b'\n_START_ARTICLE_\nMonty Hall problemi\n_START_SECTION_\nKafa kar\xc4\xb1\xc5\x9f\xc4\xb1kl\xc4\xb1\xc4\x9f\xc4\xb1n\xc4\xb1n kaynaklar\xc4\xb1\n_START_PARAGRAPH_\nMonty Hall problemi ilk a\xc3\xa7\xc4\xb1kland\xc4\xb1\xc4\x9f\xc4\xb1 zaman insanlar\xc4\xb1n b\xc3\xbcy\xc3\xbck \xc3\xa7o\xc4\x9funlu\xc4\x9fu t\xc3\xbcm kap\xc4\xb1lar\xc4\xb1n e\xc5\x9fit olas\xc4\xb1l\xc4\xb1kta oldu\xc4\x9funu ve kap\xc4\xb1 de\xc4\x9fi\xc5\x9ftirmenin \xc3\xb6nemli olmad\xc4\xb1\xc4\x9f\xc4\xb1n\xc4\xb1 d\xc3\xbc\xc5\x9f\xc3\xbcnmekteydi (Mueser and Granberg, 1999). Yap\xc4\xb1lan bir \xc3\xa7al\xc4\xb1\xc5\x9fmada 228 denekten yaln\xc4\xb1zca %13\'\xc3\xbc kap\xc4\xb1 de\xc4\x9fi\xc5\x9ftirmeyi se\xc3\xa7ti (Granberg and Brown, 1995:713). The Power of Logical Thinking isimli kitab\xc4\xb1nda vos Savant (1996:15) bili\xc5\x9fsel psikolog Massimo Piattelli-Palmarini\'nin  "... tarihte di\xc4\x9fer hi\xc3\xa7bir istatistiksel bulmaca insanlar\xc4\xb1 kand\xc4\xb1rmaya bu kadar yakla\xc5\x9fmad\xc4\xb1" ve "Nobel \xc3\x96d\xc3\xbcll\xc3\xbc fizik\xc3\xa7iler bile bu probleme sistematik \xc5\x9fekilde yanl\xc4\xb1\xc5\x9f cevap verir ve bunda \xc4\xb1srar eder. Ayr\xc4\xb1ca do\xc4\x9fru yan\xc4\xb1t\xc4\xb1 verenleri bas\xc4\xb1l\xc4\xb1 bir eserle paylamaya haz\xc4\xb1rd\xc4\xb1rlar." s\xc3\xb6zleri ile bu durumu anlatt\xc4\xb1._NEWLINE_Problemin \xc3\xa7o\xc4\x9fu a\xc3\xa7\xc4\xb1klamas\xc4\xb1, \xc3\xb6zellikle de Parade Magazine\'deki, as\xc4\xb1l yar\xc4\xb1\xc5\x9fma program\xc4\xb1n\xc4\xb1n kurallar\xc4\xb1na uymaz (Krauss and Wang, 2003:9) ve sunucunun davran\xc4\xb1\xc5\x9f\xc4\xb1n\xc4\xb1 ya da araban\xc4\xb1n yerinin rastgele se\xc3\xa7ildi\xc4\x9fini a\xc3\xa7\xc4\xb1k\xc3\xa7a belirtmez (Granberg and Brown, 1995:712). Krauss ve Wang, insanlar\xc4\xb1n a\xc3\xa7\xc4\xb1k\xc3\xa7a belirtilmi\xc5\x9f olmasa da baz\xc4\xb1 noktalara ili\xc5\x9fkin standart varsay\xc4\xb1mlarda bulundu\xc4\x9funu ileri s\xc3\xbcrd\xc3\xbc (2003:10). Bu meseleler matematiksel a\xc3\xa7\xc4\xb1dan \xc3\xb6nemli olsa da, bu etmenleri ele al\xc4\xb1rken dahi neredeyse t\xc3\xbcm insanlar h\xc3\xa2l\xc3\xa2, a\xc3\xa7\xc4\xb1lmam\xc4\xb1\xc5\x9f iki kap\xc4\xb1n\xc4\xb1n olas\xc4\xb1l\xc4\xb1\xc4\x9f\xc4\xb1n\xc4\xb1n e\xc5\x9fit oldu\xc4\x9funu ve dolay\xc4\xb1s\xc4\xb1yla kap\xc4\xb1 de\xc4\x9fi\xc5\x9ftirmenin bir \xc5\x9feyi de\xc4\x9fi\xc5\x9ftirmeyece\xc4\x9fini d\xc3\xbc\xc5\x9f\xc3\xbcn\xc3\xbcr (Mueser and Granberg, 1999). Bu "e\xc5\x9fit olas\xc4\xb1l\xc4\xb1k" varsay\xc4\xb1m\xc4\xb1, derin \xc5\x9fekilde k\xc3\xb6kle\xc5\x9fmi\xc5\x9f bir sezgidir (Falk 1992:202). \xc4\xb0nsanlar \xc3\xb6yle olsa da olmasa da, olas\xc4\xb1l\xc4\xb1klar\xc4\xb1n o anda mevcut t\xc3\xbcm bilinmeyenler aras\xc4\xb1nda e\xc5\x9fit \xc5\x9fekilde da\xc4\x9f\xc4\xb1ld\xc4\xb1\xc4\x9f\xc4\xb1n\xc4\xb1 g\xc3\xbc\xc3\xa7l\xc3\xbc bi\xc3\xa7imde d\xc3\xbc\xc5\x9f\xc3\xbcnme e\xc4\x9filimindedir (Fox and Levav, 2004:637)._NEWLINE_Monty Hall problemi a\xc3\xa7\xc4\xb1s\xc4\xb1ndan k\xc3\xb6kle\xc5\x9fmi\xc5\x9f bir di\xc4\x9fer nokta ise, zaten bilinmekte olan bilgilerin ortaya \xc3\xa7\xc4\xb1kar\xc4\xb1lmas\xc4\xb1n\xc4\xb1n olas\xc4\xb1l\xc4\xb1klar\xc4\xb1 etkilemeyece\xc4\x9fine duyulan inan\xc3\xa7t\xc4\xb1r (Falk 1992:207). Bu y\xc3\xb6ndeki sezgi, sunucunun bir kap\xc4\xb1y\xc4\xb1 