Ziegenproblem

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Das Ziegenproblem, Drei-Türen-Problem, Monty-Hall-Problem oder Monty-Hall-​Dilemma ist eine Aufgabe zur Wahrscheinlichkeitstheorie. Es geht dabei um die. Das Ziegenproblem, Drei-Türen-Problem, Monty-Hall-Problem oder Monty-Hall-Dilemma ist eine Aufgabe zur Wahrscheinlichkeitstheorie. Das Ziegenproblem ist auch als "Monty Hall Problem" bekannt. Monty Hall moderierte bereits in den 60er Jahren die Show "Let's make a deal". Ziegenproblem-Simulator. Klicken Sie auf eines der Tore, um das Ziegenproblem spielerisch zu entdecken. Weitere Information unten auf dieser Seite. Am so genannten Ziegenproblem bissen sich sogar Nobelpreisträger die Zähne aus. Deutsche Forscher haben endlich einen Weg gefunden.

Ziegenproblem

Das Ziegenproblem ist auch als "Monty Hall Problem" bekannt. Monty Hall moderierte bereits in den 60er Jahren die Show "Let's make a deal". Am so genannten Ziegenproblem bissen sich sogar Nobelpreisträger die Zähne aus. Deutsche Forscher haben endlich einen Weg gefunden. Das Ziegenproblem: Denken in Wahrscheinlichkeiten | Randow, Gero von | ISBN​: | Kostenloser Versand für alle Bücher mit Versand und. Ziegenproblem Was ist? Der Kandidat Graj W Sizzling Hot demnach in diesem Fall also ebenso gut bei Tor 1 bleiben wie zu Tor 2 wechseln. Es kommt wohl kaum zu Missverständnissen. Allerdings können durch einen asymmetrischen Spielverlauf Entscheidungssituationen Book Of Ra Online Spielen Mit Geld, bei denen ein Torwechsel gegenüber dem Durchschnitt aussichtsreicher beziehungsweise weniger aussichtsreich ist. Ich würde den Diskussionsfaden hier gerne beenden, wenigsten solange kein wirklich neuer Gedanke auftaucht. Man sieht, dass in zwei dieser drei Fälle der Kandidat durch Wechseln gewinnt. Tatsächlich ist es aber weitaus günstiger All Slots Casino Codes wechseln als zu bleiben was sich mathematisch mittels bedingter Wahrscheinlichkeiten auch zeigen lässt. Nicht mehr und nicht Gute Spiele Ipad Kostenlos. Startseite Leitlinien Lies mich Kontakt Links. Setze ein Lesezeichen auf den Permalink. In diesem Sinne, vielen Dank Pharao Gold Kostenlos Spielen die erneute Diskussion zu diesem Thema! Für die Interessierten hier noch meine — im Free Slot Zorro Artikel enthaltene — Formulierung der Aufgabe, die tatsächlich eine Zwei-Drittel-Lösung hat; auch zum Vergleich mit dem, was man ihnen bisher erzählt hat:.

Verwandte Themen. Ziegenproblem: Die Lösung einfach erklärt Ziegenproblem: Was ist das eigentlich? Das Ziegenproblem ist auch als "Monty Hall Problem" bekannt.

Monty Hall moderierte bereits in den 60er Jahren die Show "Let's make a deal". In der Show hat der Kandidat die Möglichkeit zwischen drei Toren zu wählen.

Der Kandidat wählt in diesem Glücksspiel ein Tor aus. Das Verhalten des Moderators ist Teil der Show und geschieht ebenfalls, wenn sich der unwissende Spieler bereits auf eine Niete festgelegt hat.

Der Moderator öffnet eines der anderen beiden Tore mit einer Ziege dahinter und fragt den Kandidaten zum letzten Mal, ob er das Tor nicht wechseln möchte.

Die kontrovers diskutierte Frage lautet: Sollte man das Tor wechseln, oder nicht? Warum wird das Ziegenproblem so sehr diskutiert?

Erstmalig brach die Diskussion über das Ziegenproblem aus. Die Lösung für das Problem, ob sich der Wechsel der Tore wirklich lohnt, war für die meisten Menschen jedoch nicht verständlich.

Im ersten Moment scheint es komplett egal zu sein, welches Tor man nimmt. Wir wissen, dass hinter einem Tor die zweite Ziege und hinter dem anderen Tor ein Auto steht.

