Kogorous Riddle Thread and eworms Deduction Thread

[eworm]

Spoiler:

It's impossible because

THERE CAN BE ONLY ONE!!!

king of each color. Same with the queens.

[Akonyl]

silly eworm, chess isn't Highlander :V

Spoiler:

..QQ.
K....
....Q
...QQ
KK...

[Ankor]

silly eworm, chess isn't Highlander :V

Spoiler:

..QQ.
K....
....Q
...QQ
KK...

That's correct

[Akonyl]

alright then, let's hope this one hasn't been posted yet.

A gang of 10 bandits hijacks a carriage one day, and manages to steal 50 gold coins. The bandits look at the coins, and try to decide how to split it up. The splitting procedure goes as such:
- The most important bandit will propose how to split up the coins. He may give any amount to any bandit.
- The bandits then vote on the split, if 50% or more agree to it, it passes and everyone lives.
- If less than 50% agree to it, they gang up on the proposer and kill him. The process then proceeds again with the next most important bandit.

A few more things about the bandits:
- No two bandits are equally important.
- The bandits are greedy, and they want to maximize their own profit from the split.
- The bandits are perfectly rational and don't band together for revenge or anything. They simply try to maximize their profit.
- The bandits like to kill people, including each-other (but not themselves). In the case where a bandit is offered a sum of money which he knows he will be able to get even if he votes to kill the current most important bandit, he'll vote to kill him anyway because it's fun.
- Above all else, the bandits don't want to die. Because if you die, you don't get any money.

The most important bandit looks at his 50 coins and at his 9 subordinates. What split does he propose and why?

[eworm]

Woah, that looks possibly brain-frying. Hm, let's see...

Spoiler:

If less than 50% bandits agree, the less important ones shall profit, because of the more important ones dying. Therefore, six bandits will probably disagree, whatever the split will be. Then the leader would die, the coins would be split again. But even then, five of nine (the least important ones) will disagree again - to maximize their profit. Eight bandits remain. Five disagree. Seven remain. Four disagree. Six remain. Four disagree. Five remain. Three disagree. Four remain. Three disagree. Three remain. Two disagree. Two remain. They fight among themselves, because both want to have all the money.

Now, what can the leader do in order not to let that happen? He has to split the coins in a way that at max four of the bandits would disagree... But even if he split all the coins among five of them, giving the other five none... I think six bandits will protest anyway. So for me it seems the leader is dead anyhow...

Although... Well, yeah. He split the coins equally among five of the bandits (including himself). The least important bandit from the profiting five thinks about disagreeing, but then realizes, that he would die too eventually if he starts that chain of events. So he agrees to the splitting. That way 5 of 10 (50%) agrees but that's enough.

So yeah. Leader gets 10 coins and so do the four other most important bandits of the group. The rest gets nothing.

[Akonyl]

I'm gonna avoid commenting on your answer too much, but no, that's not the answer. Also:

He has to split the coins in a way that at max four of the bandits would disagree...

keep in mind, there are 10 bandits and for the vote to pass, 50% or more must agree. Thus, with 10 bandits, a maximum of 5 can disagree and have the vote still pass, not 4 like you're thinking.

[eworm]

Yeah, I realized that after posting and I edited the numbers but it seems I missed one.

But I'd be glad to hear what's incorrect about my answer... Or is it that it's correct but not the answer?

[VQ]

alright then, let's hope this one hasn't been posted yet.

A gang of 10 bandits hijacks a carriage one day, and manages to steal 50 gold coins. The bandits look at the coins, and try to decide how to split it up. The splitting procedure goes as such:
- The most important bandit will propose how to split up the coins. He may give any amount to any bandit.
- The bandits then vote on the split, if 50% or more agree to it, it passes and everyone lives.
- If less than 50% agree to it, they gang up on the proposer and kill him. The process then proceeds again with the next most important bandit.

A few more things about the bandits:
- No two bandits are equally important.
- The bandits are greedy, and they want to maximize their own profit from the split.
- The bandits are perfectly rational and don't band together for revenge or anything. They simply try to maximize their profit.
- The bandits like to kill people, including each-other (but not themselves). In the case where a bandit is offered a sum of money which he knows he will be able to get even if he votes to kill the current most important bandit, he'll vote to kill him anyway because it's fun.
- Above all else, the bandits don't want to die. Because if you die, you don't get any money.

The most important bandit looks at his 50 coins and at his 9 subordinates. What split does he propose and why?

Finally! a decent puzzle.

Spoiler:

First we have to order the bandits by importance.
10 ,9, 8, 7, 6, 5, 4, 3, 2, 1    10 being the most important and 1 the less important.

