Logic Puzzles

Kit- · 05-6-2007, 07:26 PM

New thread because I have no idea where to find the old one, and I couldn't be bothered to search for it and bump it.

1. If you knew the answer beforehand, please don't spoil it for the rest of us.
2. Don't Google for answers, that defeats the whole purpose.
3. Don't bother with repost reposts like "Why did the man shoot himself after drinking repost soup?"

That being said, I'll kick it off with a really hard one. I don't think I've seen this anywhere else. It took me two hours to figure it out with a fairly big hint.

prty heug hint: Has something to do with a topic covered in group theory.

There are 100 people in prison with ID numbers from 1-100, and the warden makes them a proposal. There is a room with 100 boxes in a line, each with one of their ID numbers inside, and each person is allowed to open up to 50 boxes. If they can each find their own number within these 50 guesses, they're all set free. However, if even one person fails, they all die. They are only allowed to speak to each other beforehand; no information transfer takes place at any time after. So, for example, you can't tell the next person in line that his box is the 27th one by coming out after 27 minutes have passed, even though no actual speaking is involved. The room is reset to its original state after each person exits. What is the only strategy with at least a 30% success rate given an equal probability of any configuration?

sammo123 · 05-6-2007, 07:35 PM

Everyone takes the numbers out of the box and puts them in there pocket then wehn they all get out of the room they switch to get the right id's then theyy win

or

they all decide not to go in there for never having the challenge of finding there number so technically never lose for never tryi9ng and they stay in jail and dont die

Kit- · 05-6-2007, 07:44 PM

Quote:

Originally Posted by sammo123

Everyone takes the numbers out of the box and puts them in there pocket then wehn they all get out of the room they switch to get the right id's then theyy win

Quote:

Originally Posted by Kit-

The room is reset to its original state after each person exits.

Quote:

Originally Posted by sammo123

or

they all decide not to go in there for never having the challenge of finding there number so technically never lose for never tryi9ng and they stay in jail and dont die

Quote:

Originally Posted by Kit-

What is the only strategy with at least a 30% success rate given an equal probability of any configuration?

PS: The solution is really elegant and cool.

Risking · 05-6-2007, 08:15 PM

100 people give 3 people 33(one with 34) ID numbers; That way, they have a 100% chance of not failing.

Kit- · 05-6-2007, 08:18 PM

Quote:

Originally Posted by Risking

100 people give 3 people 33(one with 34) ID numbers; That way, they have a 100% chance of not failing.

ID numbers are nontransferable.

**Doug31** · 05-6-2007, 08:21 PM

I don't really understand, is only one person in the room at a time, or are they all there?

Kit- · 05-6-2007, 08:22 PM

One at a time. One person allowed at a time. When they come out, they're sequestered and can't do anything else except hope other people don't mess up.

**TheRapingDragon** · 05-6-2007, 08:25 PM

Quote:

Originally Posted by Doug31

I don't really understand, is only one person in the room at a time, or are they all there?

One at a time. Each person gets 50 attempts (with 100 boxes) to get their number. Each of the 100 prisoners has to do this in turn.

**Doug31** · 05-6-2007, 08:29 PM

Then unless they're something you're still not telling us, or there's something hidden extremely deep within the meaning of those words, then it can't be done.

Kit- · 05-6-2007, 08:32 PM

Oh no, it's very possible. There are no hidden meanings, twists, or tricks hidden in the problem. It means exactly what you think it means.

thecheat372 · 05-6-2007, 08:33 PM

the first guy checks box 1-50
the second checks 2-51
the third 3-52
and so on...

**Doug31** · 05-6-2007, 08:35 PM

If you do it like that, then each person has a 1/2 chance and it all comes out to a (1/2)^100 chance just like any other possible way, the way I see it.

Kit- · 05-6-2007, 08:37 PM

Quote:

Originally Posted by thecheat372

the first guy checks box 1-50
the second checks 2-51
the third 3-52
and so on...

1 has a 50% chance of being in 1-50
If it is, then:
2% chance that 1 was in 1, in which case there is a 50/99 chance that 2 is in 2-51
98% chance that 1 was in 2-49, in which case there is a 49/99 chance that 2 is in 2-51.
.02*50/99+.98*49/99=49.52% chance that 2 is in 2-51.
50*49.52% = 24.76% < 30% for just the first two inmates.

Quote:

Originally Posted by Doug31

If you do it like that, then each person has a 1/2 chance and it all comes out to a (1/2)^100 chance just like any other possible way, the way I see it.

As much sense as this seems to make, there is a method that can get over a 30% success rate. Read the hint.

**Doug31** · 05-6-2007, 08:40 PM

No, that's wrong from the way I understood it, since I thought it returned to its original state, which would have meant #2 still would have had to check a box even if someone else had already found theirs. Now I think I know how to do this...I'll type it up in a sec.

