The exchange paradox (or envelope paradox ) describes a special mathematical situation in which the naive calculation with expected values , in particular the application of the principle of indifference , leads to a contradiction to common sense . It is similar to the two-slip game and the goat problem , but this has a different probabilistic background.

history

The exchange paradox can be traced back at least to 1953 and was formulated as follows in a book by the Belgian mathematician Maurice Kraitchik :

“Two equally wealthy people meet and want to compare the contents of their wallets. Nobody knows how much money the other has in the exchange. They agree on the following game: The one who has less money in the purse wins the contents of the other's purse. If both have the same amount, everyone keeps their money. Now one of the two could make the following thought: “Suppose I have the amount A in my purse. Then that's the maximum I can lose. However, with probability 0.5 I win and have then more than 2 A . So the game is good for me. ”The other could argue the same way. For reasons of symmetry, the game has to be fair. What is the fallacy of this argument? "

Martin Gardner spread the riddle in 1982 in his book Aha! Gotcha , also in the form of a wallet game. The current form with the two envelopes was formulated in 1989 by Barry Nalebuff .

The exchange situation

Mr Lemke would like to give Mr Schmidt a present and gives him two envelopes with the words “I'm giving you one of these envelopes. In both there is an amount of money, in one twice as much as in the other. You can open an envelope and then decide which of the two envelopes to take. "

Mr. Schmidt opens a randomly selected one of the two envelopes, finds 100 euros, for example, and thinks: “I have 100 euros in this envelope. If I trade, there is a 50% probability that I have 200 euros and the same probability that I have 50 euros. This makes an expected value of 125 euros. "

${\ displaystyle 125 = 0 {,} 5 \ cdot 50 + 0 {,} 5 \ cdot 200}$

After this consideration, the swap would be worthwhile.

The following consideration allegedly takes Mr. Schmidt's bill ad absurdum : If Mr. Schmidt's bill for any amount yielded the result that exchanging is worthwhile, he would not have to open the envelope at all, but could take the other envelope straight away. But it cannot be that the other envelope is always better, since both envelopes are obviously of equal value before opening.

The thought trap

A possible thought trap is that Mr. Schmidt uses either the principle of indifference or the conditional probability incorrectly, i.e. assumes that the 100 euros represent half or double the amount with a 50-50 probability. Depending on Mr Lemke's selection process, this may be correct for this amount, but not for all amounts. First of all, there is actually a 50% probability that the envelope with the smaller or larger amount will be opened. It is therefore either the 50/100 or the 100/200 euro combination. However, it cannot be concluded from this that the probabilities for the two combinations are the same under the condition that 100 euros have been found. Nothing is known about the probabilities of these cases, and the principle of indifference, based on a revealed amount, cannot be  applied to the events “double amount” ( ) and “half amount” ( ) for fundamental reasons. In the Think Trap Collection, this is illustrated by a few calculation examples. ${\ displaystyle x}$${\ displaystyle 2x}$${\ displaystyle {\ tfrac {1} {2}} x}$

On the other hand, it is entirely possible that, based on the value of the opened envelope, the conditional expected value of the unopened envelope is always higher; but only if the expected value of the unopened envelope is greater than the expected value of the opened envelope or if both expected values ​​are infinite.

Conditional Probability Analysis

The calculation with a 50-50 probability thus arises from an inadmissible application of the principle of indifference. However, the calculation of the expected value can also lead to the apparent contradiction with other probabilities that an exchange would always be indicated. In order to generally make a meaningful exchange decision, Mr. Schmidt must keep in mind that he must use conditional probabilities when including the amount in the opened envelope in his invoice. The following section contains a formal analysis of whether, given a given probability distribution, an exchange can make sense for certain amounts and whether there can even be a probability distribution in which an exchange is always indicated. Whether this probability distribution is a subjective assessment by Mr. Schmidt or whether a probability distribution of the amounts is actually known is irrelevant for the analysis. For example, the following notation can be used for this:

• the random variable denotes the smaller amount in the envelopes (the amount is then in the other envelope ).${\ displaystyle Z> 0}$${\ displaystyle 2Z}$
• the random variable denotes the amount that Mr. Schmidt finds in the envelope that was opened first.${\ displaystyle X}$
• the random variable denotes the amount that is in the other, still unopened envelope.${\ displaystyle Y}$

Since both envelopes are chosen equally likely, have and the same distribution , but are due to each other stochastically dependent . The following applies to the expected values . Mr. Schmidt obviously wants to calculate the expectation of if he knows, i.e. the conditional expectation . ${\ displaystyle X}$${\ displaystyle Y}$${\ displaystyle X + Y = 3Z}$${\ displaystyle EX = EY = {\ tfrac {3} {2}} EZ}$${\ displaystyle Y}$${\ displaystyle X}$ ${\ displaystyle E (Y \ mid X)}$

Discrete distributions

First of all, the case should be treated that (and thus also and ) has a discrete distribution . In this case, denote the probability that the smaller amount in the envelopes is the same . For the distribution of and then follows ${\ displaystyle Z}$${\ displaystyle X}$${\ displaystyle Y}$${\ displaystyle p_ {n} = P (Z = n)}$${\ displaystyle n}$${\ displaystyle X}$${\ displaystyle Y}$

${\ displaystyle P (X = n) = P (Y = n) = {\ frac {p_ {n}} {2}} + {\ frac {p_ {n / 2}} {2}} \ ,.}$

In the discrete case, the required conditional expectation applies

${\ displaystyle E (Y \ mid X = n) = \ sum _ {j} y_ {j} P (Y = y_ {j} \ mid X = n) \ ,.}$

The conditional probability is only different, according to Problem of zero if either or . In these cases, the following applies to the probability that Mr. Smith will find twice the amount in the other envelope: ${\ displaystyle P (Y = y_ {j} \ mid X = n)}$${\ displaystyle y_ {j} = 2n}$${\ displaystyle y_ {j} = n / 2}$

{\ displaystyle {\ begin {aligned} P \ left (Y = 2n \ mid X = n \ right) & = P \ left (Z = n \ mid X = n \ right) \\ & = {\ frac {P \ left (Z = n {\ text {and}} X = n \ right)} {P \ left (X = n \ right)}} \\ & = {\ frac {{\ frac {1} {2} } P \ left (Z = n \ right)} {{\ frac {P \ left (Z = n \ right)} {2}} + {\ frac {P \ left (Z = n / 2 \ right)} {2}}}} \\ & = {\ frac {p_ {n}} {p_ {n} + p_ {n / 2}}}. \ End {aligned}}}

The conditional probability that Mr. Schmidt will find half the amount in the other envelope is complementary to this, so the following applies:

${\ displaystyle P \ left (Y = n / 2 \ mid X = n \ right) = 1 - {\ frac {p_ {n}} {p_ {n} + p_ {n / 2}}} = {\ frac {p_ {n / 2}} {p_ {n} + p_ {n / 2}}}.}$

If the expected value of the distribution exists with the correct probabilities, one would get

${\ displaystyle E: = E (Y \ mid X = n) = 2n {\ frac {p_ {n}} {p_ {n} + p_ {n / 2}}} + {\ frac {n} {2} } {\ frac {p_ {n / 2}} {p_ {n} + p_ {n / 2}}} = {\ frac {4p_ {n} + p_ {n / 2}} {2 \ left (p_ { n} + p_ {n / 2} \ right)}} n.}$

Accordingly, swapping would pay off if and only if ; this is the case if and only if applies. Distributions that fulfill this condition for all possible ones can be constructed, but then have no finite expectation value. For such a priori distribution, the given advantage of the exchange decision for each value found in the opened envelope is not intuitive, but not paradoxical. ${\ displaystyle E> n}$${\ displaystyle p_ {n}> {\ tfrac {p_ {n / 2}} {2}}}$${\ displaystyle n}$${\ displaystyle Z}$