Reasoning:

If there were only 2 bandits, bandit 2 being the most important, bandit 2 would vote for himself (50%) and keep all the money for himself.

If there were just 3 bandits, bandit 3 would have to give bandit 1 only one coin, bandit 1 being perfectly rational knows that if he doesn't vote for 3 and 3 dies then he would get nothing.

If there were 4 bandits, bandit 4 would have to give only one coin to bandit 2 to get 50% of the votes because bandit 2 knows that if he doesn't vote for bandit 4, he gets nothing.

If there were 5 bandits, bandit 5 just have to give bandit 1 and 3 one coin each, because they now they'll get nothing if 5 dies.

You follow the same reasoning until you reach 10, then bandit number 10 could keep 46 coins and give one coin to bandits 8, 6, 4 and 2 each.

[eworm]

Spoiler:

If there were 4 bandits, bandit 4 would have to give only one coin to bandit 2 to get 50% of the votes because bandit 2 knows that if he doesn't vote for bandit 4, he gets nothing.

And why would that be? Seriously, I don't get it.
Bandit 4 gives bandit 2 one coin and the rest gets nothing. None of them are satisfied. None of them votes for bandit 4. Bandit 4 dies and there is 50 coins for three people, instead of four. Profit. Profit also for bandit 2 because he'll surely get some coins - new leader, number 3, doesn't want to die and so he gets some coins himself and gives some to number 2. Number 1 gets nothing of course. And disagrees with the split. And SO DOES NUMBER 2. Number 3 dies. Number 2 profits a lot.

Am I missing something?

[VQ]

Spoiler:

If there were 4 bandits, bandit 4 would have to give only one coin to bandit 2 to get 50% of the votes because bandit 2 knows that if he doesn't vote for bandit 4, he gets nothing.

And why would that be? Seriously, I don't get it.
Bandit 4 gives bandit 2 one coin and the rest gets nothing. None of them are satisfied. None of them votes for bandit 4. Bandit 4 dies and there is 50 coins for three people, instead of four. Profit. Profit also for bandit 2 because he'll surely get some coins - new leader, number 3, doesn't want to die and so he gets some coins himself and gives some to number 2. Number 1 gets nothing of course. And disagrees with the split. And SO DOES NUMBER 2. Number 3 dies. Number 2 profits a lot.

Am I missing something?

Bandit 4 only needs 50% of the votes to not die, so he votes for himself and gives 1 coin to bandit 2.
2 votes= 50%

[eworm]

Yeah, riight... And bandit 2 is so very satisfied with getting one of fifty coins...

[VQ]

Yeah, riight... And bandit 2 is so very satisfied with getting one of fifty coins...

Remember all the bandits always try to maximize their profit and if 50% or more agree with the splitting procedure, it passes and everyone lives.

Bandit 2 doesn't have a choice, he knows that if bandit 4 dies, bandit 3 would give bandit 1 just one coin and keep 49 coins for himself and bandit 2 would get nothing, bandit 1 would accept because he knows that if 3 dies, bandit 2 could keep all the coins voting for himself (50%).

[eworm]

Bandit 2 doesn't have a choice, he knows that if bandit 4 dies, bandit 3 would give bandit 1 just one coin and keep 49 coins for himself and bandit 2 would get nothing, bandit 1 would accept because he knows that if 3 dies, bandit 2 could keep all the coins voting for himself (50%).

And how about he gave bandit 2 some coins instead of bandit 1? After all bandit 2 is more important and that would make more sense. I just don't see why it would have to be bandit 1.

[kidSherlock]

Bandit 2 doesn't have a choice, he knows that if bandit 4 dies, bandit 3 would give bandit 1 just one coin and keep 49 coins for himself and bandit 2 would get nothing, bandit 1 would accept because he knows that if 3 dies, bandit 2 could keep all the coins voting for himself (50%).

And how about he gave bandit 2 some coins instead of bandit 1? After all bandit 2 is more important and that would make more sense. I just don't see why it would have to be bandit 1.

because if bandit 3 give coins only to bandit 2, bandit 2 will know that bandit 1 will decline. and because all of them will want to maximize their own profit bandit 2 will also decline meaning bandit 3 will be dead.

am i right VQ?

[eworm]

Hm, okay, that makes sense...
So bandit 3 gets 1 coin, bandit 2 gets nothing and the rest goes to the leader, bandit 3...
And bandit 1 agrees because if he doesn't, there will be only two bandits left, with him being the less important one...
.
.
.
You know, I can't help but think that if there were two bandits left - bandit 2 and bandit 1 - they'd abandon the rules and just fight over the coins.