Imagine there are 100 prisoners in a prison yard. The warden comes out, holding a lightbulb. He says to the prisoners, "you are all about to be placed into solitary confinement indefinetely. But I will let you all go if you can solve this riddle.
Here in my hand I have a lightbulb that you can turn off and on. You can turn it off and on as many times as you want, but you can do nothing else to the lightbulb. You may not break it, you make not paint it with anything. Each prisoner has the possibility of seeing the lightbulb once a day, but only once a day. Now I will let you all go free if at any time one of you can say that all prisoners have seen the lightbulb. However, if you do make the claim that all have seen it and you are wrong then I will kill all of you. Therefore, you only get one chance to answer."
Then all the prisoners are placed in solitary confinement, how will a prisoner be able to find out if everyone has seen the lightbulb?
2006-12-08
06:41:11
·
12 answers
·
asked by
jazzman1127
2
in
Entertainment & Music
➔ Jokes & Riddles
Hint: Because of solitary confinement the prisoners will not be able to talk to each other. No notes will be able to be passed either. A prisoner cannot hear another prisoner from his cell. There is no communication between the prisoners.
Secondly, all prisonersview death as worse than solitary confinement so I don't want any answers looking like "death is better than prison anyhow."
Good luck.
2006-12-08
06:43:21 ·
update #1
Clarification: I feel I wasn't clear yet again, by "see the lightbulb" I meant that each prisoner had seen it inside their cells. Or in better words, the lightbulb had been inside each prisoners cell. This is not a simple a semantic riddle. And remember all prisoners are insolitary confinement, meaning they cannot see anythign outside their cells. Sorry for the confusion.
2006-12-08
06:49:59 ·
update #2
Revision: There are 100 prisoners in different cells. The prisoners cannot hear eachother, see each other or have any form of communication between one another. Each day a lightbulb that can be turned on and off will come to a prisoner's cell.
The prisoners will all be set free if any prisoner can say that every other prisoner has had the lightbulb in his cell. However, if he is wrong all the prisoners will be killed. How can a prisoner find out whether or not all the other prisoners have had the lightbulb in their cells?
Important notes: The lightbulb has no set order to which cell it is brought to. It is possible that the same prisoner can see the lightbulb ten days in a row. But the law of probabilities tells us that eventually every prisoner will see it. (We are assuming that the prisoners won't die of natural causes before the answer is reached).
2006-12-08
08:28:26 ·
update #3
this is more of a math problem than a riddle in all fair ness.
I wrote a program to perform the binary dynamic method with 7 stages on 100 prisoners. The trick with finding the optimal time is that you're optimizing over all stage lengths simultaneously (including those in later rounds) and there will be local extrema that are not the global extrema. My numbers are certainly not perfectly optimal, but they were derived from the properties of the distribution and should be at least close. The couple lines of code (for determining stage boundaries) I posted above yield close to 4500 days as the expected incarceration time under the dynamic binary method with 100 prisoners. The chance of success in each round with these numbers is in the ballpark of 2/3-3/4. In batches of 2000 trials generally the worst case takes about 5 rounds to complete.
I think its slightly faster to get to 128 by just starting 1 players counter at 29 and all the others at 1, though the difference is very small.
Here is the c++ code for determing the expected mean/variation of each stage's completion time.
// Set up data used for determining stage sizes
// slightly inefficient computation method
meanEstimate = new double[stages];
varianceEstimate = new double[stages];
k=numPrisoners; // k is anticipated number of message passers in current stage
for(i=0 ; i
varianceEstimate[i] = 0.0;
meanEstimate[i] = 0.0;
for(j=1;j<=k;j++){
meanEstimate[i] += // computing H(k) here
1.0 / (double) j;
varianceEstimate[i] += // summing variances of geometric variables
(double) (numPrisoners - j) * (double) numPrisoners / (double) (j * j);
}
meanEstimate[i] *= numPrisoners; // mean = Harmonic(k) * numPrisoners
k = (int) ceil((double) k / 4.0) * 2;
so about 4500 days.
2006-12-08 08:56:29
·
answer #1
·
answered by Anonymous
·
0⤊
0⤋
I get the feeling you are spelling the answer out for us. I also think that it has to do with a lot of complicated math, which means after about 2 minutes I gonna be asleep. But on the other hand I have read this about a million times and have come up with a series of crazy answers. The one I am gonna tell you really requires a genius to be in each cell: Every person who glimpses the lightbulb can turn it on for 3 minutes, then turn it off. Each person can determine by the amount of heat the lightbulb has, how many people have turned it on and off. Not even sure this is possible but I was just throwin' it out there. Another possibilty is that the warden is just toying with the prisoners' minds and it is completely impossible.
Keep posting hints or we'll never get it!!!!!:0)
2006-12-08 07:06:46
·
answer #2
·
answered by ♥Petlover♥ 4
·
0⤊
1⤋
One person is chosen as the counter. If the light switch is on, no prisoner will touch it. If it is off, and a prisoner who has never flicked the switch enters, then he flicks the switch on. It stays like that until the counter returns. The counter never flicks it up, only flicks it down. Once he notices that it's been flicked up 99 times, he says that 100 prisoners have visited, and they go free. or he just counts 100 days and on the 100th day tells the warden everyone has seen it !!!
2006-12-08 08:42:05
·
answer #3
·
answered by LISA P 2
·
0⤊
0⤋
Faith is trusting and believing in the One who is Loving, Faithful, Wise, and True. If we truly trust His love toward us, we believe His will is always best even if the outcome of a request may not be what we hoped for. Example, I believed God could and would heal my son. Instead, he died. Could He have done it? No question. However, He had His reasons for not restoring his life, here. At the time, I had the faith to believe God loved us enough to raise my son back to life just as He did, Lazarus. What I did not know at the time, was that my trust was being tested. Would I turn my back on God as a result? Did God's character change because my son died? No and No. Since that 'trial' , God has revealed to me the 'why' about my son's physical death. His reasons healed my broken heart and gave me deeper revelation about His ways, which are higher than our own. Believing in God in spite of it all is a risk I'm willin' to take because He has honored my faith with miracles many times. Gambling is a selfish ambition, while believing in God is so much more than puttin' in your 'prayer requests'. I wonder, do you have a genuine desire to know the Truth?, or are you simply just being cynical?
2016-05-23 07:11:02
·
answer #4
·
answered by Anonymous
·
0⤊
0⤋
He held the lightbulb up in front of everyone.
2006-12-08 06:45:03
·
answer #5
·
answered by deadsolidcenter 2
·
3⤊
1⤋
They all just saw it when he brought it to the prison yard
or
the next day they would have all seen it the day before.
~HD
2006-12-08 06:46:38
·
answer #6
·
answered by mrsdycus 2
·
1⤊
1⤋
i have no idea , u lost me..ooo i know it, when he held the lightbulb the first time, everyone saw it, lol so the first one to say it gets free+
2006-12-08 06:54:41
·
answer #7
·
answered by LittleBit 3
·
0⤊
2⤋
didn't everyone c the lightbulb when he was telling them about it?
2006-12-08 06:46:15
·
answer #8
·
answered by ? 3
·
2⤊
1⤋
when the warden was holding up the light bulb..they all saw the light bulb then
2006-12-08 06:51:12
·
answer #9
·
answered by Michael D 5
·
1⤊
1⤋
You didn't tell this very well. I imagine your answer will be just as confusing as your question.
2006-12-08 06:58:42
·
answer #10
·
answered by Neerdowellian 6
·
1⤊
2⤋