Of course, the assumption that there can be any amount of money in the envelope contradicts the practical restriction that no one, including Mr. Lemke, has any amount of money available.

example

If one assumes a probability distribution with which Mr. Lemke distributes the money in the envelopes, the situation can be simulated very well. For example, suppose he determines the amount by rolling a fair dice. If the dice shows eyes, he puts euros in one envelope and euros in the other. Mr. Schmidt will then find the amount 25 euros in the envelope with probability, one of the amounts 50, 100, 200, 400 or 800 euros each and again with the probability 1600 euros. If he does not exchange, the expected value of the money gift is therefore ${\ displaystyle k}$${\ displaystyle 2 ^ {k-1} \ cdot 25}$${\ displaystyle 2 ^ {k} \ cdot 25}$${\ displaystyle {\ tfrac {1} {12}}}$${\ displaystyle {\ tfrac {2} {12}}}$${\ displaystyle {\ tfrac {1} {12}}}$

${\ displaystyle {\ tfrac {1} {12}} \ left (25 + 2 \ cdot 50 + 2 \ cdot 100 + 2 \ cdot 200 + 2 \ cdot 400 + 2 \ cdot 800 + 1600 \ right) = 393 { ,} 75 {\ text {Euro.}}}$

If Mr. Schmidt exchanges in any case, his expected value does not change, since he also exchanges the amount of 1600 euros in particular, although in this case he cannot win anything. However, if Mr. Schmidt suspects that there is hardly more than 1000 euros in the envelope and therefore decides to exchange if and only if there is a maximum of 500 euros in the envelope, the probabilities change: After the exchange, Mr. Schmidt then still has with probability the amount of 25 euros in the envelope, also with probability each of the amounts 50, 100 or 200 euros, the amount of 400 euros however only with probability (since Mr. Schmidt no longer exchanges at 800 euros), but with probability the Amount of 800 euros and again with a probability of 1600 euros. The expected value of the money gift is now ${\ displaystyle {\ tfrac {1} {12}}}$${\ displaystyle {\ tfrac {2} {12}}}$${\ displaystyle {\ tfrac {1} {12}}}$${\ displaystyle {\ tfrac {3} {12}}}$${\ displaystyle {\ tfrac {1} {12}}}$

${\ displaystyle {\ frac {1} {12}} \ left (25 + 2 \ cdot 50 + 2 \ cdot 100 + 2 \ cdot 200 + 1 \ cdot 400 + 3 \ cdot 800 + 1600 \ right) = 427 { ,} 08 {\ text {Euro.}}}$

If Mr. Schmidt assesses the situation better and decides not to swap until 1000 euros or more, he can even increase the expected value to 460.62 euros; but if it becomes too greedy and exchanges up to 2000 euros, for example, it falls back to the initial value of 393.75 euros.

It is of course difficult for Mr Schmidt to assess Mr Lemke correctly; What is essential, however, is that the paradox disappears as soon as one assumes any concrete probability distribution. Depending on Mr. Schmidt's exchange strategy, the expected value of the monetary gift changes; however, the “always exchange” strategy is just as good (or bad) as the “never exchange” strategy.

Continuous distributions

In the event that the smaller amount of money (and thus also and ) has a continuous distribution, the qualitative results are essentially the same, but the above formulas cannot simply be adopted analogously for the discrete case. ${\ displaystyle Z}$${\ displaystyle X}$${\ displaystyle Y}$

In addition, it must be noted that in the continuous case the conditional event has the probability zero for all , so that the elementary definitions for the conditional probability and the conditional expectation can no longer be used, but more abstract versions must be used. ${\ displaystyle \ {X = x \}}$${\ displaystyle x}$

Let it denote the density function of . The variables and then both have the same density and it applies ${\ displaystyle f}$${\ displaystyle Z}$${\ displaystyle X}$${\ displaystyle Y}$${\ displaystyle g}$

