The Two-Envelope Fallacy- A probability paradox

Introduction

A paradox is a statement that is self-contradictory in nature but might seem to be correct or true at first. Paradoxes may also arise in probability. This happens because of the ambiguous nature in which the policy has been framed. Such a situation opens the door to numerous interpretations of the same problem which might seem paradoxical.

There are many paradoxes in probability theory that include the Newcomb’s paradox, the betting crowd paradox, the open box problem, the doomsday argument, the Hadron collider card game, Berkson’s paradox, sleeping beauty problem, two envelopes problem, Bertrand’s box paradox, the Monty Hall problem, the two boys problem or the boy-girl problem, three prisoner’s problem, etc.

Understanding the problem and the Paradox

Let us understand the two envelopes problem in detail and let us try to solve it. The two envelopes problem is one of the most popular problems or paradoxes from probability theory. According to the two envelopes problem, a person is given the liberty of choosing an envelope from the two given envelopes. It is given that one envelope contains double the money than in the other envelope. The person who is asked to pick one envelope is unaware of the amount of the money contained in the closed envelopes. But before choosing an envelope, the person is again provided with an opportunity or an option to reconsider his decision of which envelope to choose. Here comes the dilemma as a person may not be sure if he must consider his first choice or should he alter it. Let us consider this situation using variables to enhance our understanding. Let us assume that one envelope contains ‘a’ amount of rupees. Therefore, according to our assumption, the other envelope will contain either ‘2a’ or ‘a/2’ amount of money as we started with the assumption that one envelope contains double the amount of money than in the other. The probability that the chosen envelope contains ‘a/2’ amount of money is ½ and likewise is the probability for the chosen envelope to contain ‘2a’ amount of money. Therefore, the average expected amount of money the person will get for switching will be ½ (2a + a/2) which will be equal to 5a/4. Hence, the amount of money obtained is 5a/4 which is obviously larger than ‘a’ amount of money. Going by this philosophy or belief, the person must swap the envelopes before picking one. But here is where the confusing part starts as according to the belief stated above, the person must swap the envelopes again before choosing which one to select and open. Going by the same argument, the person may be tempted to swap the envelope a third time and so on. Going by this thought process, a person will get caught in an infinite loop of swapping and may end up getting no money at all.

What makes the Two-Envelop Paradox counterintuitive?

Where the fallacy comes in our reasoning? Where we go wrong? Well, the answer is that the belief or the philosophy as discussed above makes this condition counterintuitive because an ordinary individual will believe in swapping again and again as according to their belief, they must get ‘5a/4’ amount of money which is clearly higher than ‘a’ and so on and so forth. Now, this is the correct time for us to try and explore the possible solutions to this fallacy.

Possible solutions to the Paradox

                                    

                                            

In order to simplify the issue, let us consider that the envelope that the mentioned person selected at first was ‘X’ and the other one was ‘Y’. Now, let us assume that the ‘X’ letter contained ‘a’ amount of money. As we are unaware of the money in envelope ‘X’ as we haven’t the envelope, ‘a’ won’t be a fixed amount. It would rather be a random variable. Hence, it can take two values, either the lesser amount of money or the larger amount of money which is hidden in the two envelopes. Let’s assume ‘2b’ to be the larger amount and ‘b’ to be the lesser amount of money. As the person picks up the ‘X’ envelope randomly, there is a fifty-fifty chance that ‘X’ contains either of the two amounts. This implies that the expected money in envelope ‘X’ is E(X) = ½ (b + 2b) = 3b/2. It has been mentioned above that the amount of money in envelope ‘Y’ is E(Y) = ½ (2a + a/2).

As ‘a’ isn’t a fixed which was mentioned above; therefore, one of the two values can be taken. In the case that the ‘Y’ envelope contained ‘2a’, envelope ‘X’ would contain the smaller amount; therefore, a = 2b. Hence, in the equation E(Y) = ½ (2a + a/2), the first ‘a’ stands for ‘b’ and the second ‘a’ stands for ‘2b’. In reality, the two ‘a’ which has been used are unlike each other and cannot be simply added.

Similarly, E(Y) = ½ (2b + b).

Therefore, E(X) = E(Y). Hence, there is no added benefit or incentive in switching the envelopes and hence the paradox is solved and is no longer a paradox. Now let us try and examine what mistakes we all were making initially taking clues from the probability theory. 

Correcting the misapplication of the conditional probability theory

In our initial workings, we had assumed that the second choice is independent of the first choice, but in reality, this is not the case and the paradox arises because of this.

If the second envelope chosen by the player contains less money than the first it is guaranteed that the person would be swapping it for the one containing double the money (say ‘2a’ rupees) at a loss of ‘a’ rupee.

Similarly, if the second envelope chosen by the person contains more money than the first, it is guaranteed to be swapped with the one containing half the amount (say ‘a’ rupee) at a gain of ‘a’ amount of rupee.

Mathematically the expectation value of the second envelope will be represented as,

 E (Second envelope = Y)  = E (Y/ X<Y)P(X<Y) + E(Y/X>Y) P(X>Y)

Taking a numerical example to make understanding clearer.

Let’s consider that either of envelope X or Y contains 100 rupees and the other contains 200 rupees. We don’t know exactly which one contains how much amount of money. Therefore, if X contains 100 rupees Y necessarily contains 200 rupees, not 50 rupees.

