Logic Puzzles

[Kit-]

Originally Posted by Doug31

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.

You're letting them know a number. That's transferring information. Besides, I already made it quite clear that timing counts as information transfer.

Originally Posted by Doug31

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.

Your encoding function isn't injective, so there's no way to figure out the sequence of numbers from the single number for all cases.

[thecheat372]

Edit:
ahah, nvm
just reread it
i got it now =]


we're just missing something...

[Doug31]

Well then, I still don't see anything you can even do to increase your odds. Even with that knowing your not dead thing, you could still only know that it wasn't 1 of 50, and even if the second person just tried the second 50, it'd still already be a 1/2*50/99 chance, which is already less than 30%. Is there some way of affecting your chances that I still don't see?

[Kit-]

I'll give you guys a bigger hint. Using this method, if one person doesn't find his/her ID, if the ordeal continued, at least 50 other people wouldn't find their IDs either.

[thecheat372]

Ive got it!

Each box has 1 of the numbers of the ID in it

if the id number is 56
then theyd need to find numbers 5 and 6 to get their ID number

[probably wrong]

so then that means that of the 100 boxes, every 10 has numbers 0-9

so the first person [number one]
has to find just 1,
since 10 of the 100 have the number 1
hes got a 10% chance to find it...
nvm then
that doesnt work =[

[Kit-]

Nope, an inmate with ID 56 would have to find the one box with the ID 56 in it.

[Doug31]

Can you at least tell us at least one way to even get 2 people through and still be at at least 30%?

[Kit-]

That would kind of give it away, actually.

[Doug31]

So, there's really only one way to even get 2 people through still over 30%?

[Kit-]

I'm really not sure, but I'm positive that this method is the only way to get 100 people through.

[Doug31]

Well, there's still obviously something I'm missing. What can the first person do? They obviously have to try at least 30 boxes. Even if they do 30, that would still only be a 30% chance, which would mean everyone else would need a 100% chance, would obviously wouldn't happen, so they need to try more than 30. If they try 40, then that's a 40% chance, and even if the second person know it was in one of the first 40, they'd still barely have more than a 1/2 chance of finding theirs, which would mean it's already below 30%. If the first person tries 50, then they're still at 50%, but then the next person now knows the first person's was one of the first 50, but still is limited to a 50/99 chance by choosing the second 50, which is still a microscopic difference over 1/2, which means they still lose. Let's see, what else can they do. Person one could open different boxes than just 1-50, but it really makes no difference. Hmm, are they any other options. The first person could open 50, and the second person could guess that the first person probably found it about half way through, which would be about 25, and start from there, and keep repeating that, but they'd still already be under 30% after even the second person. I really don't get this. There must be something I don't understand.

[Kit-]

From your posts, you perfectly understand the problem. You're just approaching the problem in the wrong way.

[~HentaiXXX~]

One Prisoner does all the Even Numbered boxes which equals 50.
Then another would do all Odds. which equals 50 also.
Then they'd do that back and forth...i hope?

[madmatt621]

~HentaiXXX~ : Change your damn avatar.

[Doug31]

Yea, and this other approach is probably something that I completely don't understand you're allowed to do.

If there were 2 people and 2 boxes, and each person got to pick one box, then the first person tries a box, and has a 50-50 chance. Then the second person would know if they were still alive that it must be the second box, so that would be 50%.

Now, if there were 3 people, 3 boxes, and each got 2 choices, then the first person could try the first 1, so they'd be at 33.33~%. But, then the remaining two would know they both had to be in the last 2, which would be a 100% chance, so I can see how it could be done like that, as well.

Now, if there were 4 people, 4 boxes, and 2 choices each, I can't even see how to do that. What can the first person do? They have to pick 2 boxes or it's already less than 30%, so they pick 2 boxes...50%. Then the next person now knows the first person's was in one of the first two, so theirs is much more likely to be in the second 2. So the only way they can stay about 30% from that point is to pick the second 2. Even still, there was only a 2/3 chance that it would have been in one of the second 2, so they're down to 33.33~%. Now, the third person would only have a 50% chance, so it's over.

What am I missing here?

Edit: And ~HentaiXXX~, that would give the first person a 50/100 chance, and the second person a 50/99 chance, which is already only 25.2525~%, so it's already too low of a chance after only 2 people, just like all the ways I can come up with.

[thecheat372]

its a trick question?
i mean
theres an answer
but its not going to be like that...
itll be something that youd have to think up...
but its hidden in the text

[Doug31]

Ok, I got it, it is a trick question. The people want to die, so it's a success if they don't find their ID numbers, which they can make have a 100% chance of happening by all picking the first 50 boxes.

[thecheat372]

lol,
i like that one

[~HentaiXXX~]

good enough fer me doug =P

[Doug31]

And it even meets the 50 other people wouldn't get theirs if it continued clue, as well.