Egal, welches Tor man wählt, die Wahrscheinlichkeit liegt stets bei So logisch das im ersten Moment auch klingt, korrekt ist es nicht.

Einfach erklärt: Die Lösung für das Ziegenproblem Das Ziegenproblem wurde nur so populär, weil kaum einer die komplexen und unverständlichen Erklärungsversuche verstand.

Since he does not know how the car is hidden nor how the host makes choices, he may be able to make use of his first choice opportunity, as it were to neutralize the actions of the team running the quiz show, including the host.

Following Gill, a strategy of contestant involves two actions: the initial choice of a door and the decision to switch or to stick which may depend on both the door initially chosen and the door to which the host offers switching.

For instance, one contestant's strategy is "choose door 1, then switch to door 2 when offered, and do not switch to door 3 when offered".

Twelve such deterministic strategies of the contestant exist. Elementary comparison of contestant's strategies shows that, for every strategy A, there is another strategy B "pick a door then switch no matter what happens" that dominates it Gnedin, No matter how the car is hidden and no matter which rule the host uses when he has a choice between two goats, if A wins the car then B also does.

For example, strategy A "pick door 1 then always stick with it" is dominated by the strategy B "pick door 1 then always switch after the host reveals a door": A wins when door 1 conceals the car, while B wins when one of the doors 2 and 3 conceals the car.

Similarly, strategy A "pick door 1 then switch to door 2 if offered , but do not switch to door 3 if offered " is dominated by strategy B "pick door 3 then always switch".

Dominance is a strong reason to seek for a solution among always-switching strategies, under fairly general assumptions on the environment in which the contestant is making decisions.

In particular, if the car is hidden by means of some randomization device — like tossing symmetric or asymmetric three-sided die — the dominance implies that a strategy maximizing the probability of winning the car will be among three always-switching strategies, namely it will be the strategy that initially picks the least likely door then switches no matter which door to switch is offered by the host.

Strategic dominance links the Monty Hall problem to the game theory. In the zero-sum game setting of Gill, , discarding the non-switching strategies reduces the game to the following simple variant: the host or the TV-team decides on the door to hide the car, and the contestant chooses two doors i.

The contestant wins and her opponent loses if the car is behind one of the two doors she chose. A simple way to demonstrate that a switching strategy really does win two out of three times with the standard assumptions is to simulate the game with playing cards Gardner b ; vos Savant , p.

Three cards from an ordinary deck are used to represent the three doors; one 'special' card represents the door with the car and two other cards represent the goat doors.

The simulation can be repeated several times to simulate multiple rounds of the game. The player picks one of the three cards, then, looking at the remaining two cards the 'host' discards a goat card.

If the card remaining in the host's hand is the car card, this is recorded as a switching win; if the host is holding a goat card, the round is recorded as a staying win.

As this experiment is repeated over several rounds, the observed win rate for each strategy is likely to approximate its theoretical win probability, in line with the law of large numbers.

Repeated plays also make it clearer why switching is the better strategy. After the player picks his card, it is already determined whether switching will win the round for the player.

If this is not convincing, the simulation can be done with the entire deck. Gardner b ; Adams In this variant, the car card goes to the host 51 times out of 52, and stays with the host no matter how many non -car cards are discarded.

A common variant of the problem, assumed by several academic authors as the canonical problem, does not make the simplifying assumption that the host must uniformly choose the door to open, but instead that he uses some other strategy.

The confusion as to which formalization is authoritative has led to considerable acrimony, particularly because this variant makes proofs more involved without altering the optimality of the always-switch strategy for the player.

The variants are sometimes presented in succession in textbooks and articles intended to teach the basics of probability theory and game theory.

A considerable number of other generalizations have also been studied. The version of the Monty Hall problem published in Parade in did not specifically state that the host would always open another door, or always offer a choice to switch, or even never open the door revealing the car.

I personally read nearly three thousand letters out of the many additional thousands that arrived and found nearly every one insisting simply that because two options remained or an equivalent error , the chances were even.

Very few raised questions about ambiguity, and the letters actually published in the column were not among those few. The table below shows a variety of other possible host behaviors and the impact on the success of switching.

Determining the player's best strategy within a given set of other rules the host must follow is the type of problem studied in game theory.

For example, if the host is not required to make the offer to switch the player may suspect the host is malicious and makes the offers more often if the player has initially selected the car.

In general, the answer to this sort of question depends on the specific assumptions made about the host's behavior, and might range from "ignore the host completely" to "toss a coin and switch if it comes up heads"; see the last row of the table below.