a\xc3\xa7mas\xc4\xb1n\xc4\xb1n, yar\xc4\xb1\xc5\x9fmac\xc4\xb1n\xc4\xb1n en ba\xc5\x9fta 1/3 olan arabay\xc4\xb1 bulma olas\xc4\xb1l\xc4\xb1\xc4\x9f\xc4\xb1n\xc4\xb1 de\xc4\x9fi\xc5\x9ftirmeyece\xc4\x9fini \xc3\xb6ne s\xc3\xbcren \xc3\xa7\xc3\xb6z\xc3\xbcmlerin de temelini olu\xc5\x9fturur. Tam problem i\xc3\xa7in bu sezgi, se\xc3\xa7imi de\xc4\x9fi\xc5\x9ftirmenin 2/3 olas\xc4\xb1l\xc4\xb1kla kazand\xc4\xb1raca\xc4\x9f\xc4\xb1 do\xc4\x9fru sonucuna g\xc3\xb6t\xc3\xbcr\xc3\xbcr; ancak, do\xc4\x9fru yan\xc4\xb1t\xc4\xb1n bu olmad\xc4\xb1\xc4\x9f\xc4\xb1 k\xc4\xb1smen de\xc4\x9fi\xc5\x9ftirilmi\xc5\x9f problemlerde de g\xc3\xb6t\xc3\xbcrece\xc4\x9fi sonu\xc3\xa7 ayn\xc4\xb1d\xc4\xb1r (Falk 1992:207)._NEWLINE_According to Morgan et al. (1991) "The distinction between the conditional and unconditional situations here seems to confound many."  That is, they, and some others, interpret the usual wording of the problem statement as asking about the conditional probability of winning given which door is opened by the host, as opposed to the overall or unconditional probability._NEWLINE_These are mathematically different questions and can have different answers depending on how the host chooses which door to open when the player\'s initial choice is the car (Morgan et al., 1991; Gillman 1992).  For example, if the host opens Door 3 whenever possible then the probability of winning by switching for players initially choosing Door 1 is 2/3 overall, but only 1/2 if the host opens Door 3.  In its usual form the problem statement does not specify this detail of the host\'s behavior, nor make clear whether a conditional or an unconditional answer is required, making the answer that switching wins the car with probability 2/3 equally vague.  Many commonly presented solutions address the unconditional probability, ignoring which door was chosen by the player and which door opened by the host; Morgan et al. call these "false solutions" (1991).  Others, such as Behrends (2008), conclude that "One must consider the matter with care to see that both analyses are correct."\n_START_SECTION_\nOlas\xc4\xb1l\xc4\xb1k neden 1/2 de\xc4\x9fil?\n_START_PARAGRAPH_\nYar\xc4\xb1\xc5\x9fmac\xc4\xb1n\xc4\xb1n ilk turda arabal\xc4\xb1 kap\xc4\xb1y\xc4\xb1 se\xc3\xa7mesi i\xc3\xa7in 3\'te 1 \xc5\x9fans\xc4\xb1 vard\xc4\xb1r. Sonra Monty Hall se\xc3\xa7ilmeyen iki kap\xc4\xb1dan i\xc3\xa7inde ke\xc3\xa7i oldu\xc4\x9funu bildi\xc4\x9fi birini rastgele olmayan \xc5\x9fekilde a\xc3\xa7ar. E\xc4\x9fer yar\xc4\xb1\xc5\x9fmac\xc4\xb1 ba\xc5\x9flang\xc4\xb1\xc3\xa7ta arabal\xc4\xb1 kap\xc4\xb1y\xc4\xb1 se\xc3\xa7mi\xc5\x9fse (ba\xc5\x9flang\xc4\xb1\xc3\xa7ta \xc5\x9fans\xc4\xb1 1/3) geriye kalan kap\xc4\xb1da bir ke\xc3\xa7i olacakt\xc4\xb1r. E\xc4\x9fer yar\xc4\xb1\xc5\x9fmac\xc4\xb1 ke\xc3\xa7ili kap\xc4\xb1y\xc4\xb1 se\xc3\xa7mi\xc5\x9fse (ba\xc5\x9flang\xc4\xb1\xc3\xa7taki \xc5\x9fans\xc4\xb1 2/3) geri kalan kap\xc4\xb1da bir araba olacakt\xc4\xb1r._NEWLINE_Buradaki kritik nokta Monty\'nin bir kap\xc4\xb1y\xc4\xb1 rastgele se\xc3\xa7medi\xc4\x9fidir - yar\xc4\xb1\xc5\x9fmac\xc4\xb1 tercihini yapt\xc4\xb1\xc4\x9f\xc4\xb1nda Monty i\xc3\xa7inde ke\xc3\xa7i oldu\xc4\x9funu bildi\xc4\x9fi bir kap\xc4\xb1y\xc4\xb1 se\xc3\xa7er. Bunun anlam\xc4\xb1 Monty\'nin se\xc3\xa7iminin, ba\xc5\x9flang\xc4\xb1\xc3\xa7ta