10 Dec '06 15:10>
Originally posted by dmnelson84I said it involved switching the CONTENTS of the boxes
The google solution I found had nothing to do with switching boxes. You found a faulty answer.
Originally posted by XanthosNZYes, I realize you were saying they couldn't mark the boxes...but the answer had already been given on google.... I was just suggesting that from a practical standpoint it is easy to mark a cardboard box by bending one of the flaps. It would leave a crease and would be easy to spot.
Uzeless, your reasoning is wrong. The first prisoner would only be able to determine the odd/evenness of half the boxes (and would have a 50% chance of finding his own number). So unless he managed to find all odds or all evens (very low odds) the second prisoner would need to find odds/evens as well and would have a 50% chance of finding his own number ...[text shortened]... ou broke the rules of the puzzle and your solution still isn't as good as the optimal solution.
Originally posted by uzlessFirstly you are making the assumption that a prisoner always gets to open 50 boxes (even if their first box is their own number). Fine.
Yes, I realize you were saying they couldn't mark the boxes...but the answer had already been given on google.... I was just suggesting that from a practical standpoint it is easy to mark a cardboard box by bending one of the flaps. It would leave a crease and would be easy to spot.
Anyway, if the first person was even and found 40 evens and 10 odds, a ...[text shortened]... show the box had been identified and therfore should not be looked in. Yes, that would work
Originally posted by XanthosNZYes, I was assuming the first person looked in 50 boxes as per your example. For sure though, he may only look in 1 box, or any other number up to 50 before he finds his number.
Firstly you are making the assumption that a prisoner always gets to open 50 boxes (even if their first box is their own number). Fine.
So let's look at your example.
Prisoner A opens 50 boxes, finding his number in the process, and uncovers 40 evens and 10 odds.
Prisoner B enters the room. He is even so he opens the 40 known evens. There is a 4/5 nario is).
The best way of working out the overall probability would be millions of trials.
Originally posted by uzlessWhy must there be a formula? The complex nature of the problem as stated means that if one were to write out the probability then the sum would go on for a hell of a long time.
Yes, I was assuming the first person looked in 50 boxes as per your example. For sure though, he may only look in 1 box, or any other number up to 50 before he finds his number.
There must be a formula. The math guys here should be able to figure it out. PBE6 knows this stuff better than i do. Fabian seems good also.
Originally posted by XanthosNZThe answer would be simple. X% chance the prisoners would find their numbers using my method.
Why must there be a formula? The complex nature of the problem as stated means that if one were to write out the probability then the sum would go on for a hell of a long time.
Originally posted by XanthosNZHALF 1-50 OTHER HALF 51-100
100 prisoners are given the sort of crazy deal they always get in these problems. Each prisoner gets a number. There's a room with 100 identical boxes in a row, each containing a random number from 1-100, with no repeats. Each prisoner, in turn and isolated from all the others, will get a chance to open 50 boxes of his choosing, one at a time. If each pris ...[text shortened]... ly zero, actually it's ~8*10^-31). What is the optimal strategy that the prisoners should use?