${\ displaystyle g (x) = {\ frac {1} {2}} f (x) + {\ frac {1} {4}} f \ left ({\ frac {x} {2}} \ right) \ ,.}$

In comparison to the discrete case, the factor may initially appear surprising. However, the function defined by the “analog” expression would not even be a probability density. The correct formula can be obtained, for example, by looking at the distribution functions and . Since the selection of the envelopes is made with the same probability and independently of , the following applies ${\ displaystyle {\ tfrac {1} {4}}}$${\ displaystyle {\ tilde {g}} (x): = {\ tfrac {1} {2}} f (x) + {\ tfrac {1} {2}} f (x / 2)}$ ${\ displaystyle F (x) = P (Z \ leq x)}$${\ displaystyle G (x) = P (X \ leq x)}$${\ displaystyle Z}$

${\ displaystyle P (X \ leq x) = {\ frac {1} {2}} P (Z \ leq x) + {\ frac {1} {2}} P (2Z \ leq x) \ ,,}$

so

${\ displaystyle G (x) = {\ frac {1} {2}} F (x) + {\ frac {1} {2}} F (x / 2) \ ,.}$

Differentiating according to results because and the above formula for . ${\ displaystyle x}$${\ displaystyle G '(x) = g (x)}$${\ displaystyle F '(x) = f (x)}$${\ displaystyle g}$

As a conditional probability that Mr. Schmidt will find twice the amount in the other envelope can now

${\ displaystyle P (Y = 2x \ mid X = x) = {\ frac {2f (x)} {2f (x) + f (x / 2)}}}$

and accordingly for the other case

${\ displaystyle P (Y = x / 2 \ mid X = x) = 1 - {\ frac {2f (x)} {2f (x) + f (x / 2)}} = {\ frac {f (x / 2)} {2f (x) + f (x / 2)}}}$

be set.

So you get

${\ displaystyle E (Y \ mid X = x) = 2x {\ frac {2f (x)} {2f (x) + f (x / 2)}} + {\ frac {x} {2}} \ cdot {\ frac {f (x / 2)} {2f (x) + f (x / 2)}} = {\ frac {8f (x) + f (x / 2)} {4f (x) + 2f ( x / 2)}} \ cdot x \ ,.}$

as a possible version of the conditional expectation. So if and only if is. ${\ displaystyle E (Y \ mid X = x)> x}$${\ displaystyle f (x)> {\ tfrac {1} {4}} f (x / 2)}$

If, for example, a constant uniform distribution is assumed over the interval , the result is ${\ displaystyle Z}$${\ displaystyle [25,800]}$

${\ displaystyle E (Y \ mid X = x) = {\ begin {cases} 2x & {\ mbox {for}} \ quad 25 \ leq x <50, \\ {\ frac {3} {2}} x & { \ mbox {for}} \ quad 50 \ leq x \ leq 800, \\ {\ frac {x} {2}} & {\ mbox {for}} \ quad 800

The first and third cases are clear: if there is less than 50 euros in the opened envelope, it must be the smaller of the two amounts, if it is more than 800 euros it must be the larger. In the middle case, on the other hand, a comparison with the discrete case is interesting, because a discrete uniform distribution of on the set results in this area only for even , but for odd as a conditional expected value. ${\ displaystyle Z}$${\ displaystyle \ {25,26,27, \ ldots, 800 \}}$${\ displaystyle {\ tfrac {5} {4}} x}$${\ displaystyle x}$${\ displaystyle 2x}$${\ displaystyle x}$

There are also continuous distributions, so that formally applies to all . As in the discrete case, however , it then has no finite expectation value. An example is the distribution of with the density for (and otherwise). Formally, this applies to everyone${\ displaystyle E (Y \ mid X = x)> x}$${\ displaystyle x}$${\ displaystyle Z}$${\ displaystyle Z}$${\ displaystyle f (z) = {\ tfrac {1} {2z ^ {3/2}}}}$${\ displaystyle z \ geq 1}$${\ displaystyle f (z) = 0}$${\ displaystyle x \ geq 2}$

