Envelope Problem

[London Colin]

Ok, I have re-read the articles at the links that Colin provided. The first link very clearly points out the flaw in the formula that the EV of switching would be

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

The flaw is that the first term is correct only when A is the smaller amount and the second term is correct only when A is the larger amount. That is, two different values of A are assumed within the same formula.

The correct approach is to realize that the envelopes contain the amounts x and 2x and that, if you are holding the envelope with the smaller amount, you gain x by switching (you go from x to 2x) and if you are holding the envelope with the larger amount, you lose x by switching (you go from 2x to x). So the EV of switching is (1/2)x - (1/2)x = 0.

Furthermore, the EV of each envelope is actually (x + 2x) / 2 = (3/2)x.

Colin is correct when he said that opening one envelope changes nothing. Replacing A by 100 in the flawed formula above changes nothing. So I stand corrected again (thanks Colin). Even if you open your envelope and find 100, there is no value in switching.

Thanks, Gronbog. I concur with all of the above, but in truth that's just a good summary of all the points that have been made multiple times already within this thread.

The article provides several other solutions, but the above is the simplest one.

It's some of the other stuff within those articles that I feel has the potential to shine greater light on the nature of this paradox, but the details of which still leave me feeling like I haven't fully understood.

Specifically, it's the way they both tackle the case when one envelope is opened.

It seems to me that your summary shows us an algebraic proof of what we intuitively already know - with no information on which to base a switching decision, there is nothing to be gained from switching. Having proved that for the general case (X, 2X), it's obviously true for any specific case we can think of.

The problem arises when we do think of a specific case and then try to 'walk through' the mechanics of what is going on. The paradox jumps out and bites us on the bum! (With Meistro cheering it on! )

The two papers tackle this in different ways -

The wonderfully named Panagiotis Tsikogiannopoulos actually goes through exactly the example Meistro chose, and says -

Suppose that both players see that the envelope of player A contains 100 euros and let's see a correct calculation using only numeric values. Once we know the amount of 100 euros, we conclude that the other envelope can contain either 50 or 200 euros. There are now two equally possible events for the two fixed amounts that the game is played with:

Event 1: Amounts of 100 and 200 euros
Event 2: Amounts of 50 and 100 euros

As we pointed out in the Introduction, the players will have to assign equal probabilities to these two events. In every variation of two fixed amounts where one of them is revealed, the players will have to weigh the return derived from each event with the average fixed amount by which the game is played in this event. In Event 1, player A will have a profit of 100 euros by exchanging his envelope whereas in Event 2 will have a loss of 50 euros. The average fixed amount in Event 1 is 150 euros while in Event 2 is 75 euros.

The formula of expected return in the case of exchange for player A that summarizes the above remarks is the following:
E(A) = 1/2 (+100/150) + 1/2(-50/75) = 0     E (1.3.1)

Similarly, player B will apply the following formula and will come to the result:
E(A) = 1/2 (-100/150) + 1/2(+50/75) = 0     E (1.3.2)

We notice that we have concluded as expected to the same result as of the Variation 1.1, i.e. that the exchange of envelopes is indifferent for both players.

Let’s clarify the need to weigh in case of exchanging envelopes: In the Event 1, the player who will switch the amount of 100 euros with the amount of 200 euros will have a profit of 100 euros in a game that shares 300 euros in total. So we can say that this player played the game with a success factor of 100 euros / 300 euros = +1/3. Similarly, the other player played the game with a success factor of -1/3. In the Event 2, the player who will switch the amount of 50 euros with the amount of 100 euros will have a profit of 50 euros in a game that shares 150 euros in total. This player played the game with a success factor of +1/3 also and his opponent with a success factor of -1/3 also.

In reality, when a player sees that his envelope contains 100 euros, he doesn’t know whether he is in the game of the Event 1 or in the game of the Event 2. If he is in the Event 1 and switches he will have a success factor of +1/3 whereas if he is in the Event 2 and switches he will have a success factor of -1/3. As we mentioned above, these two events are considered to have equal probability of 1/2 to occur, so the total success factor of player A considering both possible events is zero. This means that the decision of a player to switch or not switch his envelope is indifferent, even when he makes his calculations based on the amount of money that is revealed to him. We used this reasoning in formulas (1.3.1) and (1.3.2) with the only difference that instead of the total amount we used the average fixed amount which is more appropriate.