araban\xc4\xb1n yar\xc4\xb1\xc5\x9fmac\xc4\xb1n\xc4\xb1n kap\xc4\xb1s\xc4\xb1n\xc4\xb1n ard\xc4\xb1nda olma olas\xc4\xb1l\xc4\xb1\xc4\x9f\xc4\xb1n\xc4\xb1 etkilemeyece\xc4\x9fidir. Yar\xc4\xb1\xc5\x9fmac\xc4\xb1ya kap\xc4\xb1y\xc4\xb1 de\xc4\x9fi\xc5\x9ftirip de\xc4\x9fi\xc5\x9ftirmeyece\xc4\x9fi soruldu\xc4\x9funda, hala ba\xc5\x9flang\xc4\xb1\xc3\xa7ta arabal\xc4\xb1 kap\xc4\xb1 se\xc3\xa7ilme olas\xc4\xb1l\xc4\xb1\xc4\x9f\xc4\xb1 3\'te 1 ve ke\xc3\xa7ili kap\xc4\xb1 se\xc3\xa7ilme olas\xc4\xb1l\xc4\xb1\xc4\x9f\xc4\xb1 3\'te 2\'dir. Ancak \xc5\x9fimdi Monty di\xc4\x9fer kap\xc4\xb1lardan birini a\xc3\xa7ar ve a\xc3\xa7t\xc4\xb1\xc4\x9f\xc4\xb1 kap\xc4\xb1da araba bulunamaz, yani yar\xc4\xb1\xc5\x9fmac\xc4\xb1n\xc4\xb1n kap\xc4\xb1s\xc4\xb1nda ke\xc3\xa7i bulunma olas\xc4\xb1l\xc4\xb1\xc4\x9f\xc4\xb1 3\'te 2 ve kalan kap\xc4\xb1da araba olma olasl\xc4\xb1\xc4\x9f\xc4\xb1 da 3\'te 2\'dir._NEWLINE_Bu Monty\'nin kap\xc4\xb1s\xc4\xb1n\xc4\xb1 rastgele se\xc3\xa7mesi ve arabay\xc4\xb1 a\xc3\xa7\xc4\xb1\xc4\x9fa \xc3\xa7\xc4\xb1karmas\xc4\xb1 ihtimalinin var olmas\xc4\xb1ndan farkl\xc4\xb1 bir senaryodur. Bu durumda ke\xc3\xa7inin a\xc3\xa7\xc4\xb1\xc4\x9fa \xc3\xa7\xc4\xb1kmas\xc4\xb1, yar\xc4\xb1\xc5\x9fmac\xc4\xb1n\xc4\xb1n ba\xc5\x9flang\xc4\xb1\xc3\xa7 se\xc3\xa7imininin araba olmas\xc4\xb1 olas\xc4\xb1l\xc4\xb1\xc4\x9f\xc4\xb1n\xc4\xb1 2\'de 1\'e y\xc3\xbckseltecektir. Bu farkl\xc4\xb1l\xc4\xb1k orijinal problemin, vos Savant\'\xc4\xb1n 2006 Kas\xc4\xb1m\'\xc4\xb1ndaki k\xc3\xb6\xc5\x9fesinde g\xc3\xb6r\xc3\xbclen de\xc4\x9fi\xc5\x9ftirilmi\xc5\x9f versiyonu ile kar\xc5\x9f\xc4\xb1la\xc5\x9ft\xc4\xb1r\xc4\xb1lmas\xc4\xb1 ile ispatlanabilir. Bu versiyonda Monty Hall hangi kap\xc4\xb1da araban\xc4\xb1n sakland\xc4\xb1\xc4\x9f\xc4\xb1n\xc4\xb1 unutur. Kap\xc4\xb1lardan birini rastgele se\xc3\xa7er ve ke\xc3\xa7i a\xc3\xa7\xc4\xb1\xc4\x9fa \xc3\xa7\xc4\xb1kt\xc4\xb1\xc4\x9f\xc4\xb1nda rahatlar. Yar\xc4\xb1\xc5\x9fmac\xc4\xb1ya kap\xc4\xb1s\xc4\xb1n\xc4\xb1 de\xc4\x9fi\xc5\x9ftirip de\xc4\x9fi\xc5\x9ftirmeyece\xc4\x9fi soruldu\xc4\x9funda vos Savant do\xc4\x9fru bir \xc5\x9fekilde yan\xc4\xb1tlar "E\xc4\x9fer sunucu bilmiyorsa, de\xc4\x9fi\xc5\x9ftirmeniz ya da de\xc4\x9fi\xc5\x9ftirmemeniz hi\xc3\xa7bir \xc5\x9fey farkettirmez. E\xc4\x9fer sunucu biliyorsa de\xc4\x9fi\xc5\x9ftirin" (vos Savant, 2006)._NEWLINE_Another way of looking at the situation is to consider that if the contestant chooses to switch then they are effectively getting to see what is behind 2 of the 3 doors, and will win if either one of them has the car. In this situation one of the unchosen doors will have the car 2/3 of the time and the other will have a goat 100% of the time. The fact that Monty Hall shows one of the doors has a goat before the contestant makes the switch is irrelevant, because one of the doors will always have a goat and Monty has chosen it deliberately. The contestant still gets to look behind 2 doors and win if either has the car, it is just confirmed that one of doors will have a goat first.