Both changed the wording of the Parade version to emphasize that point when they restated the problem. They consider a scenario where the host chooses between revealing two goats with a preference expressed as a probability q , having a value between 0 and 1.

This means even without constraining the host to pick randomly if the player initially selects the car, the player is never worse off switching.

As N grows larger, the advantage decreases and approaches zero Granberg A quantum version of the paradox illustrates some points about the relation between classical or non-quantum information and quantum information , as encoded in the states of quantum mechanical systems.

The formulation is loosely based on quantum game theory. The three doors are replaced by a quantum system allowing three alternatives; opening a door and looking behind it is translated as making a particular measurement.

The rules can be stated in this language, and once again the choice for the player is to stick with the initial choice, or change to another "orthogonal" option.

The latter strategy turns out to double the chances, just as in the classical case. However, if the show host has not randomized the position of the prize in a fully quantum mechanical way, the player can do even better, and can sometimes even win the prize with certainty Flitney and Abbott , D'Ariano et al.

In this puzzle, there are three boxes: a box containing two gold coins, a box with two silver coins, and a box with one of each.

After choosing a box at random and withdrawing one coin at random that happens to be a gold coin, the question is what is the probability that the other coin is gold.

This problem involves three condemned prisoners, a random one of whom has been secretly chosen to be pardoned.

The warden obliges, secretly flipping a coin to decide which name to provide if the prisoner who is asking is the one being pardoned. The question is whether knowing the warden's answer changes the prisoner's chances of being pardoned.

The first letter presented the problem in a version close to its presentation in Parade 15 years later. The second appears to be the first use of the term "Monty Hall problem".

The problem is actually an extrapolation from the game show. As Monty Hall wrote to Selvin:. And if you ever get on my show, the rules hold fast for you — no trading boxes after the selection.

A version of the problem very similar to the one that appeared three years later in Parade was published in in the Puzzles section of The Journal of Economic Perspectives Nalebuff Nalebuff, as later writers in mathematical economics, sees the problem as a simple and amusing exercise in game theory.

Phillip Martin's article in a issue of Bridge Today magazine titled "The Monty Hall Trap" Martin presented Selvin's problem as an example of what Martin calls the probability trap of treating non-random information as if it were random, and relates this to concepts in the game of bridge.

A restated version of Selvin's problem appeared in Marilyn vos Savant 's Ask Marilyn question-and-answer column of Parade in September Tierney Due to the overwhelming response, Parade published an unprecedented four columns on the problem.

Adams initially answered, incorrectly, that the chances for the two remaining doors must each be one in two. After a reader wrote in to correct the mathematics of Adams's analysis, Adams agreed that mathematically he had been wrong.

Now you're offered this choice: open door 1, or open door 2 and door 3. In the latter case you keep the prize if it's behind either door.

You'd rather have a two-in-three shot at the prize than one-in-three, wouldn't you? If you think about it, the original problem offers you basically the same choice.

Monty is saying in effect: you can keep your one door or you can have the other two doors, one of which a non-prize door I'll open for you.

Numerous readers, however, wrote in to claim that Adams had been "right the first time" and that the correct chances were one in two.

The Parade column and its response received considerable attention in the press, including a front-page story in the New York Times in which Monty Hall himself was interviewed.

Tierney Hall understood the problem, giving the reporter a demonstration with car keys and explaining how actual game play on Let's Make a Deal differed from the rules of the puzzle.

In the article, Hall pointed out that because he had control over the way the game progressed, playing on the psychology of the contestant, the theoretical solution did not apply to the show's actual gameplay.

He said he was not surprised at the experts' insistence that the probability was 1 out of 2. By opening that door we were applying pressure.

We called it the Henry James treatment. It was ' The Turn of the Screw. Caveat emptor. It all depends on his mood. From Wikipedia, the free encyclopedia.

A probability puzzle. Since you seem to have difficulty grasping the basic principle at work here, I'll explain.

After the host reveals a goat, you now have a one-in-two chance of being correct. Whether you change your selection or not, the odds are the same.

There is enough mathematical illiteracy in this country, and we don't need the world's highest IQ propagating more. University of Florida vos Savant a.

Possible host behaviors in unspecified problem Host behavior Result The host acts as noted in the specific version of the problem. Switching wins the car two-thirds of the time.

Tierney Switching always yields a goat. Switching always wins the car. Switching wins the car half of the time. The host knows what lies behind the doors, and before the player's choice chooses at random which goat to reveal.