If instead of player A’s envelope it was player B’s envelope that would be opened, then all that would have changed in the above calculations is that player B would have to apply the formula (1.3.1) and player A would have to apply the formula (1.3.2).

Which is all well and good, but, even though he uses the phrase 'Let’s clarify the need to weigh in case of exchanging envelopes...', I don't find the justification for that step entirely clear (other than the fact that it leads us to what we already know to be the right answer).

Post was too long. Splitting in two ....

 

[London Colin]

Eric Bliss's paper takes a different approach, considering an example from the point of view of the organizer of the game, aware of the values he has placed in both envelopes -

2. A New Perspective

To see how the paradox can be resolved in its entirety, we must examine it from a different perspective. Previously the literature has focused solely on the perspective of the player who is offered the envelope, not that of the offering party. To rectify that, let us set the game up where we are offering the money. Let us suppose we know that we have two envelopes, one containing $5.00 and one containing $10.00. We offer the player the knowledge that one envelope has twice as much money as the other and the opportunity to pick one.

If the player picks the $5.00 envelope, is allowed to look inside, and then is given the opportunity to switch he could reason he has a .5 probability of losing $2.50 and a .5 probability of gaining $5.00; obviously, he reasons, it is in his best interest to switch. If there was a third party who knew the values of the envelopes but did not observe which amount the player chose, he would conclude that there was a .5 chance that switching would gain him $5.00 and a .5 chance that it would lose him $5.00. This third party would also know that both envelopes have a value under Bayesian probability of $7.50; clearly, there is no value in switching. We know that there is in fact a 0.0 probability of his losing $2.50 and a 1.0 probability of his gaining $5.00; therefore, we know that he should of course switch.

The player however, has estimated the relative utility of his options as well as he might be expected based on the probabilities he calculates from his subjective position and the use of Bayesian decision theory. In this case, even though his information was incomplete, his estimation based on subjective probability seemed to be helpful; he chose the larger amount of money, but because his reasoning was flawed it is only by luck on his part that he chose the more favorable option.

If, on the other hand, the player picks the $10.00 envelope, under the same conditions, he could reason he has a .5 probability of losing $5.00 and a .5 probability of gaining $10.00. Should he not then take the other envelope? By his estimate, it is clear he should. If again there were a third party who knew the values of the envelopes but not which one the player chose, he would arrive at the same conclusion as before, that there was a .5 chance that switching would gain him $5.00 and a .5 chance that it would lose him $5.00. We, on the other hand, know that in fact his odds off losing $5.00 are 1.0 and his odds of gaining $10.00 are 0.0.

In this case, while he still calculates the Bayesian interpretation as well as can be expected with the information afforded to him, his conclusion that he should switch was less advantageous than it would have been had he chosen the other envelope first. In essence, Bayesian decision theory was not helpful.

3. The Source of the Paradox

The reason the Two Envelope problem appears so problematic is that Bayesian Decision theory does not, on average, give good advice to the player using their subjective estimations of the probabilities of different outcomes. The third party who knew the values of the envelopes correctly applied Bayesian decision theory to determine that there was no rational reason to switch without further information. We, the controllers of the game, who had that further information, obviously knew whether or not the player should switch.

The player, however, had to rely on assumptions that, while reasonable, were wrong. If the player chose the envelope with $10.00, he believed there to be a .5 chance that the other envelope contained a $20.00 payoff. While that assumption cannot be faulted based on the information he had, it simply was wrong. There was no possibility of a $20.00 payoff. It skewed his value expectations higher. Had $20.00 been an option, had the value of the second envelope been determined by a coin toss between $5.00 and $20.00 after opening the first, it would have been in the player’s best interest to switch. The erroneous belief in the possibility of a $20.00 payoff creates the problem with subjective probability.