\n_START_SECTION_\nKap\xc4\xb1lar\xc4\xb1n say\xc4\xb1s\xc4\xb1n\xc4\xb1 art\xc4\xb1rmak\n_START_PARAGRAPH_\nAyn\xc4\xb1 problemde kap\xc4\xb1lar\xc4\xb1n say\xc4\xb1s\xc4\xb1n\xc4\xb1n \xc3\xbc\xc3\xa7 yerine 1.000.000 oldu\xc4\x9funu d\xc3\xbc\xc5\x9f\xc3\xbcnmek, \xc3\xa7\xc3\xb6z\xc3\xbcm\xc3\xbc kavramay\xc4\xb1 kolayla\xc5\x9ft\xc4\xb1rabilir (vos Savant 1990; Nesin 2009). Bu durumda ard\xc4\xb1nda ke\xc3\xa7i olan 999.999 kap\xc4\xb1 ve \xc3\xb6d\xc3\xbcl bulunan bir kap\xc4\xb1 olur. Yar\xc4\xb1\xc5\x9fmac\xc4\xb1 bir kap\xc4\xb1y\xc4\xb1 se\xc3\xa7er. Daha sonra sunucu di\xc4\x9fer 999.998 kap\xc4\xb1y\xc4\xb1 a\xc3\xa7arak ke\xc3\xa7ileri g\xc3\xb6sterir - sunucunun ilk kap\xc4\xb1dan ba\xc5\x9flayarak 1.000.000 kap\xc4\xb1l\xc4\xb1k s\xc4\xb1ray\xc4\xb1 takip etti\xc4\x9fini, her birini a\xc3\xa7t\xc4\xb1\xc4\x9f\xc4\xb1n\xc4\xb1, yaln\xc4\xb1zca yar\xc4\xb1\xc5\x9fmac\xc4\xb1n\xc4\xb1n se\xc3\xa7ti\xc4\x9fi kap\xc4\xb1 ile ba\xc5\x9fka bir kap\xc4\xb1y\xc4\xb1 daha atlad\xc4\xb1\xc4\x9f\xc4\xb1n\xc4\xb1 d\xc3\xbc\xc5\x9f\xc3\xbcnelim. Ard\xc4\xb1ndan sunucu yar\xc4\xb1\xc5\x9fmac\xc4\xb1ya a\xc3\xa7\xc4\xb1lmam\xc4\xb1\xc5\x9f di\xc4\x9fer kap\xc4\xb1y\xc4\xb1 se\xc3\xa7me \xc5\x9fans\xc4\xb1 oldu\xc4\x9funu s\xc3\xb6yler. Ortalama 1.000.000\'da 999.999 olas\xc4\xb1l\xc4\xb1kla di\xc4\x9fer kap\xc4\xb1 \xc3\xb6d\xc3\xbcle a\xc3\xa7\xc4\xb1l\xc4\xb1r, zira 1.000.000\'da 999.999 olas\xc4\xb1l\xc4\xb1kla yar\xc4\xb1\xc5\x9fmac\xc4\xb1n\xc4\xb1n ilk se\xc3\xa7ti\xc4\x9fi kap\xc4\xb1n\xc4\xb1n ard\xc4\xb1nda ke\xc3\xa7i vard\xc4\xb1r. Ak\xc4\xb1lc\xc4\xb1 bir yar\xc4\xb1\xc5\x9fmac\xc4\xb1 kap\xc4\xb1y\xc4\xb1 de\xc4\x9fi\xc5\x9ftirmelidir. Sezgisel bak\xc4\xb1l\xc4\xb1rsa, yar\xc4\xb1\xc5\x9fmac\xc4\xb1, bir milyon kap\xc4\xb1 i\xc3\xa7inden do\xc4\x9fru olan\xc4\xb1 se\xc3\xa7meyi ba\xc5\x9farma olas\xc4\xb1l\xc4\xb1\xc4\x9f\xc4\xb1n\xc4\xb1 sorgulamal\xc4\xb1d\xc4\xb1r. \xc3\x96rnek, se\xc3\xa7imi de\xc4\x9fi\xc5\x9ftirerek ba\xc5\x9far\xc4\xb1 sa\xc4\x9flama olas\xc4\xb1l\xc4\xb1\xc4\x9f\xc4\xb1n\xc4\xb1n (1 eksi ilk se\xc3\xa7imde do\xc4\x9fru kap\xc4\xb1y\xc4\xb1 se\xc3\xa7me olas\xc4\xb1l\xc4\xb1\xc4\x9f\xc4\xb1) her say\xc4\xb1da kap\xc4\xb1 i\xc3\xa7in e\xc5\x9fit oldu\xc4\x9funu ortaya koymak i\xc3\xa7in de kullan\xc4\xb1labilir. Ancak unutulmamal\xc4\xb1 ki, bu \xc3\xb6rnek, sunucunun \xc3\xb6d\xc3\xbcl\xc3\xbcn nerede oldu\xc4\x9funu bildi\xc4\x9fi varsay\xc4\xb1m\xc4\xb1na ve \xc3\xb6d\xc3\xbcl\xc3\xbc saklayan kap\xc4\xb1y\xc4\xb1 a\xc3\xa7mamas\xc4\xb1 gere\xc4\x9fine dayan\xc4\xb1r. E\xc4\x9fer yar\xc4\xb1\xc5\x9fmac\xc4\xb1 ilk se\xc3\xa7iminde \xc3\xb6d\xc3\xbcl\xc3\xbc saklayan kap\xc4\xb1y\xc4\xb1 bulmay\xc4\xb1 ba\xc5\x9farm\xc4\xb1\xc5\x9fsa, sunucu a\xc3\xa7mayaca\xc4\x9f\xc4\xb1 di\xc4\x9fer kap\xc4\xb1y\xc4\xb1 rastgele se\xc3\xa7er._NEWLINE_Bu \xc3\xb6rnek, sunucunun \xc3\xb6d\xc3\xbcl\xc3\xbcn yerini bilmedi\xc4\x9fi ve kap\xc4\xb1lar\xc4\xb1 rastgele a\xc3\xa7t\xc4\xb1\xc4\x9f\xc4\xb1 tam tersi bir durumu a\xc3\xa7\xc4\xb1klamakta da kullan\xc4\xb1labilir. 1.000.000\'da 999.999 olas\xc4\xb1l\xc4\xb1kla yar\xc4\xb1\xc5\x9fmac\xc4\xb1 en ba\xc5\x9fta yanl\xc4\xb1\xc5\x9f kap\xc4\xb1y\xc4\xb1 se\xc3\xa7ecektir, \xc3\xb6d\xc3\xbcl di\xc4\x9fer kap\xc4\xb1lardan birinin ard\xc4\xb1ndad\xc4\xb1r. E\xc4\x9fer sunucu \xc3\xb6d\xc3\xbcl\xc3\xbcn yerini bilmeksizin kap\xc4\xb1lar\xc4\xb1 rastgele a\xc3\xa7\xc4\xb1yorsa, aralar\xc4\xb1nda yeni bir se\xc3\xa7im yap\xc4\xb1lacak son iki kap\xc4\xb1ya (yar\xc4\xb1\xc5\x9fmac\xc4\xb1n\xc4\xb1n se\xc3\xa7imi ile di\xc4\x9fer