Edit: Ok, person 1 checks box 1, then person 2 checks box 2...all the way to person 100 checks box 100. Now, they should each have 49 left, and there should be 99 boxes left, on average. Now they repeat it again, and they should have 48 left except for one person with 49, and 98 boxes left, and just keep repeating that until they get to the end.

Kit- · 05-6-2007, 08:45 PM

Quote:

Originally Posted by Doug31

No, that's wrong from the way I understood it, since I thought it returned to its original state, which would have meant #2 still would have had to check a box even if someone else had already found theirs. Now I think I know how to do this...I'll type it up in a sec.

Yes, but person 2 knows that he hasn't died yet, hopefully, meaning that he knows that 1 is somewhere in the range 1-50, which in turn means that it has a 2% chance of being in 1, and a 98% chance of being in 2-50.

Quote:

Originally Posted by Doug31

Edit: Ok, person 1 checks box 1, then person 2 checks box 2...all the way to person 100 checks box 100. Now, they should each have 49 left, and there should be 99 boxes left, on average. Now they repeat it again, and they should have 48 left except for one person with 49, and 98 boxes left, and just keep repeating that until they get to the end.

They only get to go in once. Then they go off to the side and hope their fellow inmates don't screw them over. Also, it wouldn't help. The boxes aren't removed when they're gone, and they transfer information to each other once they start checking boxes.

thecheat372 · 05-6-2007, 08:50 PM

what if they all bum rushed the warden, 100:1 sounds like more than a 30% chance of sucessfulness rate to me

**Doug31** · 05-6-2007, 08:51 PM

Ok then, person 1 goes in opens boxes 1-50 and takes the number of the of IDs in each of them and makes the equation of 2^(the number in the first one)*3^(the number in the second one) and so on, and spends a total of that many seconds in the room to leave, then the other people do that math to find out which numbers are in 1-50, and then they know if theirs is in the 1-50 or not, and then the rest all get theirs. 50% success rate.

thecheat372 · 05-6-2007, 08:53 PM

or! 1 looks at box number 1-50 and shows all the other people the number in each one, giving him 50% to find his,
#2 would then know if his was in the the first fifty or not, that being the case, since hes still alive he would then be able to find his at 100%, and so on with on the other prisoners

Kit- · 05-6-2007, 08:55 PM

Quote:

Originally Posted by thecheat372

what if they all bum rushed the warden, 100:1 sounds like more than a 30% chance of sucessfulness rate to me

Pretend the warden has a dead-man's switch which gases the room when he dies.

Quote:

Originally Posted by Doug31

Ok then, person 1 goes in opens boxes 1-50 and takes the number of the of IDs in each of them and makes the equation of 2^(the number in the first one)*3^(the number in the second one) and so on, and spends a total of that many seconds in the room to leave, then the other people do that math to find out which numbers are in 1-50, and then they know if theirs is in the 1-50 or not, and then the rest all get theirs. 50% success rate.

Quote:

Originally Posted by Kit-

They are only allowed to speak to each other beforehand; no information transfer takes place at any time after. So, for example, you can't tell the next person in line that his box is the 27th one by coming out after 27 minutes have passed, even though no actual speaking is involved.

Also, it'd work better if it was primes, as 2^2 = 4^1.

Quote:

Originally Posted by thecheat372

or! 1 looks at box number 1-50 and shows all the other people the number in each one, giving him 50% to find his,
#2 would then know if his was in the the first fifty or not, that being the case, since hes still alive he would then be able to find his at 100%, and so on with on the other prisoners

I already mentioned no communication several times now.

**Doug31** · 05-6-2007, 08:56 PM

But you can't transfer information between eachother. That's why I came up with a way that they'd all figure out the information without really transferring information.

Edit: But if those were all primes, it would come out to a number of seconds bigger than the number of seconds a person lives.

05-6-2007, 07:26 PM	#1
Kit- Private College Join Date: Feb 2006 Location: Lol badger Posts: 536	Logic Puzzles New thread because I have no idea where to find the old one, and I couldn't be bothered to search for it and bump it. 1. If you knew the answer beforehand, please don't spoil it for the rest of us. 2. Don't Google for answers, that defeats the whole purpose. 3. Don't bother with repost reposts like "Why did the man shoot himself after drinking repost soup?" That being said, I'll kick it off with a really hard one. I don't think I've seen this anywhere else. It took me two hours to figure it out with a fairly big hint. prty heug hint: Has something to do with a topic covered in group theory. There are 100 people in prison with ID numbers from 1-100, and the warden makes them a proposal. There is a room with 100 boxes in a line, each with one of their ID numbers inside, and each person is allowed to open up to 50 boxes. If they can each find their own number within these 50 guesses, they're all set free. However, if even one person fails, they all die. They are only allowed to speak to each other beforehand; no information transfer takes place at any time after. So, for example, you can't tell the next person in line that his box is the 27th one by coming out after 27 minutes have passed, even though no actual speaking is involved. The room is reset to its original state after each person exits. What is the only strategy with at least a 30% success rate given an equal probability of any configuration? __________________ <img src="Bent Lines" />