${\ displaystyle E (Y \ mid X = x) = \ left ({\ frac {3} {2}} {\ sqrt {2}} - 1 \ right) \ cdot x \ approx 1 {,} 121x \, .}$

Explanation by the formula of the total expected value

The discrete case can be clearly explained by the formula of the total expected value. For this purpose, the initial situation is slightly generalized. It is only assumed in advance that the two envelopes are simultaneously filled with money via a random process and that one of the two envelopes is then selected and opened. To do this, the notation is slightly changed:

• the discrete random vector denotes the amounts of money that are in the two envelopes.${\ displaystyle Z = (Z_ {1}, Z_ {2})}$
• the random variable denotes the amount in the opened envelope, the random variable the amount in the other, unopened envelope.${\ displaystyle X}$${\ displaystyle Y}$
• ${\ displaystyle p_ {1}}$and are the probabilities that the first and second envelopes will be opened, respectively.${\ displaystyle p_ {2}}$

It should and be and the expected values and should exist. Then the expected values ​​from and to are calculated : ${\ displaystyle Z_ {1}> 0}$${\ displaystyle Z_ {2}> 0}$${\ displaystyle \ operatorname {E} (Z_ {1})}$${\ displaystyle \ operatorname {E} (Z_ {2})}$${\ displaystyle X}$${\ displaystyle Y}$

${\ displaystyle \ operatorname {E} (X) = \ operatorname {E} (Z_ {1}) \ cdot p_ {1} + \ operatorname {E} (Z_ {2}) \ cdot p_ {2} \ quad { \ text {and}} \ quad \ operatorname {E} (Y) = \ operatorname {E} (Z_ {1}) \ cdot p_ {2} + \ operatorname {E} (Z_ {2}) \ cdot p_ { 1}}$

First it is assumed that the opened envelope is chosen at random. Then the following applies and both expected values ​​are equal: ${\ displaystyle p_ {1} = p_ {2} = {\ tfrac {1} {2}}}$

${\ displaystyle \ operatorname {E} (X) = {\ frac {\ operatorname {E} (Z_ {1}) + \ operatorname {E} (Z_ {2})} {2}} = \ operatorname {E} (Y)}$

The random selection of the envelopes can always be achieved by changing the distribution of . Therefore, for the following considerations o. B. d. A. and be accepted. So the first envelope is always opened and it is , and thus . ${\ displaystyle Z}$ ${\ displaystyle p_ {1} = 1}$${\ displaystyle p_ {2} = 0}$${\ displaystyle X = Z_ {1}}$${\ displaystyle Y = Z_ {2}}$${\ displaystyle Z = (X, Y)}$

It is now assumed that for every possible amount of money in the opened envelope, the conditional expected value of the other envelope is always greater:

${\ displaystyle \ operatorname {E} (Y \ mid X = x_ {i})> x_ {i} \ quad {\ text {if}} \ quad \ operatorname {P} (X = x_ {i})> 0 }$

This results automatically if one assumes that the other envelope always contains half or double the amount and that the conditional probabilities for both event possibilities are always evenly distributed:

${\ displaystyle \ operatorname {P} (Y = {\ tfrac {1} {2}} x_ {i} \ mid X = x_ {i}) = {\ tfrac {1} {2}} \ quad {\ text {and}} \ quad \ operatorname {P} (Y = 2x_ {i} \ mid X = x_ {i}) = {\ tfrac {1} {2}}}$

Because then the conditional expected value of the other envelope is calculated as:

${\ displaystyle \ operatorname {E} (Y \ mid X = x_ {i}) = {\ tfrac {1} {2}} x_ {i} \ cdot {\ tfrac {1} {2}} + 2x_ {i } \ cdot {\ tfrac {1} {2}} = {\ tfrac {5} {4}} x_ {i}> x_ {i}}$