kap\xc4\xb1) gelmeden, b\xc3\xbcy\xc3\xbck olas\xc4\xb1l\xc4\xb1kla \xc3\xb6d\xc3\xbcl\xc3\xbc saklayan kap\xc4\xb1y\xc4\xb1 a\xc3\xa7acakt\xc4\xb1r. Bu durum bir ba\xc5\x9fka yar\xc4\xb1\xc5\x9fma program\xc4\xb1 olan Deal or No Deal\'a benzer. Bu programda yar\xc4\xb1\xc5\x9fmac\xc4\xb1 numaraland\xc4\xb1r\xc4\xb1lm\xc4\xb1\xc5\x9f bir \xc3\xa7antay\xc4\xb1 se\xc3\xa7er ve daha sonra di\xc4\x9ferlerini her defas\xc4\xb1nda bir tane olmak \xc3\xbczere a\xc3\xa7ar._NEWLINE_Stibel ve di\xc4\x9ferleri (2008) \xc3\xa7al\xc4\xb1\xc5\x9fan zihin talebinin Monty Hall problemi s\xc4\xb1ras\xc4\xb1nda zorland\xc4\xb1\xc4\x9f\xc4\xb1n\xc4\xb1 ve bunun, insanlar\xc4\xb1n se\xc3\xa7imlerinin iki e\xc5\x9fit olas\xc4\xb1l\xc4\xb1kl\xc4\xb1 se\xc3\xa7ene\xc4\x9fe d\xc3\xb6n\xc3\xbc\xc5\x9ferek \xc3\xa7\xc3\xb6kmesine yol a\xc3\xa7t\xc4\xb1\xc4\x9f\xc4\xb1n\xc4\xb1 \xc3\xb6ne s\xc3\xbcrd\xc3\xbc. Ayr\xc4\xb1ca se\xc3\xa7eneklerin (kap\xc4\xb1lar\xc4\xb1n) say\xc4\xb1s\xc4\xb1n\xc4\xb1n yedinin \xc3\xbczerine \xc3\xa7\xc4\xb1kmas\xc4\xb1n\xc4\xb1n insanlardaki se\xc3\xa7imi de\xc4\x9fi\xc5\x9ftirme e\xc4\x9filimini art\xc4\xb1rd\xc4\xb1\xc4\x9f\xc4\xb1n\xc4\xb1; ancak, yine de \xc3\xa7o\xc4\x9funlu\xc4\x9fun h\xc3\xa2l\xc3\xa2 ba\xc5\x9far\xc4\xb1 \xc5\x9fans\xc4\xb1n\xc4\xb1 hatal\xc4\xb1 olarak yar\xc4\xb1 yar\xc4\xb1ya g\xc3\xb6rd\xc3\xbc\xc4\x9f\xc3\xbcn\xc3\xbc ortaya koydu.\n_START_SECTION_\nPratik \xc3\xa7\xc3\xb6z\xc3\xbcm\n_START_PARAGRAPH_\nYap\xc4\xb1lan ara\xc5\x9ft\xc4\xb1rmalara g\xc3\xb6re, ke\xc3\xa7i se\xc3\xa7me ihtimaline asl\xc4\xb1nda arabay\xc4\xb1 kazanma ihtimaline e\xc5\x9fittir. \xc4\xb0lk se\xc3\xa7ilen kap\xc4\xb1 ke\xc3\xa7i olmas\xc4\xb1 kaidesiyle her hal\xc3\xbckarda araba kazan\xc4\xb1l\xc4\xb1r. Bu da kazanma ihtimalinin 2/3 oldu\xc4\x9funu ispatlar.\n_START_SECTION_\nSim\xc3\xbclasyon yoluyla \xc3\xa7\xc3\xb6z\xc3\xbcm\n_START_PARAGRAPH_\nSe\xc3\xa7imi de\xc4\x9fi\xc5\x9ftirme stratejisinin ger\xc3\xa7ekten ortalama \xc3\xbc\xc3\xa7te iki olas\xc4\xb1l\xc4\xb1kla kazand\xc4\xb1raca\xc4\x9f\xc4\xb1n\xc4\xb1 ispatlaman\xc4\xb1n basit bir yolu da, yar\xc4\xb1\xc5\x9fmay\xc4\xb1 oyun kartlar\xc4\xb1na uyarlamakt\xc4\xb1r (Gardner 1959b; vos Savant 1996:8). Alel\xc3\xa2de bir desteden \xc3\xbc\xc3\xa7 kart, \xc3\xbc\xc3\xa7 kap\xc4\xb1y\xc4\xb1 temsil eder. Ma\xc3\xa7a as\xc4\xb1 gibi bir \'\xc3\xb6zel\' kart arabay\xc4\xb1 saklayan kap\xc4\xb1y\xc4\xb1, iki k\xc4\xb1rm\xc4\xb1z\xc4\xb1 ikili gibi iki s\xc4\xb1radan kart da, ke\xc3\xa7ileri saklayan kap\xc4\xb1lar\xc4\xb1 temsil eder._NEWLINE_\xc5\x9eu y\xc3\xb6ntem defalarca tekrarlanarak oyunun bir\xc3\xa7ok devresi uyarlanabilir: Yar\xc4\xb1\xc5\x9fmac\xc4\xb1n\xc4\xb1n ilk se\xc3\xa7imini temsil eden kart yar\xc4\xb1\xc5\x9fmac\xc4\xb1n\xc4\xb1n \xc3\xb6n\xc3\xbcnde ters \xc5\x9fekilde durur. Daha sonra en az biri k\xc4\xb1rm\xc4\xb1z\xc4\xb1 ikili olan iki karta bakan sunucu, k\xc4\xb1rm\xc4\xb1z\xc4\xb1 ikili olan bir kart\xc4\xb1 a\xc3\xa7ar. E\xc4\x9fer sunucunun elinde kalan kart Ma\xc3\xa7a as\xc4\xb1 ise, bu, yar\xc4\xb1\xc5\x9fmac\xc4\xb1n\xc4\xb1n se\xc3\xa7imini de\xc4\x9fi\xc5\x9ftirerek kazanaca\xc4\x9f\xc4\xb1 bir devredir. E\xc4\x9fer sunucunun elinde kalan kart k\xc4\xb1rm\xc4\xb1z\xc4\xb1 ikili ise, bu, yar\xc4\xb1\xc5\x9fmac\xc4\xb1n\xc4\xb1n ilk se\xc3\xa7iminde \xc4\xb1srar