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There are 100 of bandits taken as prisoners one evening.
The prison manager gave them chance to change their destiny to be executed.
Every prisoner gets a unique number (1- 100), and in the manager office there are 100 numbered drawers. There are 100 number on papers (1-100) randomly inserted in each of the drawers.
The next morning, the manager will take prisoners one by one, and let them open 50 drawers, the prisoner has to find the drawer with his humber or all the bandits get executed.
Once he finds the number, he goes into another rooms and all drawers are closed, he can't say any information to others.

"Oh no, our chance to survive this are 1:2^100, this is practically zero" says the bandit statistic.
The combinatoric bandit tries to calm him down, actually there is chance for us to survive of more than 30%.
How can they do it?

Are prisoners allowed to shuffle the papers between the drawers they open?

Holy-Fire wrote:Are prisoners allowed to shuffle the papers between the drawers they open?

no

I don't know if this may be a spoiler, so I'll just spoiler it: I'll try to work it out when I get home tonight.

Gammro wrote:I don't know if this may be a spoiler, so I'll just spoiler it:

Does this have something to do with factorials?

I'll try to work it out when I get home tonight.

It won't help much but:

Haha, well I had to look up the term combinatorics to see what it actually means
I have no clue where to begin, and exams coming up. So I better focus on that first

Gammro wrote:I have no clue where to begin, and exams coming up. So I better focus on that first

I think so

Not understood.
Prisoner 1 takes slot 1 and in the slot, there is (what a luck) the number 2. So he opens slot 2 and there lays 3 ... and so on.
The chances that he finds his number is for the first slot 1:100, for the second 2:100, 3:100 and so on until 50:100.

And now I got lost, because when added the whole, I come to a chance of 12.75.

Nirahiel wrote:[…]

I figured something along these lines might work. Do you have any source as to how high the actual chance will be ()? I did some testing and only got to 29.9%. That might of course have to do with insufficient randomness…

wrtlprnft wrote:

Nirahiel wrote:[…]

I figured something along these lines might work. Do you have any source as to how high the actual chance will be (

or, equivalently, the chance of a random permutation not containing a 51+-cycle

)? I did some testing and only got to 29.9%. That might of course have to do with insufficient randomness…

That is very correct way

kovarex wrote:Math behind

I'm not sure i understand this (and I'll try again later since I'm tired anyways), but I think the best answer to this

i think its this : SPOILER !! edit :

Math3vv wrote:i think its this : SPOILER !!

isn't it just a 50/50 chance since they only have to find 1 paper and they can open 50 out of 100 drawers

edit :

i ment 1:2 chance

That is even mentioned in the OP. The statistics view of this is the first one has 1:2 chance, so if they all have to be correct, they have 1:2^100 (chance 1 / 2^100), the trick is, that actually there is better way.

kovarex wrote:

Math3vv wrote:i think its this : SPOILER !!

isn't it just a 50/50 chance since they only have to find 1 paper and they can open 50 out of 100 drawers

edit :

i ment 1:2 chance

That is even mentioned in the OP. The statistics view of this is the first one has 1:2 chance, so if they all have to be correct, they have 1:2^100 (chance 1 / 2^100), the trick is, that actually there is better way.

they all need to find ther own number in order to not get executed ?

edit :

Dudes, I gave the right answer..
Also Math3vv, no, each prisoner must put the papers back once he's done.

Nirahiel wrote:Dudes, I gave the right answer..
Also Math3vv, no, each prisoner must put the papers back once he's done.

im trying to find my own solution O_O