying factor now also x is varying.
5b) As a further complication we would need to be able to predict the current random factor applied to a in order to adjust x accordingly.

From this I suspect that we are mislead, that the solution cannot be found relative to x, and that the correct way is via the fraction of the total of both envelopes.

Monty Hall problem.

The priming effect.
In the Monty Hall problem there is the situation where the host has opened one of the empty doors, the guest is faced with two closed doors and now the guest is asked for a decision whether to open his initially chosen door or the remaining door. Faced with two doors the answer of a fifty-fifty situation immediately comes to mind.
The initial selection of 1/3 and the remaining 2/3 seems harder to be graspable. (Again Kahneman’s „Cognitive Ease“?)

Crashing the show.
A little change in the words spoken can alter the way the guest comes to a decision. Even better - the right answer is intuitively at hand.
I’d like to make it clear: The activities during the show remain unchanged. Only the words spoken are modified.
This of course is under the assumption that the random experiment is only depending on the activities carried out and that the words spoken do not have any influence on the progress and outcome of the underlying pure random experiment.

Here is how it works.

In the original show (assumption: guest switches) the host
1) has the guest select a door
2) opens an empty door
2a) asks whether the guest wishes to switch to the third door
3) opens the third door

In the improved version the host
1) has the guest select a door
1a) asks whether the guest wishes to open the selected door or the other two doors
2) opens an empty door
3) opens the third door

Steps 1), 2), and 3) are activities and are the same in both scenarios.
Steps 2a) and 1a) are spoken words, no activities are carried out.

You may argue that these are different experiments and therefore not comparable. But imagine the following:
The guest is really very nervous and hardly listens to what the host is saying. He/she mainly takes the visual input and has a basic knowledge of the sequence of the game.
He/she selects a door. He/she now sees the two other doors, how the host opens the empty door and now thinks: „If he opens the third door two doors will have been opened for me instead of the single one I chose - twice the chance of winning the car!“.

Again I’d like to state that we are mislead by the priming effect, here supported by a distracting host. Without the distraction the correct answer is given intuitively.