Two envelopes problem

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The two envelope problems , also known as paradoxical exchanges , are brain teasers, puzzles, or paradoxes in logic, possibilities, and recreational math. This is of particular interest in decision theory, and for the Bayesian interpretation of probability theory. Historically, it emerged as a variant of the tie paradox. The problem is usually introduced by formulating a hypothetical challenge of the following types:

It seems clear that there is no point in flipping the envelope because the situation is symmetrical. However, since you stand to earn twice as much money if you switch while risking only lose half of what you have today, it is possible to argue that it is more useful to switch. The problem is showing what is wrong with this argument.

Video Two envelopes problem

Introduction

Issues

Basic settings : You are given two indistinguishable envelopes, each containing a positive amount of money. One envelope contains twice as many as the other. You can select one envelope and save any amount. You select one envelope randomly but before you open it you are given the opportunity to take another envelope instead.

Redirect arguments : Now suppose you reasoned as follows:

Puzzle : The puzzle is to find the flaw in the very interesting line of reasoning above. This includes specifying exactly why and under what conditions is not correct, to make sure not to make this error in a more complicated situation where the error may not be so clear. In short, the problem is solving the paradox. Thus, in particular, this puzzle is not completed by a very simple task to find another way to calculate probabilities that do not lead to contradictions.

Maps Two envelopes problem

Solution

Many solutions to resolve the paradox have been presented. The underlying probability theory of this problem is well understood, and any paradox that appears generally because it treats what is actually a conditional probability as unconditional probability. A large number of similar formulations of the paradox are possible, and have produced much literature on this subject.

The issue version continues to spark interest in the field of philosophy and game theory.

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Simple resolution

Step 7 assumes that the second option does not depend on the first option. This is a mistake and this is the source of a real paradox.

The general way to resolve the paradox, both in popular literature and in academic literature, especially in philosophy, is to assume that 'A' in step 7 is intended to be the expected value in envelope A and we mean to write down the formula for the expected value in envelope B.

Step 7 states that the expected value in B = 1/2 (2A A/2)

It is shown that the 'A' in the first part of the formula is the expected value, given that the A envelope contains less than envelope B, but 'A', in the second part of the formula is the expected value in A, given that the A envelope contains more than the envelope B. The flaw in the argument is that the same symbol is used with two different meanings in both parts of the same calculation but it is assumed to have the same value in both cases.

The correct calculations are: Expected value in B = 1/2 (expected value in B, given A greater than B) (Expected value in B, given A smaller than B))

If we then take the sum in one envelope to x and the other to 2x, the calculation of the expectation value becomes:

Expected value in B = 1/2 ( x 2 x )

which is equal to the expected amount in A.

In non-technical language, what is wrong (see Necktie paradox) is that, in the provided scenario, mathematics uses the relative values ​​of A and B (ie, assumes that someone will earn more money if A is less than B than someone will lose if otherwise true). However, both values ​​are fixed (one envelope contains, say, $ 20 and the other $ 40). If envelope values ​​are restated as x and 2 x , it would be much easier to see that, if A is larger, someone will lose x by switching and, if B is larger, someone will get x by switching. One does not really get a larger amount of money by switching because the total T of A and B (3 x ) remains the same, and the difference is x fixed to T/3 .

A will become larger when A is greater than B, than when it is smaller than B. So the average value (expectation value) in both cases is different. And the average value of A is not the same as A itself. Two mistakes are being made: the writer forgets he's taking the value of hope, and he forgets he's picking up the value of hope in two different conditions.

It would be easier to calculate E (B) directly. Showing lower than two numbers with x , and picked it up for repair (even if unknown) we found that

${\ displaystyle \ operatorname {E} (B) = {\ frac {1} {2}} 2x {\ frac {1} {2}} x = {\ frac {3} {2}} x} Â Â$

We learn that 1.5 x is the expected value of the sum in Envelope B. With the same calculation it is also the expected value of the sum in Envelope A. They are the same so there is no reason to prefer one envelope to another. This conclusion, of course, is clear beforehand; The point is we identified the wrong step in the argument to switch by explaining exactly where the calculations are made there out of the rails.