ederek kazanaca\xc4\x9f\xc4\xb1 bir devredir._NEWLINE_B\xc3\xbcy\xc3\xbck say\xc4\xb1lar yasas\xc4\xb1na g\xc3\xb6re bu deney, kazanma olas\xc4\xb1l\xc4\xb1\xc4\x9f\xc4\xb1n\xc4\xb1 yakla\xc5\x9f\xc4\xb1k olarak ortaya koyar ve deneyi yeterli defa tekrarlamak yaln\xc4\xb1zca yar\xc4\xb1\xc5\x9fmac\xc4\xb1n\xc4\xb1n se\xc3\xa7imini de\xc4\x9fi\xc5\x9ftirerek \xc3\xbc\xc3\xa7te iki olas\xc4\xb1l\xc4\xb1kla kazanaca\xc4\x9f\xc4\xb1n\xc4\xb1 de\xc4\x9fil, bunun nedenini de g\xc3\xb6sterir. Yar\xc4\xb1\xc5\x9fmac\xc4\xb1ya bir kart da\xc4\x9f\xc4\xb1t\xc4\xb1ld\xc4\xb1ktan sonra, de\xc4\x9fi\xc5\x9ftirmenin kazand\xc4\xb1r\xc4\xb1p kazand\xc4\xb1rmayaca\xc4\x9f\xc4\xb1 zaten bellidir ve Ma\xc3\xa7a as\xc4\xb1, \xc3\xbc\xc3\xa7te iki olas\xc4\xb1l\xc4\xb1kla sunucunun elindedir._NEWLINE_Bu ikna edici de\xc4\x9filse, benzetme, kartlardan biri yar\xc4\xb1\xc5\x9fmac\xc4\xb1ya verilmek ve kalan 51\'i sunucuda kalmak \xc3\xbczere t\xc3\xbcm bir desteyle de yap\xc4\xb1labilir (Gardner 1959b; Adams 1990). Bu durumda Ma\xc3\xa7a as\xc4\xb1 52\'de 51 olas\xc4\xb1l\xc4\xb1kla sunucudad\xc4\xb1r ve ka\xc3\xa7 Ma\xc3\xa7a as\xc4\xb1 olmayan kart a\xc3\xa7\xc4\xb1l\xc4\xb1rsa a\xc3\xa7\xc4\xb1ls\xc4\xb1n, onda kalmaya devam eder._NEWLINE_vos Savant\'\xc4\xb1n \xc3\xb6ne s\xc3\xbcrd\xc3\xbc\xc4\x9f\xc3\xbc bir ba\xc5\x9fka benzetmede, sunucu kap\xc4\xb1lar\xc4\xb1 temsil eden \xc3\xbc\xc3\xa7 kupadan birinin alt\xc4\xb1nda arabay\xc4\xb1 temsil eden bir madeni paray\xc4\xb1 ya da deniz kabuklar\xc4\xb1n\xc4\xb1n alt\xc4\xb1nda bir bezelyeyi saklar.\n_START_SECTION_\nN say\xc4\xb1da kap\xc4\xb1\n_START_PARAGRAPH_\nD. L. Ferguson (Selvin 1975b\'deki al\xc4\xb1nt\xc4\xb1ya g\xc3\xb6re 1975\'te Selvin\'e yazd\xc4\xb1\xc4\x9f\xc4\xb1 bir mektubunda) orijinal oyunun bir N say\xc4\xb1da kap\xc4\xb1 \xc5\x9feklinde genelle\xc5\x9ftirilmi\xc5\x9fini \xc3\xb6nermektedir. Bu durumda sunucu p say\xc4\xb1da kaybeden kap\xc4\xb1y\xc4\xb1 a\xc3\xa7ar ve oyuncuya de\xc4\x9fi\xc5\x9ftirme f\xc4\xb1rsat\xc4\xb1n\xc4\xb1 verir; bu \xc5\x9fekilde de\xc4\x9fi\xc5\x9ftirmeyle kazanma olas\xc4\xb1l\xc4\xb1\xc4\x9f\xc4\xb1 _NEWLINE_(N\xe2\x88\x921)/[N(N\xe2\x88\x92p\xe2\x88\x921)] olur. Sunucu sadece bir kap\xc4\xb1y\xc4\xb1 a\xc3\xa7m\xc4\xb1\xc5\x9fsa bile oyuncunun de\xc4\x9fi\xc5\x9ftirmesi yine de yarar\xc4\xb1na olur ancak N say\xc4\xb1s\xc4\xb1 b\xc3\xbcy\xc3\xbcd\xc3\xbck\xc3\xa7e avantaj da s\xc4\xb1f\xc4\xb1ra yakla\xc5\x9f\xc4\xb1r (Granberg 1996:188). Di\xc4\x9fer u\xc3\xa7 durumda sunucu bir kaybeden kap\xc4\xb1 d\xc4\xb1\xc5\x9f\xc4\xb1nda di\xc4\x9ferlerinin tamam\xc4\xb1n\xc4\xb1 a\xc3\xa7arsa, de\xc4\x9fi\xc5\x9ftirmeyle kazanma \xc5\x9fans\xc4\xb1 1\'e yakla\xc5\x9f\xc4\xb1r._NEWLINE_Bapeswara Rao ve Rao (1992) farkl\xc4\xb1 bir N say\xc4\xb1da kap\xc4\xb1 versiyonu \xc3\xb6nermektedir. Bu versiyonda sunucu oyuncunun se\xc3\xa7iminden fakl\xc4\xb1 bir kaybeden kap\xc4\xb1 a\xc3\xa7maktad\xc4\xb1r. Sunucu, geriye yaln\xc4\xb1zca iki kap\xc4\xb1 kalana kadar her kap\xc4\xb1 a\xc3\xa7\xc4\xb1\xc5\x9f\xc4\xb1nda oyuncuya tercihini de\xc4\x9fi\xc5\x9ftirmesi i\xc3\xa7in bir f\xc4\xb1rsat verir. D\xc3\xb6rt kap\xc4\xb1 ile yap\xc4\xb1labilecek en uygun strateji \xc3\xb6nce kap\xc4\xb1y\xc4\xb1 se\xc3\xa7mek ard\xc4\xb1ndan sadece iki kap\xc4\xb1 kal\xc4\xb1nca tercihi de\xc4\x9fi\xc5\x9ftirmek olacakt\xc4\xb1r. N say\xc4\xb1da kap\xc4\xb1 ile bu stratejinin kazanma olas\xc4\xb1l\xc4\xb1\xc4\x9f\xc4\xb1 (N\xe2\x88\x921)/N\'dir ve bunun en uygunu oldu\xc4\x9fu iddia edilmektedir.