He offers the option to switch only when the player's choice happens to differ from his. Four-stage two-player game-theoretic Gill, , Gill, The player is playing against the show organizers TV station which includes the host.

First stage: organizers choose a door choice kept secret from player. Second stage: player makes a preliminary choice of door.

Third stage: host opens a door. Fourth stage: player makes a final choice. The player wants to win the car, the TV station wants to keep it.

This is a zero-sum two-person game. By von Neumann's theorem from game theory , if we allow both parties fully randomized strategies there exists a minimax solution or Nash equilibrium Mueser and Granberg Minimax solution Nash equilibrium : car is first hidden uniformly at random and host later chooses uniform random door to open without revealing the car and different from player's door; player first chooses uniform random door and later always switches to other closed door.

As previous, but now host has option not to open a door at all. Minimax solution Nash equilibrium : car is first hidden uniformly at random and host later never opens a door; player first chooses a door uniformly at random and later never switches.

Deal or No Deal case: the host asks the player to open a door, then offers a switch in case the car hasn't been revealed. Topics in game theory.

Cooperative game Determinacy Escalation of commitment Extensive-form game First-player and second-player win Game complexity Graphical game Hierarchy of beliefs Information set Normal-form game Preference Sequential game Simultaneous game Simultaneous action selection Solved game Succinct game.

Nash equilibrium Subgame perfection Mertens-stable equilibrium Bayesian Nash equilibrium Perfect Bayesian equilibrium Trembling hand Proper equilibrium Epsilon-equilibrium Correlated equilibrium Sequential equilibrium Quasi-perfect equilibrium Evolutionarily stable strategy Risk dominance Core Shapley value Pareto efficiency Gibbs equilibrium Quantal response equilibrium Self-confirming equilibrium Strong Nash equilibrium Markov perfect equilibrium.

Arrow's impossibility theorem Aumann's agreement theorem Folk theorem Minimax theorem Nash's theorem Purification theorem Revelation principle Zermelo's theorem.

Albert W. Levine David M. Kreps Donald B. All-pay auction Alpha—beta pruning Bertrand paradox Bounded rationality Combinatorial game theory Confrontation analysis Coopetition Evolutionary game theory First-move advantage in chess Game mechanics Glossary of game theory List of game theorists List of games in game theory No-win situation Solving chess Topological game Tragedy of the commons Tyranny of small decisions.

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Namespaces Article Talk. Views Read Edit View history. Help Community portal Recent changes Upload file. Download as PDF Printable version.

Wikimedia Commons. On those occasions when the host opens Door 2, switching wins twice as often as staying cases versus

Hier nochmal die Zusammenfassung des Ziegenproblems:. Dezember um Ansonsten bleibe Your Beat bei meiner Einschätzung, dass durch Gerhard Keller und andere eine sehr schöne Denksportaufgabe so lange zerredet wird, bis sie einem öde erscheint. In der Folge erhielt vos Savant nach ihrer eigenen Schätzung zehntausend Briefe, die ganz überwiegend die Richtigkeit ihrer Antwort bezweifelten. Das liegt sicher daran, dass bei richtig gestellter Aufgabe auch die Begründung leichtfällt …. Hearts Kostenlos Online Spielen Ohne Anmeldung sehen Sie, dass es eine unüberbietbar lächerliche Unterstellung Pharao Gold Kostenlos Spielen, ich wolle einen Irrtum nicht wahrhaben. Game Zynga vos Savant hat vor nunmehr 28 Jahren Play Lucky Lady Charm Online Free schon alles gesagt, was Gutscheincode Ovo Casino dazu zu sagen gibt.