Kami juga dapat melanjutkan dari yang benar tetapi sulit untuk menginterpretasikan hasil pengembangan di baris 7:

${\ displaystyle \ operatorname {E} (B) = \ operatorname {E} (A \ pertengahan A & lt; B) {\ frac {1} {4}} \ operatorname {E} (A \ mid A & gt; B) = x {\ frac {1} {4}} 2x = {\ frac {3} {2}} x}  ÂÂ$

so (of course) different routes to calculate the same thing all give the same answer.

Tsikogiannopoulos (2012) presents a different way of doing this calculation. Of course, this is the correct definition to assign the same probability to an event where another envelope contains two or a half of that number in envelope A. So the "switching argument" is correct until step 6. Given that the envelope of the player contains an A number, it distinguishes the situation actual in two different games: The first game will be played with the number (A, 2A) and the second game with the amount (A/2, A). Only one really played but we did not know which one. Both of these games should be treated differently. If the player wants to calculate the expected return (gain or loss) in exchange, he/she should weigh the returns coming from each game by the average amount in the two envelopes in a particular game. In the first case, the profit would be A with an average amount of 3A/2, whereas in the case of both losses would be A/2 with an average amount of 3A/4. So the expected return formula in terms of exchange, seen as a proportion of the total amount in two envelopes, are:

${\ displaystyle E = {\ frac {1} {2}} \ cdot {\ frac {A} {3A/2}} {\ frac {1 } {2}} \ cdot {\ frac {-A/2} {3A/4}} = 0}$

This result means again that the player should expect no profit or loss by redeeming the envelope.

We can actually open our envelopes before deciding to switch or not and the formula above will still give us the expected results correctly. For example, if we open our envelope and see that it contains 100 euros then we will set A = 100 in the above formula and the expected return in case of transition is:

${\ displaystyle E = {\ frac {1} {2}} \ cdot {\ frac {100} {150}} {\ frac {1} { 2}} \ cdot {\ frac {-50} {75}} = 0}$

Nalebuff asymmetric variant

As indicated by many authors, the mechanism by which the number of two envelopes is determined is crucial for a player's decision to switch or not to envelope. Suppose that the sums in the two envelopes A and B are not determined by first fixing the contents of the two envelopes E1 and E2, and then naming A and B randomly (eg, with a fair coin toss, Nickerson and Falk, 2006). Instead, we start from scratch by entering a number in Envelope A, and then fill B in a way that is dependent on coincidence (a coin toss) and on what we put into A. Suppose that first the amount of a in Envelope A remains in some way or other, and then the sum in Envelope B is fixed, depending on what is already in A, corresponding to the result of a fair coin. ? f coins fall Head then 2 a put in Envelope B, if coins fall Tail then a /2 inserted in Envelope B. If player aware of this mechanism, and know that they hold Envelope A, but do not know the result of the coin toss, and do not know a , then the switching argument is correct and he recommended to switch the envelope. This version of the problem was introduced by Nalebuff (1988) and is often called the Ali-Baba problem. Note that there is no need to look in Envelope A to decide whether to switch or not.

Many more variants of the problem have been introduced. Nickerson and Falk (2006) systematically surveyed a total of 8.

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Bayesian Resolution

The above simple resolution assumes that the person who finds the argument to switch tries to calculate the expected number value in Envelope A, thinks of two numbers in the envelope as fixed ( x and 2 x ). The only uncertainty is which envelope has a smaller number of x . Yet many mathematicians and statisticians interpret the argument in an attempt to calculate the expected number in Envelope B, given the actual or hypothetical number "A" in Envelope A. (A mathematician would also prefer to use the symbol a to stand for a possible value, leave the symbol A for a random variable). One does not need to look inside the envelope to see how many there are, to do the calculations. If the calculation result is a suggestion to replace the envelope, whatever the number may be there, it will appear that someone has to switch, without looking. In this case, in Steps 6, 7 and 8 of the reason, "A" is the fixed value of the sum of money in the first envelope.

