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The player has an equal chance of initially selecting the car, Goat A, or Goat B. Simulations[edit source editbeta] Simulation of 30 outcomes of the Monty Hall problem A simple way to demonstrate that a switching strategy really does win two out of three times on the average is to simulate the game with playing cards Gardner b; vos Savantp.

Three cards from an ordinary deck are used to represent the three doors; one 'special' card such as the Ace of Spades should represent the door with the car, and ordinary cards, such as the two red twos, represent the goat doors. The simulation, using the following procedure, can be repeated several times to simulate multiple rounds of the game.

One card is dealt face-down at random to the 'player', to represent the door the player picks initially. Then, looking at the remaining two cards, at least one of which must be a red two, the 'host' discards a red two. If the card remaining in the host's hand is the Ace of Spades, this is recorded as a round where the player would have won by switching; if the host is holding a red two, the round is recorded as one where staying would have won.

As this experiment is repeated over several rounds, the observed win rate for each strategy is likely to approximate its theoretical win probability. Repeated plays also make it clearer why switching is the better strategy.

After one card has been dealt to the player, it is already determined whether switching will win the round for the player; and two times out of three the Ace of Spades is in the host's hand. If this is not convincing, the simulation can be done with the entire deck, dealing one card to the player and keeping the other 51 Gardner b; Adams In this variant the Ace of Spades goes to the host 51 times out of 52, and stays with the host no matter how pajūrio nh pažintys non-Ace cards are discarded.

Another simulation, suggested by vos Savant, employs the "host" hiding a penny, representing the car, under one of three cups, representing the doors; or hiding a pea under one of greitasis pažintys melbourne fl shells.

Solutions using conditional probability[edit source editbeta] The player is only asked whether or not he would like to switch after the host has opened a particular door, different from the player's initial choice, and revealed a goat behind it.

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In order to be sure of having the best possible chance of getting the car, the player should clearly make his decision greitasis pažintys melbourne fl whether or not to switch on the basis of the probability that the car is behind the other closed door, given all the information which he has at this moment.

That information consists, in the example given by Marilyn vos Savant, of the fact that it was door 1 which was initially chosen by himself, and that it was door 3 which was opened by the host to reveal a goat.

Refining the simple solution[edit source editbeta] Given the player's initial choice, door 1, the host might also have opened door 2 to reveal a goat. Suppose we assume that the host is equally likely to open either door 2 or 3 if the car is behind door 1.

We already assumed that the car is equally likely behind any of the three doors, and this remains so after the player has made his initial choice. It follows that the probability that the car is behind door 1 given the host opens door 3 and the player initially chose door 1 must equal the probability that the car is behind door 1 given the host opens door 2 and the player initially chose door 1, since all elementary probabilities in the problem probabilities of locations of car given the player initially chose door 1, and probabilities of which door the host will open if the car is behind door 1, the door chosen by the player are unchanged by interchanging the door numbers 2 and 3.

In other words, given that the player initially chose door 1, whether the host opens door 2 or door 3 gives us no information at all as to whether or not the car is behind door 1.

Moreover, the host is certainly going to open a different door to reveal a goat, so opening a door which door, door 2 or door 3, unspecified does not change this probability. But these two probabilities are the same. This more refined analysis which can be found in the published discussion following the paper of Morgan et al.

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For example, if we imagine repetitions of the show, in all of greitasis pažintys melbourne fl the player initially chooses door 1, a strategy of "always switching" will win about times — namely every time the car is not behind door 1.

The solutions in this section consider just those cases in which not only the player picked door 1, but moreover the host went on to open door 3. After all, it is only after a door was opened to reveal a goat that the player was asked if he wants to switch or not. Of our total of repetitions, door 3 will be opened by the host about times: the other times the host will open door 2. The following solutions show that when we restrict attention to the first mentioned cases player initially picked door 1, host went on to reveal a goat behind door 3the car will be behind door 2 about times, but behind door 1 only about 50 times.

Tree showing the pažintys prižiūrėtoją darbe of every possible outcome if the player initially picks Door 1 By definition, the conditional probability of winning by switching given the contestant initially picks door 1 and the host opens door 3 is the probability the car is behind door 2 and the host opens door 3 divided by the probability the host opens door 3.

