Posers and Puzzles
07 Dec 06
11 Dec 06
9 edits
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.
Anyway, if the first person was even and found 40 evens and 10 odds, and the next person was even, then they would have a 10/39 chance of not finding their number by looking at just the evenly marked boxes....which means they have a 75% chance of finding their number. They would then have a 11/50 chance of finding their number if they didn't find it in the even marked boxes since there would still be 10 even boxes that weren't marked and 11 chances to still look into boxes and 50 boxes that hadn't been identified yet
The combined total odds would be 48.5% (75% + 22% divided by 2) for the 2nd prisoner to find his box using your example, not 25%.
EDIT: although, my solution depends on the prisoners being able to tell which boxes had already been correctly identified..hmm, maybe a 2nd bend would be required to show the box had been identified and therfore should not be looked in. Yes, that would work
12 Dec 06
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
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 chance that during this process he will uncover his number. As soon as he does so the remaining openings should be directed to the currently unknown boxes, labeling between 49 and 10 (equal probability of 49 to 11 inclusive [1/50 each] and the remaining probability being 10) of those.
And it just gets worse as Prisoner C won't have 100 boxes known either and in theory the 100th Prisoner may not have all boxes known (as unlikely as that scenario is).
The best way of working out the overall probability would be millions of trials.
12 Dec 06
1 edit
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.
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.
12 Dec 06
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.
12 Dec 06
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.
The equation is the tough part.
14 Dec 06
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?