This interpretation of two envelope problems appears in the first publication in which paradoxes are introduced in present-day form, Gardner (1989) and Nalebuff (1989). It is common in the more mathematical literature on the subject. This also applies to modification of the problem (which seems to have started with Nalebuff) where the owner of Envelope A actually looks inside the envelope before deciding whether to switch or not; although Nalebuff also stressed that it is not necessary to have Envelope A owner in its envelope. If he imagines looking in it, and if for whatever amount he can imagine being there, he has an argument to switch, then he will decide to switch too. Finally, this interpretation is also the essence of an earlier version of two envelope problems (Littlewood's switching paradox, SchrÃÆ'¶dinger, and Kraitchik); see the concluding section, on TEP history.

This kind of interpretation is often called "Bayesian" because it assumes the author also combines the previous probability distribution of the possible amount of money in the two envelopes in the switching argument.

Simple form of Bayesian resolution

A simple resolution depends on a particular interpretation of what the author is trying to calculate: that is, he assumes that he is pursuing the (unconditional) value of what is in Envelope B. In the mathematical literature on Two Problems of Envelope, a different interpretation. more generally, involves a conditional expectation value (depending on what's possible in Envelope A). To overcome this and related interpretations or versions of the problem, most authors use Bayesian interpretations of probabilities, meaning that probability reasoning applies not only to random events such as random sampling of an envelope but also to our knowledge (or lack of knowledge) of fixed but unknown things, such as two numbers originally placed in two envelopes, before one is selected randomly and is called "Envelope A". In addition, according to the long tradition of going back at least to Laplace and its principle of reason which is not enough one should give the same probability when one has no knowledge at all about the possible values ​​of some quantity. Thus the fact that we are not told anything about how filled envelopes can already be converted into probability statements about these numbers. No information means the same probability.

In steps 6 and 7 of the switching argument, the author envisages that Envelope A contains a certain amount of a , and then seems to believe that given that information, another envelope would have the same possibility to contain twice or half of that amount. The assumption can only be true, if before knowing what is in Envelope A, the author will consider the following two pairs of values ​​for both possible envelopes: number of a /2 and a ; and the number of a and 2 a . (This follows from Bayes' rule in the form of odds: posterior odds equal to the odds of the previous likelihood probability ratio). But now we can apply the same reason, imagine not a but a/2 in Envelope A. And also, for 2 a . Similarly, ad infinitum, repeatedly dividing or repeatedly doubling as many times as you like. (Falk and Konold, 1992).

Suppose for the sake of an argument, we begin by imagining the number 32 in Envelope A. For the reasons in steps 6 and 7 true whatever amount occurs in Envelope A, we seem to believe in Advancing that all the following ten sums are all equally likely to be smaller than two numbers in two envelopes: 1, 2, 4, 8, 16, 32, 64, 128, 256, 512 (possibly power 2: Falk and Konold, 1992). However, even larger or even smaller quantities, the "same" assumption is starting to seem rather unreasonable. Suppose we stop, with just ten possible possibilities for smaller amounts in two envelopes. In this case, the reasons in steps 6 and 7 are absolutely correct if the envelope A contains a sum of 2, 4,... 512: the switching envelope will give an expected (average) gain of 25%. If the envelope A contains the number 1, then the actual expected gain is 100%. But if it happens to contain the amount of 1024, a big 50% loss (of a rather large amount) will occur. It only happens once in twenty times, but it's pretty right to balance the expected earnings in 19 others out of 20 times.

As an alternative, we are not continuing the infinitum but now we are working with pretty ludicrous assumptions, implying for example, that it is much more likely to amount in envelopes A to smaller than 1, and much more likely than larger 1024, rather than between the two values. This is called an improper previous distribution: the probability calculus is broken; expectations are not even defined; see Falk and Konold and (1982).

