汽车-门 كيف يمكنني تدريب كلبي المشي على المقود؟

Empty boxesAt first, Gál and Miltersen considered in their paper the case that the number of boxes is twice the number of team members while half of the boxes are empty. This is a more difficult problem since empty boxes lead nowhere and thus the cycle-following strategy cannot be applied. It is an open problem if in this case the winning probability tends to zero with growing number of team members. In 2005, Navin Goyal and Michael Saks developed a strategy for team B based on the cycle-following strategy for a more general problem in which the fraction of empty boxes as well as the fraction of boxes each team member is allowed to open are variable. The winning probability still tends to zero in this case, but slower than suggested by Gál and Miltersen. If the number of team members and the fraction of boxes which are opened is fixed, the winning probability stays strictly larger than zero when more empty boxes are added. David Avis and Anne Broadbent considered in 2009 a quantum theoretical variant in which team B wins with certainty. The malicious directorIn case the prison director does not have to distribute the numbers into the drawers randomly, he can foil the prisoners' strategy if he knows the numbering of the drawers. To this end, he just has to ensure that his assignment of prisoners' numbers to drawers constitutes a permutation with a cycle of length larger than 50. The prisoners in turn can counter this by choosing their own numbering of the drawers randomly. One prisoner may make one changeIn the case that one prisoner may enter the room first, inspect all boxes, and then switch the contents of two boxes, all prisoners will survive. This is so since any cycle of length larger than 50 can be broken, so that it can be guaranteed that all cycles are of length at most 50. Monty Hall problemIn 2009, Adam S. Landsberg proposed the following simpler variant of the 100 prisoners problem which is based on the well-known Monty Hall problem: Behind three closed doors a car, the car keys and a goat are randomly distributed. There are two players: the first player has to find the car, the second player the keys to the car. Only if both players are successful they may drive the car home. The first player enters the room and may consecutively open two of the three doors. If he is successful, the doors are closed again and the second player enters the room. The second player may also open two of the three doors, but he cannot communicate with the first player in any form. What is the winning probability if both players act optimally?If the players select their doors randomly, the winning probability is only 4/9 (about 44%). The optimal strategy is, however, as follows: Player 1 first opens door 1. If the car is behind the door, he is successful. If the keys were behind the door, he next opens door 2; if instead the goat was behind the door, he next opens door 3. Player 2 first opens door 2. If the keys are behind the door, he is successful. If the goat was behind the door, he next opens door 3; whereas if the car was behind the door, he next opens door 1.In the six possible distributions of car, keys and goat behind the three doors, the players open the following doors (in the green cases, the player was successful): Car − Keys − Goat Car − Goat − Keys Keys − Car − Goat Keys − Goat − Car Goat − Car − Keys Goat − Keys − Car Player 1 Door 1: Car Door 1: Car Door 1: Keys Door 2: Car Door 1: Keys Door 2: Goat Door 1: Goat Door 3: Keys Door 1: Goat Door 3: Car Player 2 Door 2: Keys Door 2: Goat Door 3: Keys Door 2: Car Door 1: Keys (Door 2: Goat) (Door 3: Car) (Door 2: Car) (Door 1: Goat) Door 2: Keys The success of the strategy is based on building a correlation between the successes and failures of the two players. Here, the winning probability is 2/3, which is optimal since the first player cannot have a higher winning probability than that. In a further variant, three prizes are hidden behind the three doors and three players have to independently find their assigned prizes with two tries. In this case the winning probability is also 2/3 when the optimal strategy is employed.