The conversion paradox (or envelope paradox) describes the paradoxes situation that it with knowledge of the between two alternatives and after the value of one of the alternatives was opened appears always meaningful to make from the offered right to exchange use. The paradox analyzes naive counting on expectancy values. It is related, but not logically identical to the goat problem.

The paradox

We assume, Mr. Lemke are a sponsor of Mr. Schmidt. The secretary of Mr. Lemke took and into a the money in-did two equivalent envelopes. In the other envelope she in-did the double amount. Both envelopes look from the outside completely directly. In the evening Mr. Lemke and Mr. Schmidt on a party meet. Mr. Lemke puts both envelopes down on a table. Other party guests bring the envelopes in disorder. To the advanced hour - one already drank something - Mr. Lemke seizes one of the two envelopes and gives him to Mr. Schmidt with the words: "“In the other one, completely equal looking envelope, which lies still on the table, the double money is and with a probability of 50% the half money with a probability of 50%. They may open and then decide this envelope whether you would like to exchange the two envelopes and take the others."” Mr. Schmidt opens the envelope, finds 100 euro and considers: "“I have 100 euro in this envelope. If I exchange, I have 50 euro with a probability of 50% 200 euro and with the same probability. This makes an expectancy value of 125 EUR."”

125 EUR = 0,5 50 + 0,5 200

Is exchanging worthwhile

What is paradoxical to

Paradoxes at the situation it is that it seems just as meaningful to change, if first the other envelope had been opened. In reality the change would be meaningful whenever one selected the worse alternative - it is however unknown whether that is the case or not.

The solution

Here wrong counting on expectancy values is present. The fallacy lies in the fact that for any money "„with a probability of 50% the double money cannot be and with a probability of 50% the half money "“in the other envelope. That presupposes a uniform distribution on the quantity of the natural numbers, which there cannot be however. It is simplified spoken in such a way that the larger the amount is in the envelope, the smaller becomes the probability that in the other envelope a still larger amount is; exchange brings then thus no profit or becomes even the loss.

For an accurate computation one needs the conditioned probability distribution of the money in the other envelope, if one knows the amount in an envelope. Designates one also

• S_n the event that the secretary of Mr. Lemke n euro puts 2n into some and euro into the other envelope, also
• p_n=P (S_n) the probability of this event, also
• B_n the event that Mr. Schmidt n euro in the envelope finds, and also
• A_n the event that Mr. Schmidt n euro in the other envelope finds,

thus it applies for the probability that Mr. Schmidt n euro in the envelope finds

P \ left (B_n \ right) =P \ left (B_n|S_n \ right) P \ left (S_n \ right) +P \ left (B_n|S_ {n/2} \ right) P \ left (S_ {n/2} \ right) = \ frac {p_n} {2} + \ frac {p_ {n/2}} {2},

for the conditioned probability that Mr. Schmidt finds the double amount in the other envelope

P \ left (A_ {2n} |B_n \ right) = \ frac {P \ left (A_ {2n} \ cap B_n \ right)}{P \ left (B_n \ right)}= \ frac {\ frac {1} {2} p_n} {\ frac {p_n} {2} + \ frac {p_ {n/2}} {2}} = \ frac {p_n} {p_n+p_ {n/2}},

and for the conditioned probability that Mr. Schmidt finds the half amount in the other envelope

P \ left (A_ {n/2} |B_n \ right) = \ frac {P \ left (A_ {n/2} \ cap B_n \ right)}{P \ left (B_n \ right)}= \ frac {\ frac {1} {2} p_ {n/2}} {\ frac {p_n} {2} + \ frac {p_ {n/2}} {2}} = \ frac {p_ {n/2}} {p_n+p_ {n/2}}.

The fallacy of Mr. Schmidt consists thus of it that it assumes, p_n=p_ {n/2} for all n; such a uniform distribution does not give it however on the natural numbers.

