[Slow Ben]

Take away my first letter and I am unchanged;
Take away my second letter and I am unchanged;
Take away all my remaining letters and I am still unchanged!

What am I?

[Sir Thursday]

A postman? I don't think he gets a different pair of clothes if you take all the letters out of his mailbag...

This post has been edited by Sir Thursday: 16 February 2009 - 11:54 PM

[Slow Ben]

Most postman. I'm not sure about postman across the pond.

[frookenhauer]

Its STD time:) . Better make it reeeaaaaallly hard, cos slow ben seems quick on the uptake

[Sir Thursday]

5 Robbers (Angus, Ben, Charles, Daniel and Edward) steal 50 gold pieces from a rich lord. They decide to play a game to divide up the loot. The game works as follows:

The robbers line up in alphabetical order. On their turn, the robber proposes a method of splitting up the loot. If they can propose a method that gains a majority vote, then the treasure will be divided based on that method. If their method does not get a majority approval, then they are forced to eat a poisonous apricot and die. The next robber in the line then proposes an alternative method. The robbers are not allowed to make deals amongst themselves, and each of them is rational and selfish - they will vote against any method that doesn't maximise their share.

In the case of an even number of voters, 50% of the vote is regarded as a majority. The robbers do not have the means to cut the gold pieces in half, so there is no way to subdivide the treasure into more than 50 pieces.

So the question is, what method of dividing the loot should Angus propose to stay alive and maximise his winnings?


Sir Thursday

This post has been edited by Sir Thursday: 17 February 2009 - 12:23 AM

[Giles]

Id say something like, 17 for charles, 17 for ben and 16 for angus but thats a very quick guess

[Sir Thursday]

baudin, on Feb 17 2009, 01:43 AM, said:

Id say something like, 17 for charles, 17 for ben and 16 for angus but thats a very quick guess

Ah, but why should Ben settle for 17? If he says no, then Angus dies and he can try and get more for himself.

[Giles]

also what are you defining as the majoirty vote for an even numbe rof people? half or half+1?

[Sir Thursday]

baudin, on Feb 17 2009, 02:08 AM, said:

also what are you defining as the majoirty vote for an even numbe rof people? half or half+1?

Read the riddle again, fool! It's clearly stated there.

[Obdigore]

Sir Thursday, on Feb 16 2009, 06:21 PM, said:

5 Robbers (Angus, Ben, Charles, Daniel and Edward) steal 50 gold pieces from a rich lord. They decide to play a game to divide up the loot. The game works as follows:

The robbers line up in alphabetical order. On their turn, the robber proposes a method of splitting up the loot. If they can propose a method that gains a majority vote, then the treasure will be divided based on that method. If their method does not get a majority approval, then they are forced to eat a poisonous apricot and die. The next robber in the line then proposes an alternative method. The robbers are not allowed to make deals amongst themselves, and each of them is rational and selfish - they will vote against any method that doesn't maximise their share.

In the case of an even number of voters, 50% of the vote is regarded as a majority. The robbers do not have the means to cut the gold pieces in half, so there is no way to subdivide the treasure into more than 50 pieces.

So the question is, what method of dividing the loot should Angus propose to stay alive and maximise his winnings?


Sir Thursday

Angus gets 48, both Ben and Edward get 1.

This post has been edited by Obdigore: 17 February 2009 - 02:50 PM

[Sir Thursday]

Obdigore, on Feb 17 2009, 02:46 PM, said:

Angus gets 48, both Ben and Edward get 1.

Incorrect, I'm afraid.

This post has been edited by Sir Thursday: 17 February 2009 - 02:52 PM

[Obdigore]

Sir Thursday, on Feb 17 2009, 08:51 AM, said:

Obdigore, on Feb 17 2009, 02:46 PM, said:

Angus gets 48, both Ben and Edward get 1.

Incorrect, I'm afraid.

Nope, it is correct

[Slow Ben]

Why a poisoned apricot?

[frookenhauer]

He proposes that "Ben Charles and I get 16 gold coins and Daniel and Edward get 1 each."

This post has been edited by Frookenhauer : 17 February 2009 - 10:36 PM

[Giles]

no he'd obviously say dan and edward get 0 as Edward is always going to vote no, same with Dan basically

[frookenhauer]

but you can't divide the pieces. splitting 2 coins 3 ways is not an option.

[Giles]

give 1 each to ben and charlie i think would be the way of maximising hteir share

[Sir Thursday]

Obdigore, on Feb 17 2009, 03:26 PM, said:

Sir Thursday, on Feb 17 2009, 08:51 AM, said:

Obdigore, on Feb 17 2009, 02:46 PM, said:

Angus gets 48, both Ben and Edward get 1.

Incorrect, I'm afraid.

Nope, it is correct

If you think so, explain your reasoning.

Sir Thursday

[Sir Thursday]

Frookenhauer, on Feb 17 2009, 10:35 PM, said:

He proposes that "Ben Charles and I get 16 gold coins and Daniel and Edward get 1 each."

While he might get enough votes that way, he could get more gold than that.

@Obdigore: If Angus tries to propose the method you're suggesting, why would Ben vote for it? If Angus dies, Ben will get to propose his own method, one that will surely get him more than 1 gold piece. And if he's not giving Charles and Daniel anything they aren't going to vote for him. So he'd lose the vote and die.


Sir Thursday

[Giles]

wait the most i can see either of the first 3 getting is 16 or 17. D and E will always vote no to A unless he give them more than 25. So i cant see how 16,17,17 is wrong