The hypothetical player who was fortunate enough to pick the $5.00 envelope first used the same problematic reasoning. It was simply his good luck that in fact the other envelope did contain $10.00. He was mistaken about the possibility of losing $2.50, but the essential element is that a player who expects the possibility of a lower payoff that doesn’t exist is off by less than a player who expects the possibility of a higher payoff. This disparity is the reason the Bayesian interpretation is skewed too high, leading the player to believe he should switch.

Just as Bruss demonstrated why we cannot assign a single variable to represent the value of the first envelope, this reasoning shows what goes wrong when the game itself tells us the value of first envelope, forcing us to have a single definite value.

4. The Mathematics behind the Paradox

If this theory is correct, we would expect that it would be able to predict the factor of 1.25 by which one naively might believe switching would be better than not switching, and in fact it is able to do so. Consider two envelopes with X and 2X dollars. [....]

When I read the above (and the rest of it that I cut out for brevity), I feel tantalizingly close to a deeper understanding, but then I put myself back in the shoes of the player, faced with what seems like (but evidently, somehow isn't) a 50:50 choice and I still struggle to reconcile the two viewpoints.

 

[gronbog]

Listen to Joe or even better, just go and read the online solutions. There are several approaches there all of which show that the EV of switching is zero. There is even a very clear explanation of why your exact model is not correct. It baffles me why you refuse to do this.

Several posters have pointed out the paradox that is created by your model, as does the online research. You haven't addressed these points in any way. You just keep insisting that your roulette model is equivalent, which it is not. The reason that they cannot be equivalent is that your roulette model does not create a paradox, where your application of it to the envelope problem does.

Even if you don't want to accept the existing models and solutions for this problem, surely you can admit that a model which creates a paradox cannot be correct.

 

[Meistro]

"you just keep insisting that your roulette model is equivalent, which it is not. "

I see, so in what way is the roulette model not 100% identical to the envelope problem as posed in my OP?

The probability of each payout is the same.

The payouts are the same.

The bet is the same.

You cannot logically say that switching in the envelope game has an EV of 0 and that playing the roulette game has an EV of 5/4. That is an illogical and contradictory position.

 

[sagefr0g]

..............

Optimal strategy is to switch some finite amount of times, that gives you a decent chance of an extreme result but not so many that you start to get bogged down by the law of large numbers.

I found this statement interesting with respect to other types of bets that are near break even expectation in the long run, initially, but alas, decided it's fundamentally flawed.
the law of large numbers allows for initial extreme results for both beginners luck newbie betters and beginners unlucky newbie betters, as well as bland luck newbie betters. even if one somehow knew one was a beginner luck newbie better (which is not possible to surmise) there is no way to determine the wished for optimal finite amount of times to push ones luck. but let's say the beginner luck newbie better did guess a nice finite number of times and landed up doing pretty well. hmm, then what about tomorrow for which he wants to repeat the process, and the next day and the next day,..... ? would he not become ever bogged down by the law of large number?

 

[gronbog]

"you just keep insisting that your roulette model is equivalent, which it is not. "

I see, so in what way is the roulette model not 100% identical to the envelope problem as posed in my OP?

The probability of each payout is the same.

The payouts are the same.

The bet is the same.

You cannot logically say that switching in the envelope game has an EV of 0 and that playing the roulette game has an EV of 5/4. That is an illogical and contradictory position.

The flaw above is your assertion that "the payouts are the same".

Once again you choose not to address or refute the arguments being made against your position and yet I have done so every time you made an argument or asked me a question. There is now enough information in this thread for intelligent people to read, understand and decide.

I have one final task here which is to present the necessary information to you directly, since you either stubbornly refuse to read it, or you have no cogent response to it.