05-6-2007, 08:15 PM	#4
Risking FFR Player Join Date: Sep 2006 Location: At my cardboard computer desk. Age: 33 Posts: 681	Re: Logic Puzzles 100 people give 3 people 33(one with 34) ID numbers; That way, they have a 100% chance of not failing. __________________

05-6-2007, 08:21 PM	#6
Doug31 Falcon Paaaauuuunch!!!!!! Join Date: Jun 2004 Location: Washington Age: 36 Posts: 6,811	Re: Logic Puzzles I don't really understand, is only one person in the room at a time, or are they all there? __________________

05-6-2007, 08:22 PM	#7
Kit- Private College Join Date: Feb 2006 Location: Lol badger Posts: 536	Re: Logic Puzzles One at a time. One person allowed at a time. When they come out, they're sequestered and can't do anything else except hope other people don't mess up. __________________ <img src="Bent Lines" />

05-6-2007, 08:29 PM	#9
Doug31 Falcon Paaaauuuunch!!!!!! Join Date: Jun 2004 Location: Washington Age: 36 Posts: 6,811	Re: Logic Puzzles Then unless they're something you're still not telling us, or there's something hidden extremely deep within the meaning of those words, then it can't be done. __________________

05-6-2007, 07:35 PM	#2
sammo123 FFR Player Join Date: Jan 2007 Age: 31 Posts: 1,339	Re: Logic Puzzles Everyone takes the numbers out of the box and puts them in there pocket then wehn they all get out of the room they switch to get the right id's then theyy win or they all decide not to go in there for never having the challenge of finding there number so technically never lose for never tryi9ng and they stay in jail and dont die

05-6-2007, 08:32 PM	#10
Kit- Private College Join Date: Feb 2006 Location: Lol badger Posts: 536	Re: Logic Puzzles Oh no, it's very possible. There are no hidden meanings, twists, or tricks hidden in the problem. It means exactly what you think it means. __________________ <img src="Bent Lines" />

05-6-2007, 08:33 PM	#11
thecheat372 FFR Player Join Date: Feb 2005 Posts: 25	Re: Logic Puzzles the first guy checks box 1-50 the second checks 2-51 the third 3-52 and so on...

05-6-2007, 08:35 PM	#12
Doug31 Falcon Paaaauuuunch!!!!!! Join Date: Jun 2004 Location: Washington Age: 36 Posts: 6,811	Re: Logic Puzzles If you do it like that, then each person has a 1/2 chance and it all comes out to a (1/2)^100 chance just like any other possible way, the way I see it. __________________

05-6-2007, 08:40 PM	#14
Doug31 Falcon Paaaauuuunch!!!!!! Join Date: Jun 2004 Location: Washington Age: 36 Posts: 6,811	Re: Logic Puzzles No, that's wrong from the way I understood it, since I thought it returned to its original state, which would have meant #2 still would have had to check a box even if someone else had already found theirs. Now I think I know how to do this...I'll type it up in a sec. Edit: Ok, person 1 checks box 1, then person 2 checks box 2...all the way to person 100 checks box 100. Now, they should each have 49 left, and there should be 99 boxes left, on average. Now they repeat it again, and they should have 48 left except for one person with 49, and 98 boxes left, and just keep repeating that until they get to the end. __________________

05-6-2007, 08:50 PM	#16
thecheat372 FFR Player Join Date: Feb 2005 Posts: 25	Re: Logic Puzzles what if they all bum rushed the warden, 100:1 sounds like more than a 30% chance of sucessfulness rate to me

05-6-2007, 08:51 PM	#17
Doug31 Falcon Paaaauuuunch!!!!!! Join Date: Jun 2004 Location: Washington Age: 36 Posts: 6,811	Re: Logic Puzzles Ok then, person 1 goes in opens boxes 1-50 and takes the number of the of IDs in each of them and makes the equation of 2^(the number in the first one)*3^(the number in the second one) and so on, and spends a total of that many seconds in the room to leave, then the other people do that math to find out which numbers are in 1-50, and then they know if theirs is in the 1-50 or not, and then the rest all get theirs. 50% success rate. __________________

05-6-2007, 08:53 PM	#18
thecheat372 FFR Player Join Date: Feb 2005 Posts: 25	Re: Logic Puzzles or! 1 looks at box number 1-50 and shows all the other people the number in each one, giving him 50% to find his, #2 would then know if his was in the the first fifty or not, that being the case, since hes still alive he would then be able to find his at 100%, and so on with on the other prisoners

05-6-2007, 08:56 PM	#20
Doug31 Falcon Paaaauuuunch!!!!!! Join Date: Jun 2004 Location: Washington Age: 36 Posts: 6,811	Re: Logic Puzzles But you can't transfer information between eachother. That's why I came up with a way that they'd all figure out the information without really transferring information. Edit: But if those were all primes, it would come out to a number of seconds bigger than the number of seconds a person lives. __________________

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