Now the expected value of the second envelope can be written as-

Probability(X=100) *100 + probability of (X=200) *200

OR expectation value= ½ * 100+ ½ * 200 = 150

Hence the expectation value is exactly in the middle of the range between the values 100 and 200 as you would expect it to be intuitively. Hence you should not bother swapping.

We can understand it more intuitively by taking the total amount of money involved to be 300 rupees regardless of which envelope was chosen first. But we can say for sure that the larger envelope contains 200 rupees [2(Total)/3] and the smaller envelope contains 100 rupees.

Hence the expected return from switching is equal to ½ (200-100) + ½ (100-200), making it equal to zero. Therefore, there is no advantage or disadvantage in switching the envelopes which is exactly as expected. 

Other Strategies to Solve the Two-Envelope Paradox

                                     

                                       

Randomized switching

The key to the problem occurs when the contestant opens one of the envelopes, after this information is broken out then these envelopes are not identical anymore. Researchers use the ‘cover strategy’ to explain the phenomenon which proves the players’ chance of choosing the envelope with the greater amount say 2a if played repeatedly.

The idea here is that the player would continue to switch keeping in mind the money in the first envelope. This means that a lesser amount of random switches will lead to a larger amount even though he does not know how high or low the amount is distributed among the envelopes. This cover strategy report proved that after 20000 simulations, people who followed the cover strategy had an increase in their payoff than other players who just did simple switching.

The scientists clarified that the procedure rises up out of late advances in two-state exchanging marvels that are rising in the fields of material science, designing, and financial matters. It is this thought which is behind the rule of the notable 'Parrondo's oddity, which shows that you can blend two losing games but then win.

 This answer for the two-envelope issue is a forward leap in the field of Parrondo's mystery." between two insecure states can bring about a steady condition. 

As D.Abbott stated that a Brownian ratchet is a physical gadget that can arrange arbitrary particles to stream a specific way. He said,

"The stunt with a Brownian ratchet is that again it utilizes breaking balance, It is this thought behind the rule of the notable 'Parrondo's oddity,' which shows that you can blend two losing games but then win. This answer for the two-envelope issue is a forward leap in the field of Parrondo's mystery."

Winning while losing (Strategy to solve the new envelope problem)

Although a player can utilize the arbitrary changing methodology to win cash while having earlier information on the measurable circulation of the envelopes' qualities, the huge point is that this information isn't vital. It is astounding that the examination shows that one can generally improve their benefit utilizing Cover's technique with the obliviousness of 'as far as possible' (the most noteworthy estimation of cash permitted) and of the factual conveyance the numbers comply. Also, the explanation it is of significance is that engineers regularly need to consider what are called 'daze advancement' issues. Thus the answer may invigorate a new workaround there.

Another kind of improvement technique that imparts likenesses to the two-envelope issue is monetary putting resources into the securities exchange. For example, in "instability siphoning," exchanging between helpless ventures can bring about winning an exponentially expanding measure of cash. 

Instability siphoning is a 'toy model' that one cannot utilize precisely in its current structure on the financial exchange. It is a toy model that shows hidden systems that are valuable. It proposes the intensity of changing your arrangement of stocks occasionally, purchasing low, and selling high. Both the two-envelope process in addition to unpredictability siphoning show up firmly identified with Brownian ratchet marvels. The two of them misuse the communication of asymmetry with irregularity.

This understanding likewise carries with it various open inquiries. For instance, when playing a group of games, a player could adjust the subtleties of the technique by ceaselessly refreshing the evaluated circulation from which the envelopes' qualities are picked. Likewise, since the system identifies with two-state exchanging in different fields, maybe it might be conceivable to clarify every one of these wonders with a typical scientific structure.

Conclusion

Well, quite frankly we managed to cover the most basic, simple, and lucid explanation of the paradox which is otherwise quite complex and greatly discussed and deliberated upon in academia. There are numerous explanations and possible solutions to this celebrated paradox of the probability theory called 'The- two Envelope Fallacy'. 

Our discussion also persuaded us to realize the importance of probability theory concepts such as Bayes' Theorem, Conditional Probability, Prior and Posterior concepts of the probability theory. If anyone of you got motivated and the inner curiosity got ignited, please read more on these probability paradoxes. As it is evident from our discussion, there are numerous applications of these paradoxes, so it becomes very important to know and enjoy these beautiful and tricky concepts. 💡

Authors-

Aditya Nariyal 1933406

Aditya Pratap Verma 1933407

G. M. Shibi 1933416

John Joseph 1933420

References

Agnew, R. (2004). On the two box paradox . Math Magazine, 302- 308.

F. Jackson, P. M. (1994). The two Envelopes paradox. 43-45.

M. Clark, N. S. (2000). The two envelope paradox. 415- 442.

Paradox. (n.d.). Retrieved from https://www.merriam-webster.com/dictionary/paradox

Probability, P. i. (n.d.). Retrieved from https://brilliant.org/wiki/paradoxes-in-probability/

solved, T. t. (2018, 06 22). Retrieved from https://plus.maths.org/content/two-envelopes-problem- resolution

Looking Glass Universe. (2017, January 29). Resolution of the two envelope fallacy. [Video file]. Retrieved from https://www.youtube.com/watch?v=FXNKWxwcX9U.