For others, this information, and more, can be found (thanks to Colin) at

https://en.m.wikipedia.org/wiki/Two_envelopes_problem

First a summary of the switching argument and the paradox it creates for the general case of NOT looking at your envelope:

"The switching argument: Now suppose you reason as follows:

1. I denote by A the amount in my selected envelope.
2. The probability that A is the smaller amount is 1/2, and that it is the larger amount is also 1/2.
3. The other envelope may contain either 2A or A/2.
4. If A is the smaller amount, then the other envelope contains 2A.
5. If A is the larger amount, then the other envelope contains A/2.
6. Thus the other envelope contains 2A with probability 1/2 and A/2 with probability 1/2.
7. So the expected value of the money in the other envelope is:
   
   
8. This is greater than A, so I gain on average by swapping.
9. After the switch, I can denote that content by B and reason in exactly the same manner as above.
10. I will conclude that the most rational thing to do is to swap back again.
11. To be rational, I will thus end up swapping envelopes indefinitely.
12. As it seems more rational to open just any envelope than to swap indefinitely, we have a contradiction.

The puzzle: The puzzle is to find the flaw in the very compelling line of reasoning above. This includes determining exactly why and under what conditions that step is not correct, in order to be sure not to make this mistake in a more complicated situation where the misstep may not be so obvious. In short, the problem is to solve the paradox. Thus, in particular, the puzzle is not solved by the very simple task of finding another way to calculate the probabilities that does not lead to a contradiction."
---------------------------------

Now here is the simple resolution to the general problem.

"The total amount in both envelopes is a constant

, with

in one envelope and

in the other.
If you select the envelope with

first you gain the amount

by swapping. If you select the envelope with

first you lose the amount

by swapping. So you gain on average

by swapping.
Swapping is not better than keeping. The expected value

is the same for both the envelopes. Thus no contradiction exists."
----------------------------------------

And finally, we can address the variant in which you open your envelope and find $100 or, in this case, 100 euros:

Tsikogiannopoulos (2012)[10] presented a different way to do these calculations. Of course, it is by definition correct to assign equal probabilities to the events that the other envelope contains double or half that amount in envelope A. So the "switching argument" is correct up to step 6. Given that the player's envelope contains the amount A, he differentiates the actual situation in two different games: The first game would be played with the amounts (A, 2A) and the second game with the amounts (A/2, A). Only one of them is actually played but we don't know which one. These two games need to be treated differently. If the player wants to compute his/her expected return (profit or loss) in case of exchange, he/she should weigh the return derived from each game by the average amount in the two envelopes in that particular game. In the first case the profit would be A with an average amount of 3A/2, whereas in the second case the loss would be A/2 with an average amount of 3A/4. So the formula of the expected return in case of exchange, seen as a proportion of the total amount in the two envelopes, is:

This result means yet again that the player has to expect neither profit nor loss by exchanging his/her envelope.

We could actually open our envelope before deciding on switching or not and the above formula would still give us the correct expected return. For example, if we opened our envelope and saw that it contained 100 euros then we would set A=100 in the above formula and the expected return in case of switching would be:

-------------------------------------

This last section explains in detail every point I have made about both the general problem and the case of opening your envelope, which are:

1. You cannot use A, the value in your envelope, as both the large and small amount with the same value in your calculation of the EV of the other envelope
2. You need to analyse each possibility separately (two separate possible games) and weight each result by the average value (EV) of each envelope for that game
3. This is why the situation is not the same as that of the altered roulette game.

I have no more to say about this.

 

[Meistro]

Both payouts are the same, for $200 and for $50. The amount of sophistry you are willing to engage in to hide this rather elementary fact from yourself is quite impressive.

 

[Joe Mama]

Welcome to the flat earth society

 

[Meistro]

Let's review.

Roulette game, you have a 50% probability of gaining $100, and a 50% probability of losing $50.

In the envelope game you have a 50% probability of gaining $100, and a 50% probability of losing $50.

And yet, despite this, you would have me believe that one game is +EV and the other game is break even.

Your ad hom attacks are not an argument.

 

[gronbog]

Meisto, this is not my work, and it's not an attack. It's a presentation of the work of other minds on this well known problem.

It's not my goal to convince you. That will never happen. I just wanted the accepted solution, math and conclusion to be presented so that others can decide for themselves.

That has now been done.

 

[sagefr0g]

Let's review.

Roulette game, you have a 50% probability of gaining $100, and a 50% probability of losing $50.

In the envelope game you have a 50% probability of gaining $100, and a 50% probability of losing $50.

And yet, despite this, you would have me believe that one game is +EV and the other game is break even.