Many authors also point out that if the maximum amount that can be incorporated into envelopes with smaller numbers exists, then it is very easy to see that Step 6 is broken, because if the player holds more than the maximum amount that can be put into their "smaller" envelope should hold an envelope that contains a larger amount, and thus definitely lose by switching. This may not happen often, but when it does, the huge losses incurred by players means that, on average, there is no advantage in switching. Some authors assume that this solves all the practical cases of the problem.

But problems can also be solved mathematically without assuming maximum numbers. Nalebuff (1989), Christensen and Utts (1992), Falk and Konold (1992), Blachman, Christensen and Utts (1996), Nickerson and Falk (2006), suggest that if the amount of money in two envelopes has a probability distribution representing the player's previous confidence about the amount of money in two envelopes, it is unlikely that whatever amount of A = a in the first envelope is possible, it will be equally possible, according to previous belief, that the second contains a /2 or 2 a . So step 6 of the argument, which leads to always switch , is a non-sequitur, also when there is no maximum number in the envelope.

Introduction to further developments in relation to Bayesian probability theory

The first two resolutions discussed above ("simple resolution" and "Bayesian resolution") correspond to two possible interpretations of what happens in step 6 of the argument. They both consider that step 6 is indeed a "bad move". But the description in step 6 is ambiguous. Is the author after the unconditional (as a whole) hope value of what is in the envelope B (possibly - depending on the smaller number, x ), or is he after the conditional expectation of what is inside envelope B, given possible amount of a in envelope A? Thus, there are two main interpretations of the composer's intention of the paradoxical argument for switching, and two main resolutions.

The great literature has evolved about the variant of the problem. The standard assumption on how to install envelopes is that some money is in one envelope, and twice that amount is in another envelope. One of the two envelopes is randomly assigned to the player ( envelope A ). The original problem does not explain clearly how the smaller number of two numbers is determined, what value may be taken and, in particular, whether there is a minimum or maximum number that it may contain. However, if we use the Bayesian probability interpretation, then we begin by revealing our previous beliefs about the smaller number in the two envelopes through the probability distribution. Lack of knowledge can also be expressed in terms of probability.

The first variant in the Bayesian version is to generate precise precise probability distributions of smaller amounts of money in two envelopes, so that when Step 6 is done correctly, the suggestion still prefers Envelope B, whatever may be in Envelope A. So even if the specific calculations performed in step 6 are incorrect (no precise previous distribution so, given what is on the first envelope A, the other envelope always tends to be greater or smaller) the correct calculation, depending on what we previously use, does it lead to ${\ displaystyle E (B | A = a) & gt; a}$ for all possible values ​​of a .

In this case it can be shown that the expected number in both envelopes is unlimited. No gain, on average, in swapping.

Second math variation

Although Bayesian probability theory can solve the first mathematical interpretation of the above paradox, it turns out that examples can be found from the proper probability distribution, so the expected value of the sum in the second envelope recalls that in the first one does not exceed the number in the first, whatever it is. The first such example was given by Nalebuff (1989). See also Christensen and Utts (1992).

Write again the amount of money in the first envelope by A and the second by B . We consider this to be random. Let X be smaller than the two numbers and Y = 2X become larger. Note that after we fix the probability distribution for X then the combined probability distribution A, B is fixed, since A, B = X, Y or Y, X each with a 1/2 probability, regardless of X, Y .

The bad step 6 in the "always switch" argument brings us to the findings of E (B | A = a) & gt; a for all a , and hence to the recommendation to switch, do we know a . Now, it turns out that one can easily find the right probability distribution for X , the smaller of the two amounts of money, so that this poor conclusion is still true. One example is analyzed in more detail, one day.

As mentioned previously, it is not possible to be true that anything a , given A = a , B /i> i>/2 or 2 a , but it may be true that anything a is given A = a is greater than the expected value of a .