If one computes the expectancy value with the correct probabilities, then one receives

E=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,

Exchanges disburses itself, if E > n; that is exactly then the case if p_n > \ frac {p_ {n/2}} {2}. Distributions, which fulfill this condition for all n \ in \ N, can be designed, have then however expectancy value infinitely, so that the argumentation with the expectancy value is not permissible also in this case. Beyond that one can probably assume Mr. Lemke does not have infinitely much money available.

Example

If one accepts a probability distribution, with which the secretary of Mr. Lemke distributes the money into the envelopes, can history very well be simulated. For example is accepted, the secretary determines the amount with a fair cube. If the cube k eyes shows, then it puts for 2^ {k-1} \ cdot 25 euro into and 2^k \ cdot 25 euro into the other envelope. Mr. Schmidt finds then with probability \ frac {1} {to 12} the amount 25 euro in the envelope, with probability ever \ frac {2} {12} one of the amounts 50, 100, 200, 400 or 800 euro, and again with probability \ frac {1} {12} the amount 1600 euro. If it does not exchange, then the expectancy value of the gratuity amounts to thus

\ frac {1} {12} \ left (25+2 \ 50+2 \ cdot 100+2 \ cdot 200+2 \ cdot 400+2 \ cdot 800+ 1600 cdot \ right) =393,75 euro.

If Mr. Schmidt in each case exchanges, then its expectancy value changes not, since it exchanges in particular also the amount of 1600 euro, although it cannot win anything in this case. If Mr. Schmidt assumes however that probably hardly more than 1000 euro are in the envelope, and decides therefore, then and to only exchange, if at the most 500 euro are in the envelope, then the Wahrscheinlickheiten changes: After the exchange Mr. Schmidt has then further with probability \ frac {1} {12} the amount 25 euro in the envelope, just as with probability ever \ frac {2} {12} one of the amounts 50, 100, or 200 euro, the amount of 400 euro however only more with probability \ frac {1} {12} (there Mr. Schmidt with 800 euro no more does not exchange), but however with probability \ frac {3} {12} the amount of 800 euro, and again with probability \ frac {1} {12} the amount 1600 euro. The expectancy value of the gratuity is now thus

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

If Mr. Schmidt estimates the situation better and decides to only do starting from 1000 euro without exchanging it can increase the expectancy value even to 460,62 euro; if it becomes however too greedy and exchanges for example only starting from 2000 euro, then it drops back the initial value 393.75 euro.

For Mr. Schmidt it is naturally difficult to correctly estimate the secretary of Mr. Lemke; substantial it is however that the paradox disappears, as soon as one accepts any concrete probability distribution for the behavior of the secretary of Mr. Lemke. Depending upon exchange strategy of Mr. Lemke the expectancy value of the gratuity changes; the strategy of "“exchanges always"” is however equal well (or badly) as the strategy "“exchanges never"”.

Application of the two-note play

A general profit strategy for Mr. Schmidt consists of the fact that it, before it opens the envelope selects a random number Z, which can infinitely take all values between 0 and, whose distribution is otherwise arbitrary however. Then it opens the envelope and finds the amount n. If the found amount n is smaller than Z, then it exchanges the envelope; if the amount n is larger than Z, then it keeps the envelope. As in the article two-note play explains, it increases its chances to receive the larger amount.

Cent - or the discrete nature of the money

If one finds an amount, which means an odd number in cent (or the respective smallest monetary unit) in an envelope, then accurately half cannot be in the other envelope. In all these cases exchanging is always worthwhile itself. (Mathematics calls numbers, which are not as desired divisible, discretely.)

On the other hand money is also finite. Therefore it is not worthwhile itself exchanges reliably if in an envelope an amount is, which is so large that the donor can not possibly double it. Then one has obviously already the larger amount in the hand.

To the reality purchase

The conversion paradox often occurs in the reality, since humans are bent with two coincidental possibilities to accept a 50%-Wahrscheinlichkeit.

See also

Used topics, with which one can make the optimal decision of the remainder problem from partial information:

• Odds strategy
• Secretary problem
• Goat problem

Articles in category "Conversion paradox"

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