Ziegenproblem Video

Das Ziegenproblem (Monty-Hall-Problem, Drei-Türen-Problem) Der Zwist um das Drei-Türen-Problem (Ziegenproblem) wurde im Jahr von Marilyn vos Savant in einer ihrer Kolumnen angestoßen. Das „Ziegenproblem“. Monty open pc-boeken.nl In einer Quizshow kann sich der Kandidat zwischen drei Türen entscheiden. Hinter einer wartet. Das Ziegenproblem ist eine Aufgabe aus dem Feld der Wahrscheinlichkeitstheorie, die auf einen Leserbrief im American Statistician von Steve Selvin. Das Ziegenproblem: Denken in Wahrscheinlichkeiten | Randow, Gero von | ISBN​: | Kostenloser Versand für alle Bücher mit Versand und. Ein noch stärkeres Argument für den Kandidaten, nie das anfangs gewählte Tor beizubehalten, ergibt sich aus Gnedins Dominanz -Analysen für Strategien. In particular, vos Savant defended herself vigorously. If this is not Free Mobile Casino Games Download, the simulation can be done with the entire Valentinstag Tipps. Konkrete Ursache dafür ist, dass bei Pro7 Online Spiele hinter Tor 3 verborgenen Auto der Moderator gezwungen ist, Tor 2 zu öffnen. Der Moderator muss in diesem Fall Tor 2 öffnen, da er nicht die Wahl des Kandidaten und auch nicht das Auto enthüllen darf. Der Cable Capers kann demnach in diesem Fall also William Huill gut bei Tor 1 bleiben wie zu Tor 2 wechseln. Morgan et al. Pharao Gold Kostenlos Spielen sie allerdings meint, dass ihr der Moderator nicht gut gesinnt Glory Of Rome und sie nur von ihrer ersten, richtigen Wahl ablenken möchte, dann sollte sie bei Tor 1 bleiben. Whitaker aus Columbia, Maryland, erhalten hatte: [1].

According to Bayes' rule , the posterior odds on the location of the car, given that the host opens door 3, are equal to the prior odds multiplied by the Bayes factor or likelihood, which is, by definition, the probability of the new piece of information host opens door 3 under each of the hypotheses considered location of the car.

Given that the host opened door 3, the probability that the car is behind door 3 is zero, and it is twice as likely to be behind door 2 than door 1.

Richard Gill analyzes the likelihood for the host to open door 3 as follows. Given that the car is not behind door 1, it is equally likely that it is behind door 2 or 3.

In words, the information which door is opened by the host door 2 or door 3? Consider the event Ci , indicating that the car is behind door number i , takes value Xi , for the choosing of the player, and value Hi , the opening the door.

Then, if the player initially selects door 1, and the host opens door 3, we prove that the conditional probability of winning by switching is:.

Going back to Nalebuff , the Monty Hall problem is also much studied in the literature on game theory and decision theory , and also some popular solutions correspond to this point of view.

Vos Savant asks for a decision, not a chance. And the chance aspects of how the car is hidden and how an unchosen door is opened are unknown.

From this point of view, one has to remember that the player has two opportunities to make choices: first of all, which door to choose initially; and secondly, whether or not to switch.

Since he does not know how the car is hidden nor how the host makes choices, he may be able to make use of his first choice opportunity, as it were to neutralize the actions of the team running the quiz show, including the host.

Following Gill, a strategy of contestant involves two actions: the initial choice of a door and the decision to switch or to stick which may depend on both the door initially chosen and the door to which the host offers switching.

For instance, one contestant's strategy is "choose door 1, then switch to door 2 when offered, and do not switch to door 3 when offered".

Twelve such deterministic strategies of the contestant exist. Elementary comparison of contestant's strategies shows that, for every strategy A, there is another strategy B "pick a door then switch no matter what happens" that dominates it Gnedin, No matter how the car is hidden and no matter which rule the host uses when he has a choice between two goats, if A wins the car then B also does.

For example, strategy A "pick door 1 then always stick with it" is dominated by the strategy B "pick door 1 then always switch after the host reveals a door": A wins when door 1 conceals the car, while B wins when one of the doors 2 and 3 conceals the car.

Similarly, strategy A "pick door 1 then switch to door 2 if offered , but do not switch to door 3 if offered " is dominated by strategy B "pick door 3 then always switch".

Dominance is a strong reason to seek for a solution among always-switching strategies, under fairly general assumptions on the environment in which the contestant is making decisions.

In particular, if the car is hidden by means of some randomization device — like tossing symmetric or asymmetric three-sided die — the dominance implies that a strategy maximizing the probability of winning the car will be among three always-switching strategies, namely it will be the strategy that initially picks the least likely door then switches no matter which door to switch is offered by the host.

Strategic dominance links the Monty Hall problem to the game theory. In the zero-sum game setting of Gill, , discarding the non-switching strategies reduces the game to the following simple variant: the host or the TV-team decides on the door to hide the car, and the contestant chooses two doors i.

The contestant wins and her opponent loses if the car is behind one of the two doors she chose. A simple way to demonstrate that a switching strategy really does win two out of three times with the standard assumptions is to simulate the game with playing cards Gardner b ; vos Savant , p.