\n_START_SECTION_\nQuantum version\n_START_PARAGRAPH_\nA 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 2002, D\'Ariano et al. 2002).\n_START_SECTION_\nProblemin tarihi\n_START_PARAGRAPH_\nMonty Hall problemine benzer birka\xc3\xa7 olas\xc4\xb1l\xc4\xb1k bulmacas\xc4\xb1ndan en eskisi Bertrand\'\xc4\xb1n kutu paradoksudur. Bu 1889\'da Joseph Bertrand taraf\xc4\xb1ndan Calcul des probabilit\xc3\xa9s adl\xc4\xb1 kitab\xc4\xb1nda ortaya at\xc4\xb1ld\xc4\xb1 (Barbeau 1993). Bu bulmacada \xc3\xbc\xc3\xa7 kutu bulunur: bir kutu iki alt\xc4\xb1n madeni para, bir kutu iki g\xc3\xbcm\xc3\xbc\xc5\x9f madeni para ve bir kutu da her birinden birer tane madeni para i\xc3\xa7erir. Bir kutunun rastgele se\xc3\xa7ilmesinin ve bir madeni paran\xc4\xb1n rastgele \xc3\xa7ekilmesinin ard\xc4\xb1ndan, \xc3\xa7ekilen paran\xc4\xb1n alt\xc4\xb1n olmas\xc4\xb1 durumunda geriye kalan paran\xc4\xb1n alt\xc4\xb1n olma olas\xc4\xb1l\xc4\xb1\xc4\x9f\xc4\xb1 nedir. Monty Hall probleminde oldu\xc4\x9fu gibi burada da sezgisel yan\xc4\xb1t 1/2\'dir, halbuki olas\xc4\xb1l\xc4\xb1k 2/3\'t\xc3\xbcr._NEWLINE_Martin Gardner\'\xc4\xb1n Scientific American\'daki Mathematical Games k\xc3\xb6\xc5\x9fesinde 1959\'da yay\xc4\xb1nlanan \xc3\xbc\xc3\xa7 mahk\xc3\xbbm problemi (1959a, 1959b), Monty Hall probleminin izomorfudur. Problem, rastgele i\xc3\xa7lerinden biri gizlice se\xc3\xa7ilip affedilecek olan \xc3\xbc\xc3\xa7 mahk\xc3\xbbmu i\xc3\xa7erir. Mahk\xc3\xbbmlardan biri, di\xc4\x9fer mahk\xc3\xbbmlardan idam edilecek olan\xc4\xb1n ad\xc4\xb1n\xc4\xb1 s\xc3\xb6ylemesi i\xc3\xa7in gardiyana yalvar\xc4\xb1r. Bu mahk\xc3\xbbmun kendi kaderi hakk\xc4\xb1nda hi\xc3\xa7bir bilgiyi a\xc3\xa7\xc4\xb1\xc4\x9fa \xc3\xa7\xc4\xb1karmasa da kendi affedilme \xc5\x9fans\xc4\xb1n\xc4\xb1 1/3\'ten 1/2\'ye \xc3\xa7\xc4\xb1kar\xc4\xb1r. 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. This problem is equivalent to the Monty Hall problem; the prisoner asking the question still has a 1/3 chance of being pardoned but his unnamed cohort has a 2/3 chance._NEWLINE_Steve Selvin 1975\'te American Statistician dergisine g\xc3\xb6nderdi\xc4\x9fi bir \xc3\xa7ift mektup ile Monty Hall problemini ortaya att\xc4\xb1 (1975a, 1975b). \xc4\xb0lk mektup, 15 y\xc4\xb1l sonra Parade dergisinde yay\xc4\xb1nlanan g\xc3\xb6sterime yak\xc4\xb1n bir problem i\xc3\xa7eriyordu. \xc4\xb0kici mektup "Monty Hall problemi" teriminin ilk kullan\xc4\xb1m\xc4\xb1 gibi g\xc3\xb6r\xc3\xbcnmektedir. Problem asl\xc4\xb1nda yar\xc4\xb1\xc5\x9fma program\xc4\xb1n\xc4\xb1n bir ekstrapolasyonudur. Monty Hall heyecan yaratmak i\xc3\xa7in yanl\xc4\xb1\xc5\x9f bir kap\xc4\xb1y\xc4\xb1 a\xc3\xa7ar ve yar\xc4\xb1\xc5\x9fmac\xc4\xb1ya, kap\xc4\xb1 de\xc4\x9fi\xc5\x9ftirme f\xc4\xb1rsat\xc4\xb1 yerine 100 $ gibi daha k\xc3\xbc\xc3\xa7\xc3\xbck bir \xc3\xb6d\xc3\xbcl teklif eder. Monty Hall\'\xc3\xbcn Selvin\'e yazd\xc4\xb1\xc4\x9f\xc4\xb1 bir mektuba g\xc3\xb6re:_NEWLINE__NEWLINE_And if you ever get on my show, the rules hold fast for you\xe2\x80\x94no trading boxes after the selection. (Hall 1975)_NEWLINE_A version of the problem very similar