Your ad hom attacks are not an argument.

what is the probability of gaining $100, and what is the probability of losing $50 if you don't switch?

edit: this is not a rhetorical question. end edit

 

[London Colin]

1. You cannot use A, the value in your envelope, as both the large and small amount with the same value in your calculation of the EV of the other envelope
2. You need to analyse each possibility separately (two separate possible games) and weight each result by the average value (EV) of each envelope for that game
3. This is why the situation is not the same as that of the altered roulette game.

Could you clarify the rationale behind the part I've emboldened in point no. 2?

I don't doubt its validity, but as I mentioned a couple of times, it's the bit I'm struggling to get my head around.

 

[gronbog]

I just wanted to post to say that I am not ignoring or avoiding the question. I am currently on vacation and engaged in a few days of higher activity levels than during the previous days when we were discussing this more actively.

I will post my thoughts soon.

As a hint, re-read the Wikipedia article about how fixing the value in your envelope as A and then calculating the expected value of the other envelope using this fixed value rather than the expected value in envelope A in each case was the cause of the paradox. Then think about how this would be corrected if we still wanted to use a fixed value for A as is the case when we get to open our envelope.

 

[sagefr0g]

@Meistro :
consider the image above:

 

[gronbog]

Ok, I have some time to try and answer Colin's question. I hope I can do this clearly. I often have difficulty explaining things that are otherwise clear in my mind.

The Wikipedia article showed that the paradox in the initial problem, where we don't look into our envelope was caused by the incorrect use of an assumed fixed value A for the amount of money in envelope A when attempting to calculate the EV of envelope B.

E(B) = (1/2)2A + (1/2)A/2 = (5/4)A

This makes it appear that you only lose 1/2 as much when you swap while holding the envelope with the larger amount.

In reality, there are fixed amounts of money in play, x and 2x and we don't know which one we hold. The EV of both envelopes is (1/2)x + (1/2)2x = (3/2)x. Sagefrog's diagram show this graphically. There is no need to switch. Furthermore, you either gain or lose the amount x when swapping , so the EV of swapping is (1/2)x - (1/2)x = 0. Also note that x is a fixed fraction (1/3) of the total money in play (3x).

Sorry to rehash the dynamics of the original problem, but these concepts and especially the last point are needed for what follows.

For the case where we look into our envelope and actually find some amount A, we need to use the same principle, namely that we must work with the expected value of each envelope and that we win or lose a fixed fraction of the total amount of money in play.

We first realize that we are in one of two potential situations. Either the other envelope contains 2A (A + 2A = 3A is in play) or it contains A/2 (A + A/2 = 3/2A is in play).

First we try to calculate the EV of swapping as (1/2)A - (1/2)(A/2) = A/4

This has the same paradox issue as the initial attempt at the general problem. The actual amounts won or lost are fractions of the total amount of money in play. In the case where A is the smaller amount, there is A + 2A = 3A in play. In the other case, there is A + A/2 = 3A/2 in play. So the actual amounts won and lost are A/(3A) and (A/2)/((3A/2)) respectively. So the EV of switching is

(1/2)(A/(3A)) - (1/2)((A/2)/(3A/2)) = 0

This explains why the terms get different weights. There are different amounts of money in play in the two possible games. Note that the formula in the original problem is a simplification of this general formula in which both terms received the same weight, which gets cancelled out. This is because there was only one possible game, namely the one where one envelope contains x and the other contains 2x.

The final step is to note that it is the proportion of the weights that is important. Then consider that the average amount in each envelope is the total amount in play divided by 2. The two averages would be in the same proportion as the total amounts. So if we were to weight the terms by these averages, we would still get the same result. I'll leave that as an exercise. I don't know why the mathematician who created this solution used the averages instead of the total. It seems more complex to me.

 

[London Colin]

Thanks, Gronbog. That's very helpful.


I think we are probably coming to the end of what can be usefully said about all this, but I do have a couple more things that might be worth sharing...

Firstly, I've been trying to come up with some way of revealing the fallacy behind the paradox which is pictorial in some way - a mental model, not making any use of any confusing variables/formulae. I think I have it.