Three cards from an ordinary deck are used to represent the three doors; one 'special' card represents the door with the car and two other cards represent the goat doors.

The simulation can be repeated several times to simulate multiple rounds of the game. The player picks one of the three cards, then, looking at the remaining two cards the 'host' discards a goat card.

If the card remaining in the host's hand is the car card, this is recorded as a switching win; if the host is holding a goat card, the round is recorded as a staying win.

As this experiment is repeated over several rounds, the observed win rate for each strategy is likely to approximate its theoretical win probability, in line with the law of large numbers.

Repeated plays also make it clearer why switching is the better strategy. After the player picks his card, it is already determined whether switching will win the round for the player.

If this is not convincing, the simulation can be done with the entire deck. Gardner b ; Adams In this variant, the car card goes to the host 51 times out of 52, and stays with the host no matter how many non -car cards are discarded.

A common variant of the problem, assumed by several academic authors as the canonical problem, does not make the simplifying assumption that the host must uniformly choose the door to open, but instead that he uses some other strategy.

The confusion as to which formalization is authoritative has led to considerable acrimony, particularly because this variant makes proofs more involved without altering the optimality of the always-switch strategy for the player.

The variants are sometimes presented in succession in textbooks and articles intended to teach the basics of probability theory and game theory.

A considerable number of other generalizations have also been studied. The version of the Monty Hall problem published in Parade in did not specifically state that the host would always open another door, or always offer a choice to switch, or even never open the door revealing the car.

I personally read nearly three thousand letters out of the many additional thousands that arrived and found nearly every one insisting simply that because two options remained or an equivalent error , the chances were even.

Very few raised questions about ambiguity, and the letters actually published in the column were not among those few.

The table below shows a variety of other possible host behaviors and the impact on the success of switching. Determining the player's best strategy within a given set of other rules the host must follow is the type of problem studied in game theory.

For example, if the host is not required to make the offer to switch the player may suspect the host is malicious and makes the offers more often if the player has initially selected the car.

In general, the answer to this sort of question depends on the specific assumptions made about the host's behavior, and might range from "ignore the host completely" to "toss a coin and switch if it comes up heads"; see the last row of the table below.

Both changed the wording of the Parade version to emphasize that point when they restated the problem. They consider a scenario where the host chooses between revealing two goats with a preference expressed as a probability q , having a value between 0 and 1.

This means even without constraining the host to pick randomly if the player initially selects the car, the player is never worse off switching.

As N grows larger, the advantage decreases and approaches zero Granberg A quantum version of the paradox illustrates some points about the relation between classical or non-quantum information and quantum information , as encoded in the states of quantum mechanical systems.

The formulation is loosely based on quantum game theory. The three doors are replaced by a quantum system allowing three alternatives; opening a door and looking behind it is translated as making a particular measurement.

The rules can be stated in this language, and once again the choice for the player is to stick with the initial choice, or change to another "orthogonal" option.

The latter strategy turns out to double the chances, just as in the classical case. However, if the show host has not randomized the position of the prize in a fully quantum mechanical way, the player can do even better, and can sometimes even win the prize with certainty Flitney and Abbott , D'Ariano et al.

In this puzzle, there are three boxes: a box containing two gold coins, a box with two silver coins, and a box with one of each. After choosing a box at random and withdrawing one coin at random that happens to be a gold coin, the question is what is the probability that the other coin is gold.

This problem involves three condemned prisoners, a random one of whom has been secretly chosen to be pardoned. The warden obliges, secretly flipping a coin to decide which name to provide if the prisoner who is asking is the one being pardoned.

The question is whether knowing the warden's answer changes the prisoner's chances of being pardoned. The first letter presented the problem in a version close to its presentation in Parade 15 years later.

The second appears to be the first use of the term "Monty Hall problem". The problem is actually an extrapolation from the game show.

As Monty Hall wrote to Selvin:. And if you ever get on my show, the rules hold fast for you — no trading boxes after the selection.

A version of the problem very similar to the one that appeared three years later in Parade was published in in the Puzzles section of The Journal of Economic Perspectives Nalebuff Nalebuff, as later writers in mathematical economics, sees the problem as a simple and amusing exercise in game theory.

Phillip Martin's article in a issue of Bridge Today magazine titled "The Monty Hall Trap" Martin presented Selvin's problem as an example of what Martin calls the probability trap of treating non-random information as if it were random, and relates this to concepts in the game of bridge.