These probabilities can be determined referring to the conditional probability table below, or to an equivalent decision tree as shown to the right Chun ; Carlton ; Grinstead and Snell — These are the only possibilities given the player picks door 1 and the host opens door 3. The conditional probability table below shows how cases, in all of which the player initially chooses door 1, would be split up, on average, according to the location of the car and the choice of door to open by the greitasis pažintys melbourne fl.

Here it is with probabilities included:Case Prob. This splits case 2 into two equally likely subcases 2a pažintys tikrai haram 2b, and case 3 into two equally likely subcases 3a and 3b:Case Prob.

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In three of the six equally likely cases, the host opens door 3. In two of those three cases, the car is behind door 2. Bayes' theorem[edit source editbeta] Many probability text books and articles in the field of probability theory and mn pažintys st paul teaching of probability theory derive the conditional probability solution through a formal application of Bayes' theorem; among them Gill, and Henze, Use of the odds form of Bayes' theorem, often called Bayes' rule, makes such a derivation more transparent Rosenthal, aRosenthal, b.

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Initially, the car is equally likely behind any of the three doors: the odds on door 1, door 2, and door 3 are This remains the case after the player has chosen door 1, by independence. According to Bayes' rule, the posterior odds on the location of the car, given 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 vakaro pašto pažintys hypotheses considered location of the car.

Thus the posterior odds become equal to the Greitasis pažintys melbourne fl factor 1 : 2 : 0. Given the host opened door 3, the probability the car is behind door 3 is zero, and it is twice more likely to be behind door 2 than door 1.

Richard Gill uses a similar argument to patch a missing step in Devlin's combined doors solution Devlindescribed above "Adams and Devlin". Devlin does not explain why the information of which door had been opened by the host did not change the odds on door 1 hiding the car.

The contestant is not just given the opportunity to choose between door greitasis pažintys melbourne fl and doors 2 plus 3 — he is also told something specific about those two doors "no car behind door 3".

Correspondents pointed out this missing step to Devlin and he later Devlin retracted his "combined doors" solution, giving a direct calculation using Bayes theorem instead, saying that though unintuitive, it did at least give one the guarantee of obtaining the right answer in an automatic way.

The justification provided by Gill is as follows. Given the car is not behind door 1 the door initially chosen by the contestantit is equally likely that it is behind door 2 or door 3.

In those cases, the host is forced to open the other door.

The information "host opens door 3" contributes a Bayes factor or likelihood ratio of 50 : 50, or 1 : 1, concerning the question whether or not the car is behind door 1. Initially, the odds against door 1 hiding the car were 2 : 1. Therefore the posterior odds against door 1 hiding the car remain the same as the prior odds, 2 : 1. In words, the information which door is opened by the host door 2 or door 3?

Strategic dominance solution[edit source editbeta] Going pažintys turinčio didžiulę skolą to Nalebuffthe Monty Hall problem is also much studied in the literature on game theory and decision theory, and also some popular solutions for instance, that were published in The Economist, see above, among the simple 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 greitasis pažintys melbourne fl 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.

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Following Gill, a strategy of contestant involves two greitasis pažintys melbourne fl 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. Elementary comparison of contestant's strategies shows that for every strategy A there is another strategy B "pick a door then switch greitasis pažintys melbourne fl matter what happens" which dominates it Gnedin, No matter how the car is hidden and no greitasis pažintys melbourne fl 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 2 then always switch after the host reveals a door": A wins when door 1 conceals the car, while Bricklerry greitis pažintys wins when one of the doors 1 and 3 conceals the car.

Similarly, strategy A "pick door 1 then switch to door 2 if offeredbut 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 which 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 nonswitching 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. Confusion and criticism[edit source editbeta] Sources of confusion[edit source editbeta] When greitasis pažintys melbourne fl presented with the Monty Hall problem an overwhelming majority of people assume that each door has an equal probability and conclude that switching does not matter Mueser and Granberg, Most statements of the problem, notably the one in Parade Magazine, do not match the rules of the actual game show Krauss and Wang,and do not fully specify the host's behavior or that the car's location is randomly selected Granberg and Brown, Krauss and Wang conjecture that people make the standard assumptions even if they are not explicitly stated.