Suppose for example (Broome 1995) that smaller envelopes actually contain 2 dollars ⁿ with probability 2 ⁿ /3 ^{n 1} where n = 0, 1, 2,... The number of these probabilities to 1, then the distribution is the prior (for the subjectivist) and the law of probability is really good also for the frequentists.

dan akibatnya probabilitasnya adalah pasangan {2, 4} adalah 2/5, karena ini adalah dua kemungkinan. Dalam derivasi ini, ${\ displaystyle P (\ {1,2 \})/2}  ÂÂ$ adalah probabilitas bahwa pasangan amplop adalah pasangan 1 dan 2, dan Amplop A terjadi mengandung 2; ${\ displaystyle P (\ {2,4 \})/2}  ÂÂ$ adalah probabilitas bahwa pasangan amplop adalah pasangan 2 dan 4, dan (lagi) Envelope A terjadi mengandung 2. Itu adalah hanya dua cara yang Envelope A dapat berakhir mengandung jumlah 2.

It turns out that this proportion applies in general unless the first envelope contains 1. Marked with a the amount we imagine find in Envelope A, if we open the envelope, and suppose it is a = 2 ⁿ for some n > = 1. In this case another envelope contains a /2 with a probability of 3/5 and 2 a with a probability of 2/5.

So either the first envelope contains 1, in this case the expected number of conditionals in the other envelope is 2, or the first envelope contains a & gt; 1, and although the second envelope is smaller than the larger, the expected number is conditionalally greater: the expected number of conditionals in Envelope B is

${\ displaystyle {\ frac {3} {5}} {\ frac {a} {2}} {\ frac {2} {5}} 2a = {\ frac {11} {10}} a}$

which is more than a . This means that the player seen in Envelope A will decide to replace whatever he sees there. Therefore it is not necessary to see Envelope A to make that decision.

This conclusion is as clear as one of the earlier interpretations of Problem Two Envelopes. But now the deficiencies mentioned above do not apply; a in the calculated expected value is constant and the conditional probability in the formula is obtained from the preceding and precise defined distribution.

Proposed resolution through mathematical economics

If we do not look into the first envelope, then obviously there is no reason to switch, because we will exchange an unknown amount of money ( A ), whose value is expected to be unlimited, for the other unknown amount money ( B ), with the same probability distribution and unlimited expected value. However, if we look into the first envelope, then for all values ​​observed ( ${\ displaystyle A = a}$ ) we want to switch because ${\ displaystyle E (B | A = a) & gt; a}$ for all a . As David Chalmers (2002) noted, this problem can be described as a failure of domination reasoning.

Di bawah penalaran dominasi, fakta bahwa kami sangat memilih A untuk B untuk semua nilai yang mungkin diamati a harus menyiratkan bahwa kami sangat memilih A ke B tanpa memperhatikan a ; namun, seperti yang telah ditunjukkan, itu tidak benar karena ${\ displaystyle E (B) = E (A) = \ infty}  ÂÂ$ . Untuk menyelamatkan penalaran dominasi sambil mengizinkan ${\ displaystyle E (B) = E (A) = \ infty}  ÂÂ$ , seseorang harus mengganti nilai yang diharapkan sebagai kriteria keputusan, sehingga menggunakan argumen yang lebih canggih dari ekonomi matematika.

Beberapa penulis lebih suka berpendapat bahwa dalam situasi kehidupan nyata, ${\ displaystyle u (W A)}  ÂÂ$ dan ${\ displaystyle u (W B)}  ÂÂ$ dibatasi hanya karena jumlah uang dalam amplop dibatasi oleh jumlah total uang di dunia ( M ), menyiratkan ${\ displaystyle u (W A) \ leq u (W M)}  ÂÂ$ dan ${\ displaystyle u (W B) \ leq u (W M)}  ÂÂ$ . Dari perspektif ini, paradoks kedua diselesaikan karena distribusi probabilitas yang dipostulasikan untuk X (dengan