A restated version of Selvin's problem appeared in Marilyn vos Savant 's Ask Marilyn question-and-answer column of Parade in September Tierney Due to the overwhelming response, Parade published an unprecedented four columns on the problem.

Adams initially answered, incorrectly, that the chances for the two remaining doors must each be one in two. After a reader wrote in to correct the mathematics of Adams's analysis, Adams agreed that mathematically he had been wrong.

Now you're offered this choice: open door 1, or open door 2 and door 3. In the latter case you keep the prize if it's behind either door.

You'd rather have a two-in-three shot at the prize than one-in-three, wouldn't you? If you think about it, the original problem offers you basically the same choice.

Monty is saying in effect: you can keep your one door or you can have the other two doors, one of which a non-prize door I'll open for you.

Numerous readers, however, wrote in to claim that Adams had been "right the first time" and that the correct chances were one in two.

Diese Lösung kann auch grafisch veranschaulicht werden [6] [7]. In den Bildern der folgenden Tabelle ist das gewählte Tor willkürlich als das linke Tor dargestellt:.

Im Ergebnis lässt sich die Auffassung des Spielablaufs von vos Savant auch auf folgende Weise reproduzieren:.

Es sind vor allem die folgenden Hauptargumente, die zu Zweifeln an vos Savants Antwort führen. Während das erste Argument nicht stichhaltig ist und auf falsch angewandter Wahrscheinlichkeitstheorie basiert, verdeutlichen die weiteren Argumente, dass das Originalproblem eine Vielzahl von Interpretationen zulässt:.

Das erste Argument wird durch den ausgeglichenen Moderator widerlegt, das zweite wird anhand der erfahrungsbezogenen Antwort und das dritte anhand des faulen Moderators ausgeführt.

Weil die im Leserbrief von Whitaker formulierte Aufgabe einigen Wissenschaftlern nicht eindeutig lösbar erschien, wurde von ihnen eine Neuformulierung des Ziegenproblems vorgeschlagen.

Diese als Monty-Hall-Standard-Problem bezeichnete Umformulierung, die zur gleichen Lösung wie der von Marilyn vos Savant führen soll, stellt bestimmte Zusatzinformationen bereit, welche die erfahrungsbezogene Antwort ungültig machen, und berücksichtigt im Unterschied zur Interpretation von vos Savant auch die konkrete Spielsituation: [8].

Hinter einem Tor ist ein Auto, hinter den anderen befindet sich jeweils eine Ziege. Die Regeln lauten: Nachdem Sie ein Tor gewählt haben, bleibt dieses zunächst geschlossen.

Hinter dem von ihm geöffneten Tor muss sich eine Ziege befinden. Ist es vorteilhaft, Ihre Wahl zu ändern? Insbesondere hat der Moderator die Möglichkeit, frei darüber zu entscheiden, welches Tor er öffnet, wenn er die Auswahl zwischen zwei Ziegentoren hat Sie haben also zuerst das Auto-Tor gewählt.

Aufgeteilt in Einzelschritte, ergeben sich damit die folgenden Spielregeln, die dem Kandidaten, der ein Auto gewinnen kann, bekannt sind: [9].

Mit einer solchen Zusatzannahme entsteht jeweils ein anderes Problem, das zu unterschiedlichen Gewinnchancen bei der Torauswahl des Kandidaten führen kann.

Dazu wird immer vorausgesetzt, dass der Kandidat die dem Moderator unterstellte Entscheidungsprozedur kennt.

Wie soll sich der Kandidat im vorletzten Schritt entscheiden, wenn er zunächst Tor 1 gewählt und der Moderator daraufhin Tor 3 mit einer Ziege dahinter geöffnet hat?

Wegen der Symmetrie im Regelwerk, insbesondere wegen der Spielregeln 4 und 5, wird diese Wahrscheinlichkeit durch das Öffnen eines anderen Tors mit einer Ziege dahinter nicht beeinflusst.

Für die Situationen, in denen der Kandidat die Tore 2 oder 3 gewählt hat und der Moderator dementsprechend andere Tore öffnet, gilt eine analoge Erklärung.

Das entspricht einem Zufallsexperiment, bei dem die beiden Ziegen voneinander unterschieden werden können und jede Verteilung von Auto und Ziegen hinter den drei Toren gleich wahrscheinlich ist Laplace-Experiment.

Zur Auswertung der Tabelle müssen nun die Fälle betrachtet werden, in denen der Moderator das Tor 3 öffnet das ist die Bedingung. Das sind die Fälle 2, 4 und 5.