From the point of greitasis pažintys melbourne fl of subjective probability, the standard assumptions can be derived from the problem statement: they follow from our total lack of information about how the car is hidden, how the player initially chooses a door, and how the host chooses a door to open if there's a choice. Although these issues are mathematically significant, even when controlling for these factors nearly all people still think each of the two unopened doors has an equal probability and conclude switching does not matter Mueser and Granberg, This "equal probability" assumption is a deeply rooted intuition Falk People strongly tend to think probability is evenly distributed across as many azijos greitis pažintys auckland as are present, whether it is or not Fox and Levav, Steinbach, The problem continues to attract the attention of cognitive psychologists.

The typical behaviour of the majority, i. Experimental evidence confirms that these are plausible explanations which greitasis pažintys melbourne fl not depend on probability intuition Morone and Fiore, Among these sources are several that explicitly criticize the popularly presented "simple" solutions, saying these solutions are "correct but Some say that these solutions answer a slightly different question — one phrasing is "you have to announce before a door has been opened whether you plan to switch" Gillmanemphasis in the original.

However, the probability of winning by always switching is a logically distinct concept from the probability of winning by switching given the player has picked door 1 and the host has opened door 3. As one source says, "the distinction between [these questions] seems to confound many" Morgan et al.

## Pavyzdžiai. Empirical data

This fact that these are different can be shown by varying the problem so that these two probabilities have different numeric greitasis pažintys melbourne fl. For example, assume the contestant knows that Monty does not pick the second door randomly among all legal alternatives but instead, when given an opportunity to pick between two losing doors, Monty will open the one on the right.

In this situation the following two questions have different answers: What is the probability of winning the car by always switching? What is the probability of winning the car given the player has picked door 1 and the host has opened door 3? For this variation, the two questions yield different answers. Morgan et al. Behrends concludes that "One must consider the matter with care to see that both analyses are correct"; which is not to say that they are the same.

One analysis for one question, another analysis for the other question.

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Several discussants of the paper by Morgan et al. One discussant William Bell considered it a matter of taste whether or not one explicitly mentions that under the standard conditionswhich door is opened by the host is independent of whether or not one should want to switch.

Among the simple solutions, the "combined doors solution" comes closest to a conditional solution, as we saw in the discussion of approaches using the concept of odds and Bayes theorem. It is based on the deeply rooted intuition that revealing information that is already known does not affect probabilities.

But knowing the host can open one of the two unchosen doors to show a goat does not mean that opening a specific door would not affect the probability that the car is behind the initially chosen door. The point is, though we know in advance that the host will open a door and reveal a goat, we do not know which door he will open.

If the host chooses uniformly at random between doors hiding a goat as is the case in the standard interpretation this probability indeed remains unchanged, but if the host can choose non-randomly between such doors then the specific door that the host opens reveals additional information. The host can always open a door revealing a goat and in the standard interpretation of the problem the probability that the car is behind the initially chosen door does not change, but it is not because greitasis pažintys melbourne fl the former that the latter is true.

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Solutions based on the assertion that the host's actions cannot affect the probability that the car is behind the initially chosen appear persuasive, but the assertion is simply untrue unless each of the host's two choices are equally likely, if he has a choice FalkThe assertion therefore needs to be justified; without justification being greitasis pažintys melbourne fl, the solution is at best incomplete.

The answer can be correct but the reasoning used to justify it is defective. Some of the confusion in the literature undoubtedly arises because the writers are using different concepts of probability, in particular, Bayesian versus frequentist probability.

For the Bayesian, probability represents knowledge.

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For us and for the player, the car is initially equally likely to be behind each of the three doors because we know absolutely nothing about how the organizers of the show decided where to place it. For us and for the player, the host is equally likely to make either choice when he has one because we know absolutely nothing about how he makes his choice.

These "equally likely" probability assignments are determined by symmetries in the problem. The same symmetry can be used to argue in advance that specific door numbers are irrelevant, as we saw above.

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Variants[edit source editbeta] 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.