to the one that appeared three years later in Parade was published in 1987 in the Puzzles section of The Journal of Economic Perspectives (Nalebuff 1987)._NEWLINE_Phillip Martin\'s article in a 1989 issue of Bridge Today magazine titled "The Monty Hall Trap" (Martin 1989) 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._NEWLINE_A restated version of Selvin\'s problem appeared in Marilyn vos Savant\'s Ask Marilyn question-and-answer column of Parade in September 1990 (vos Savant 1990). Though vos Savant gave the correct answer that switching would win two-thirds of the time, she estimates the magazine received 10,000 letters including close to 1,000 signed by PhDs, many on letterheads of mathematics and science departments, declaring that her solution was wrong (Tierney 1991). Due to the overwhelming response, Parade published an unprecedented four columns on the problem (vos Savant 1996:xv).  As a result of the publicity the problem earned the alternative name Marilyn and the Goats._NEWLINE_In November 1990, an equally contentious discussion of vos Savant\'s article took place in Cecil Adams\'s column The Straight Dope (Adams 1990). 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\' analysis, Adams agreed that mathematically, he had been wrong, but said that the Parade version left critical constraints unstated, and without those constraints, the chances of winning by switching were not necessarily 2/3. Numerous readers, however, wrote in to claim that Adams had been "right the first time" and that the correct chances were one in two._NEWLINE_The Parade column and its response received considerable attention in the press, including a front page story in the New York Times (Tierney 1991) in which Monty Hall himself was interviewed. He appeared to understand 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._NEWLINE_Over 40 papers have been published about this problem in academic journals and the popular press (Mueser and Granberg 1999).  Barbeau 2000 contains a survey of the academic literature pertaining to the Monty Hall problem and other closely related problems._NEWLINE_The problem continues to resurface outside of academia. The syndicated NPR program Car Talk featured it as one of their weekly "Puzzlers," and the answer they featured was quite clearly explained as the correct one (Magliozzi and Magliozzi, 1998). An account of the Hungarian mathematician Paul Erd\xc5\x91s\'s first encounter of the problem can be found in The Man Who Loved Only Numbers\xe2\x80\x94like many others, he initially got it wrong. The problem is discussed, from the perspective of a boy with Asperger syndrome, in The Curious Incident of the Dog in the Night-time, a 2003 novel by Mark Haddon. The problem is also addressed in a lecture by the character Charlie Eppes in an episode of the CBS drama NUMB3RS (Episode 1.13) and in Derren Brown\'s 2006 book Tricks Of The Mind. Penn Jillette explained the Monty Hall Problem on the "Luck" episode of Bob Dylan\'s Theme Time Radio Hour radio series. The Monty Hall problem appears in the film 21 (Bloch 2008). Economist M. Keith Chen identified a potential flaw in hundreds of experiments related to cognitive dissonance that use an analysis with issues similar to those involved in the Monty Hall problem (Tierney 2008).'