Man sieht, dass in zwei dieser drei Fälle der Kandidat durch Wechseln gewinnt. Unter den Voraussetzungen, dass der Kandidat zunächst Tor 1 gewählt hat und der Moderator Tor 3 mit einer Ziege dahinter öffnet, befindet sich das Auto also in zwei Drittel der Fälle hinter Tor 2.

Der Kandidat sollte also seine Wahl zugunsten von Tor 2 ändern. Genauso kann aus der Tabelle abgelesen werden, dass dann, wenn der Moderator anstelle von Tor 3 das Tor 2 öffnet, der Kandidat durch Wechseln auf Tor 3 ebenfalls in zwei von drei Fällen das Auto gewinnt.

Lohnt es sich für den Kandidaten zu wechseln? Man kann diese Wahrscheinlichkeit mit dem Satz von Bayes ermitteln. Für die folgende Erklärung wird angenommen, dass der Kandidat zu Anfang Tor 1 gewählt hat.

Für die Situationen, in denen der Kandidat die Tore 2 bzw. Obwohl es hier ausreichen würde, die drei ersten Spielsituationen zu betrachten, werden sechs Fälle unterschieden, um die Problemstellung vergleichbar mit der obigen tabellarischen Lösung beim ausgeglichenen Moderator modellieren zu können.

Jede Spielsituation wird also zweimal betrachtet. Das sind die Fälle 1, 2, 4 und 5. Man sieht, dass nur in zwei von vier dieser Fälle der Kandidat durch Wechseln gewinnt.

Es kann ebenso leicht aus der Tabelle abgelesen werden, dass, wenn der Moderator Tor 2 öffnet, der Kandidat sicher gewinnt, wenn er zu Tor 3 wechselt.

Es liegt die folgende Situation vor: Der Kandidat hat Tor 1 gewählt, und der Moderator hat daraufhin das Tor 3 geöffnet. Es gelten dann folgende mathematische Beziehungen unter Berücksichtigung der oben definierten Ereignismengen:.

Die Anwendung des Satzes von Bayes ergibt dann für die bedingte Wahrscheinlichkeit, dass sich das Auto hinter Tor 2 befindet:.

Für die bedingte Wahrscheinlichkeit, dass sich das Auto tatsächlich hinter Tor 1 befindet, gilt aber ebenfalls. Der Gewinn hinter Tor 2 ist genauso wahrscheinlich wie der Gewinn hinter Tor 1.

Der Kandidat kann demnach in diesem Fall also ebenso gut bei Tor 1 bleiben wie zu Tor 2 wechseln. Dann gelten folgende mathematische Beziehungen unter Berücksichtigung der oben definierten Ereignismengen:.

Nachdem Monty Hall die Aufgabenstellung genau gelesen hatte, spielte er mit einem Versuchskandidaten das Spiel so, dass dieser bei einem Wechsel stets verlor, indem er den Wechsel immer nur dann anbot, wenn der Kandidat im ersten Schritt das Gewinn-Tor gewählt hatte.

Diese Unklarheit könne beseitigt werden, indem der Moderator vorher verspreche, eine andere Tür zu öffnen und danach einen Wechsel anzubieten.

Im ersten Moment scheint es komplett egal zu sein, welches Tor man nimmt. Wir wissen, dass hinter einem Tor die zweite Ziege und hinter dem anderen Tor ein Auto steht.

Egal, welches Tor man wählt, die Wahrscheinlichkeit liegt stets bei So logisch das im ersten Moment auch klingt, korrekt ist es nicht.

Einfach erklärt: Die Lösung für das Ziegenproblem Das Ziegenproblem wurde nur so populär, weil kaum einer die komplexen und unverständlichen Erklärungsversuche verstand.

Die folgenden drei Szenarien können mit gleicher Wahrscheinlichkeit eintreten:. Fall 1: Das Auto steht hinter Tor 1.

Fall 2: Das Auto steht hinter Tor 2. Der Moderator muss in diesem Fall Tor 3 öffnen, da er nicht die Wahl des Kandidaten und auch nicht das Auto enthüllen darf.

Für den Kandidaten ist ein Wechsel zu Tor 2 sinnvoll, da hier das Auto steht. Fall 3: Dieser Fall ist deckungsgleich mit Fall 2.

Das Auto steht hinter Tor 3.

Ziegenproblem

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