The names of 100 prisoners are placed in 100 wooden boxes, one name to a box, and the boxes are lined up on a table in a room. One by one, the prisoners are led into the room; each may look in at most 50 boxes, but must leave the room exactly as he found it and is permitted no further communication with the others.
How are the 100 prisoners in a prison numbered?
The following version is by Philippe Flajolet and Robert Sedgewick: The director of a prison offers 100 death row prisoners, who are numbered from 1 to 100, a last chance. A room contains a cupboard with 100 drawers. The director randomly puts one prisoner’s number in each closed drawer. The prisoners enter the room, one after another.
How are the prisoners numbered in death row?
The director of a prison offers 100 death row prisoners, who are numbered from 1 to 100, a last chance. A room contains a cupboard with 100 drawers. The director randomly puts one prisoner’s number in each closed drawer. The prisoners enter the room, one after another. Each prisoner may open and look into 50 drawers in any order.
Who is the creator of the 100 prisoners problem?
Danish computer scientist Peter Bro Miltersen first proposed the problem in 2003. The 100 prisoners problem has different renditions in the literature. The following version is by Philippe Flajolet and Robert Sedgewick: The director of a prison offers 100 death row prisoners, who are numbered from 1 to 100, a last chance.
What happens if you don’t find your name in the first 50 boxes?
Before the exercise, all prisoners agree that they will each learn the name of the person going in after them. The first prisoner will search the first 50 boxes for his name and for the name of the second person. He has a 50% chance of finding his own name. If he doesn’t find it in the first 50 boxes, everyone dies.
Where can I buy a 100 breaker box?
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What is the probability of putting the last ball in an empty box?
Now, the last ball must be placed in an empty box or a non-empty box. The probability of placing it in an empty box is E_9/5 because, well, there are E_9 empty boxes out of the 5. Thus, E_9/5 of the time, you’ll have E_9 – 1 empty boxes; the rest of the time, you’ll have E_9 empty boxes, that is]
Is there a way to save 50% of the prisoners?
I argue that 50% of the prisoners can be saved: regardless of any ordering of the prisoners names in boxes, in any given string of 50 boxes (a string need not be consecutive) there are exactly 50 names which correspond to exactly 50 of the 100 prisoners.
