Here is a riddle, that I came across recently. Enjoy!
There are a hundred passengers to board a plane with a hundred seats. Each passenger has a ticket with number from 1 to 100 and enters the plane in this order. The first passenger sits on seat 1, and the second sits on seat 2 and so on. However, the first passenger, instead of using his seat, chooses a random seat of the available 100. The second passenger comes in and sits in its seat, if it is available, otherwise chooses randomly from the 99 free seats. Third passenger sits in seat 3, if it is free, otherwise chooses a random seat from the remaining 98, etc. What is the probability that 100th passenger sits in the 100th seat?
2 comments:
Merhaba Meltem,
Cok uzun zaman oldu gorusmeyeli, ama bana hala masa tenisinde gosterdigin backhand teknigi aklimda, kizlarima bile ogrettim! Umarim keyfin yerindedir...
Blog sayfana tesadufen rastladim. Bu riddle da cok hosuma gitti. Cozumu icin ufak bir dynamic programming formulu turettim. Buna gore cozum 100 yolculu bir ucak icin 0.640879566 gibi bir olasilik cikiyor; 1000 yolcuya kadar hesaplayinca 0.645383921 olasiligina ulasiyor. Enteresan bir asimptotu var sanki.
Kolay gelsin,
Onur Seref
Onur selamlar :) Nasilsin ? Sorunun cevabini 1/2 diye hatirliyorum. bir daha bakayim.
Post a Comment