From the formula of the total expected value it now follows:

${\ displaystyle \ operatorname {E} (Y) = \ sum _ {x_ {i}} {\ operatorname {E} (Y \ mid X = x_ {i}) \ operatorname {P} (X = x_ {i} )}> \ sum _ {x_ {i}} {x_ {i} \ operatorname {P} (X = x_ {i})} = \ operatorname {E} (X)}$

Finally, it is assumed that for every possible amount of money in the other envelope, the conditional expected value of the opened envelope is always greater:

${\ displaystyle \ operatorname {E} (X \ mid Y = y_ {j})> y_ {j} \ quad {\ text {if}} \ quad \ operatorname {P} (Y = y_ {j})> 0 }$

This results automatically if one assumes that the opened envelope always contains half or double the amount and that the conditional probabilities for both event possibilities are always equally distributed:

${\ displaystyle \ operatorname {P} (X = {\ tfrac {1} {2}} y_ {j} \ mid Y = y_ {j}) = {\ tfrac {1} {2}} \ quad {\ text {and}} \ quad \ operatorname {P} (X = 2y_ {j} \ mid Y = y_ {j}) = {\ tfrac {1} {2}}}$

Because then the conditional expected value of the opened envelope is calculated as:

${\ displaystyle \ operatorname {E} (X \ mid Y = y_ {j}) = {\ tfrac {1} {2}} y_ {j} \ cdot {\ tfrac {1} {2}} + 2y_ {j } \ cdot {\ tfrac {1} {2}} = {\ tfrac {5} {4}} y_ {j}> y_ {j}}$

From the formula of the total expected value it now follows completely analogously:

${\ displaystyle \ operatorname {E} (X) = \ sum _ {y_ {j}} {\ operatorname {E} (X \ mid Y = y_ {j}) \ operatorname {P} (Y = y_ {j} )}> \ sum _ {y_ {j}} {y_ {j} \ operatorname {P} (Y = y_ {j})} = \ operatorname {E} (Y)}$

The exchange paradox lives solely from the fact that these three assumptions are incompatible:

• If one assumes that the opened envelope is selected at random, then it must be.${\ displaystyle \ operatorname {E} (X) = \ operatorname {E} (Y)}$
• If one assumes that the conditional expected value of the other envelope is always greater than the amount of money in the opened envelope, then it must be.${\ displaystyle \ operatorname {E} (Y)> \ operatorname {E} (X)}$
• If one assumes that the conditional expected value of the opened envelope is always greater than the amount of money in the other envelope, then it must be.${\ displaystyle \ operatorname {E} (X)> \ operatorname {E} (Y)}$

However, the insufficient reason principle offers a solution to this problem. It just means that if no further information is available, a discrete uniform distribution for the unknown occurrence probabilities should be applied. However, there is a reason not to apply the equal distribution if one already has the information that the opened envelope will be selected randomly and that therefore is: In order not to get into the contradiction that also or is. If, instead of the uniform distribution, one assumes that the conditional probability for the smaller amount of money is always twice as large as that for the larger amount of money, this contradiction does not arise because from ${\ displaystyle \ operatorname {E} (X) = \ operatorname {E} (Y)}$${\ displaystyle \ operatorname {E} (Y)> \ operatorname {E} (X)}$${\ displaystyle \ operatorname {E} (X)> \ operatorname {E} (Y)}$

${\ displaystyle \ operatorname {P} (Y = {\ tfrac {1} {2}} x_ {i} \ mid X = x_ {i}) = {\ tfrac {2} {3}} \ quad {\ text {and}} \ quad \ operatorname {P} (Y = 2x_ {i} \ mid X = x_ {i}) = {\ tfrac {1} {3}}}$

always follows

${\ displaystyle \ operatorname {E} (Y \ mid X = x_ {i}) = {\ tfrac {1} {2}} x_ {i} \ cdot {\ tfrac {2} {3}} + 2x_ {i } \ cdot {\ tfrac {1} {3}} = x_ {i}}$

and thus also

${\ displaystyle \ operatorname {E} (Y) = \ operatorname {E} (X)}$.