Consider all the possible scenarios that involve 100 being in one of the envelopes -

There are just two possible pairs of envelopes, and four possible transitions that can occur when one envelope is swapped for another. So we can concisely list every possibility:

scenario 1 -
pair : (50,100)
50 -> 100
100->50

scenario 2 -
pair : (100,200)
100 ->200
200->100

Someone observing the game can see at a glance that there is no value in switching -

• We are in either scenario 1 or scenario 2.
• Before an envelope is opened, we can see that in scenario 1 the player has a 50:50 chance of either gaining or losing 50, and in scenario 2 the player has a 50:50 chance of either gaining or losing 100.
• After an envelope is opened to reveal 100, the player, were he to know which scenario he is in, should keep the 100 in scenario 1, and should swap it for the 200 in scenario 2.

The player, unaware of which scenario he is in, reasons that the only two possibilities are those that are highlighted in bold, and thus he should switch, for an EV of 125.

But in thinking this way the player is blending the two distinct scenarios. He is effectively stealing some of the value of scenario 2, and applying it even though he might be in scenario 1.

 

[sagefr0g]

got to say, i'm thankful to Meistro for posting the two envelope problem. i'd never been aware of it before.
doubtless it's a vicious, malicious monstrous scenario, lol.
just me maybe, even if one knows the proper solution, it's easy to get mired in the wrong approach and its darn near impossible to convince one's self that it's wrong and then even when you realize the solution it's very difficult to understand how and why it's the correct way.
what really kind of scares me about a problem such as this, is the question of if and when one might run into a similar problem in the real world and just how difficult it would be to recognize it for what it is.

 

[gronbog]

I think that Colin is on the right track with his thinking that the paradox occurs because of one situation "stealing value" from the other . I further believe that it goes both ways. That is, the amount of the win is overstated and the amount of the loss is understated. This is hinted at in the Wikipedia article when it says:

A will be larger when A is larger than B, than when it is smaller than B. So its average values (expectation values) in those two cases are different. And the average value of A is not the same as A itself, anyway​

Also, if you examine the weights on each term, you will see that one of them reduces the value of its term while the other increases it.

 

[gronbog]

Sagefrog, I would not worry too much about running across something, like this in real life. These kinds of situations are carefully crafted and are unlikely to come up by accident. You would probably smell something fishy. Having said that, the Monty Hall problem on Norm's forum is a real life example of another non intuitive result. However, it could be argued that the situation there is far from typical.

 

[gronbog]

I would like to point out that there is a variant of this problem for which the switching argument is correct and it is exactly analogous to Meistros altered roulette game. From the Wikipedia article:

As pointed out by many authors,[10][11] the mechanism by which the amounts of the two envelopes are determined is crucial for the decision of the player to switch or not his/her envelope. Suppose that the amounts in the two envelopes A and B were not determined by first fixing contents of two envelopes E1 and E2, and then naming them A and B at random (for instance, by the toss of a fair coin; Nickerson and Falk, 2006). Instead, we start right at the beginning by putting some amount in Envelope A, and then fill B in a way which depends both on chance (the toss of a coin) and on what we put in A. Suppose that first of all the amount a in Envelope A is fixed in some way or other, and then the amount in Envelope B is fixed, dependent on what is already in A, according to the outcome of a fair coin. Ιf the coin fell Heads then 2a is put in Envelope B, if the coin fell Tails then a/2 is put in Envelope B. If the player was aware of this mechanism, and knows that they hold Envelope A, but don't know the outcome of the coin toss, and doesn't know a, then the switching argument is correct and he/she is recommended to switch envelopes. This version of the problem was introduced by Nalebuff (1988) and is often called the Ali-Baba problem. Notice that there is no need to look in Envelope A in order to decide whether or not to switch.​

The key point is that the filling of the second envelope is part of the mechanics of the single game being played based on the amount placed in the first envelope and that it is not filled before presenting the two envelopes to the player. Using a fair coin or Meistro altered roulette wheel would make no difference.

I dont know why one of the conditions is that the player doesn't know 'a'. I can't see how it makes a difference.