Exactly results from

${\ displaystyle \ operatorname {P} (X = {\ tfrac {1} {2}} y_ {j} \ mid Y = y_ {j}) = {\ tfrac {2} {3}} \ quad {\ text {and}} \ quad \ operatorname {P} (X = 2y_ {j} \ mid Y = y_ {j}) = {\ tfrac {1} {3}}}$

always

${\ displaystyle \ operatorname {E} (X \ mid Y = y_ {j}) = {\ tfrac {1} {2}} y_ {j} \ cdot {\ tfrac {2} {3}} + 2y_ {j } \ cdot {\ tfrac {1} {3}} = y_ {j}}$

And also

${\ displaystyle \ operatorname {E} (X) = \ operatorname {E} (Y)}$.

A simple example illustrates how far one can be wrong with the apparently plausible assumption of an equal distribution for the conditional probabilities. The envelopes are always filled with 100 and 200 euros. A bivariate two-point distribution of : ${\ displaystyle Z = (X, Y)}$

${\ displaystyle \ operatorname {P} (Z = (100,200)) = \ operatorname {P} (X = 100, Y = 200) = {\ tfrac {1} {2}}}$
${\ displaystyle \ operatorname {P} (Z = (200,100)) = \ operatorname {P} (X = 200, Y = 100) = {\ tfrac {1} {2}}}$

Then the following conditional probabilities result:

${\ displaystyle \ operatorname {P} (Y = 50 \ mid X = 100) = 0 \ quad {\ text {and}} \ quad \ operatorname {P} (X = 50 \ mid Y = 100) = 0}$
${\ displaystyle \ operatorname {P} (Y = 200 \ mid X = 100) = 1 \ quad {\ text {and}} \ quad \ operatorname {P} (X = 200 \ mid Y = 100) = 1}$
${\ displaystyle \ operatorname {P} (Y = 100 \ mid X = 200) = 1 \ quad {\ text {and}} \ quad \ operatorname {P} (X = 100 \ mid Y = 200) = 1}$
${\ displaystyle \ operatorname {P} (Y = 400 \ mid X = 200) = 0 \ quad {\ text {and}} \ quad \ operatorname {P} (X = 400 \ mid Y = 200) = 0}$

These conditional probabilities are miles away from the assumed uniform distribution.

Application of the two-slip game

In the above examples it was assumed that the principle according to which the amounts of money are distributed is known. With this assumption, it is easy to specify winning strategies. The problem formulation does not contain any information about the distribution. However, there is also a general winning strategy for Mr. Schmidt that does not require this assumption. This strategy consists in the fact that Mr. Smith chooses a random number S before he opens the envelope. The probability distribution of S must have a density that is truly greater than 0 between 0 and infinitely, but is otherwise arbitrary. Then he opens the envelope and finds the amount n. If the amount n is less than or equal to S, he swaps the envelope; if the amount n is greater than S, it keeps the envelope. This strategy goes back to Thomas M. Cover . As explained in the two-slip game article , this theoretically increases his chances of getting the larger amount.

Assume that Mr. Schmidt decides to use the two-slip game. If the envelopes contain the amounts and and if Mr.Smith opens the envelope with its contents first , he will switch, if so . The conditional expectation of his gain is then ${\ displaystyle Z}$${\ displaystyle 2Z}$${\ displaystyle Z}$${\ displaystyle S \ geq Z}$

{\ displaystyle {\ begin {aligned} E_ {Z; Z} & = P (S \ geq Z) \ cdot 2Z + P (S

If he first opens the envelope with its contents , he changes, if so . The conditional expectation of his gain is then ${\ displaystyle 2Z}$${\ displaystyle S \ geq 2Z}$

{\ displaystyle {\ begin {aligned} E_ {2Z; Z} & = P (S \ geq 2Z) \ cdot Z + P (S <2Z) \ cdot 2Z \\ & = P (S \ geq 2Z) \ cdot Z + \ left (1-P (S \ geq 2Z) \ right) \ cdot 2Z \\ & = Z \ left (2-P (S \ geq 2Z) \ right). \ End {aligned}}}

Overall, the conditional expectation is for fixed content, but before choosing the first cover

{\ displaystyle {\ begin {aligned} E_ {Z} & = {\ frac {E_ {Z; Z} + E_ {2Z; Z}} {2}} \\ & = {\ frac {3} {2} } Z + {\ frac {Z} {2}} {\ Big (} P (S \ geq Z) -P (S \ geq 2Z) {\ Big)} \\ & = {\ frac {3} {2} } Z + {\ frac {Z} {2}} P (Z \ leq S <2Z). \ End {aligned}}}

If he always swaps or he never swaps, his expected value is

${\ displaystyle {\ frac {Z + 2Z} {2}} = {\ frac {3} {2}} Z.}$

When using the two-slip game, the expected value is up

${\ displaystyle {\ frac {Z} {2}} P (Z \ leq S <2Z)}$

higher than with the "never swap" or "always swap" approach.

example

For selected Mr. Schmidt, for example, a random variable S, which exponentially distributed with expected value in 1000, ie . If the amounts of money are evenly distributed across the amount as in the example given above , the overall expected value of the gift of money is as follows: ${\ displaystyle S}$${\ displaystyle P (S \ geq s) = \ exp (-s / 1000)}$${\ displaystyle \ {25,50,100,200,400,800 \}}$

${\ displaystyle Z}$ ${\ displaystyle 2Z}$ ${\ displaystyle p_ {Z}}$ ${\ displaystyle P (S \ geq Z)}$ ${\ displaystyle E_ {Z; Z}}$ ${\ displaystyle P (S \ geq 2Z)}$ ${\ displaystyle E_ {2Z; Z}}$ ${\ displaystyle E_ {Z}}$ ${\ displaystyle 3Z / 2}$ ${\ displaystyle p_ {Z} \ cdot E_ {Z}}$
25th 50 1/6 0.975 49.382 0.951 26,219 37,801 37.5 6,300
50 100 1/6 0.951 97.561 0.904 54.758 76.160 75.0 12.693
100 200 1/6 0.904 190.484 0.819 118.127 154,305 150.0 25.718
200 400 1/6 0.819 363.746 0.670 265,936 314,841 300.0 52.473
400 800 1/6 0.670 688.128 0.449 620.268 644.198 600.0 107.366
800 1600 1/6 0.449 1159,463 0.202 1438,483 1298.973 1200.0 216.496
total 1 421,046

The expected value of the cash gift is 421.046 euros with this approach. That is less than with the optimal strategy (swap less than 1000 euros), where the expected value is 460.62 euros; but in any case more than with the “never swap” or “always swap” procedure, in which the expected value is 393.75 euros. As can be seen from the table, is greater than in each row . The exact expected value naturally depends heavily on the choice of distribution of , but is always greater than with the "never swap" or "always swap" procedure. ${\ displaystyle E_ {Z}}$${\ displaystyle \ textstyle {\ frac {3} {2}} Z}$${\ displaystyle S}$

Related topics where you can make the optimal decision about the remaining problem from partial information :

Individual evidence

1. ^ Maurice Kraitchik: La mathématique des jeux. 1953
2. Martin Gardner: Aha! Gotcha. 1982
3. Barry Nalebuff: Puzzles: the other person's envelope is always greener. In: Journal of Economic Perspectives. Volume 3, 1989, yale.edu (PDF; 205 kB)
4. Thought Traps and Paradoxes