Merlot Mafia: Last Queen of the Amazons

Sounds good - sign me up.
 
Any more?

I don't dare, as what I said earlier still applies:

I became concerned about the Merlot team spirit at one point and stopped posting because of that. The game requires more aggressive stance that normal discussion on this team forum and while paranoia is good for the mafia game, I was not sure how well everyone could keep it inside the magia game. I didn't want to hurt my image any more - I got the feeling I was quite annoying, which was to some degree quite intentional/irrelevant for the goal. As none of you know me as the nice person I actually am (or hope to be) and we are going to spend quite a few more weeks/months playing the actual game this team was made for, I decided civ game > mafia game. Had everyone been my old friends, I would've been much less worried about it.
 
Well, no matter, five is enough for a game. (albeit a very small one)

The game will begin in 48 hours.

Entrants are still welcome until then.

Current players are:
1) landlubber
2) slaze
3) Caledorn
4) whb
5) Tboy
 
This is all I know. I don't know how the voting works and as far as I'm concerned, Winston has neither confirmed nor denied many of the things being written here. If we are forced to lynch let us do so in an organized manner, one that may ... you know ... lead us to a further understanding, or at least attempt to do so.

This seems to me like it's a binary riddle. Computer language. I could give you another one:

25 (or more) prisoners. They will each, one at a time, be led to an isolated room with no chance of communicating with their fellow prisoners except for two swiches in the room. The 2 switched each have 2 positions: up and down. The prisoners are hearded together and told of their future situation, one by one in the room, and allowed time to organize a strategy but once the first prisoner is shown inside all prisoners are separated and have no contact with each other, except for the position of the switches in the room. They are all, in any random or determined or manipulated order, led to the room where they have to move the position of a single switch and only that. Finally, the prisoners must declare that they have all been in the room at least once, and if they get the answer wrong they all die.

There's a solution to it, and it's not relying on some almost 50% odds stab at the end.

Anyway, we have to vote, so I'm just seeing if mafia votes don't count. But we'd all need to vote for that to work (if it can).

I could be totally wrong here. Maybe this is just some, personality/deception game. Either way my fate's sealed, one way or the other.



The solution is the prisoners must establish a "counter". That counter becomes the only one who deals with the guards - he declares when everyone has been in the room.

and as i re-read the above, let me re-phrase to be clear.

a bunch of prisoners, once everything has started they are, one by one, led into a room, they have to switch one of two switches, they are separated, they can't see who's going into the room, they are only allowed time before everything gets going to formulate a strategy, after things proceed they see no one, only guards when they get led to the room with the switch. And one by one the prisoners are led to the room to switch a switch until they declare that they've all been there - if they get the answer right they live, wrong, they die.

Thus, they must declare when they are SURE that all of them have been into the room at least once. and they can be led into the room (w/ the switch) in ANY order. And when they're led in to the room the HAVE to switch one and ONLY one switch.

anyway, the solution:

They agree on a switch and the position. Let's say they chose the left one switched down. Everyone (but the counter) agrees that if they go into the room and see the left switch in the up position, they switch it down (but twice only, and then the right switch only, but more on this later). Otherwise they switch the right one. And when the counter gets into the room, he switches the left switch up.

But they are not certain of the initial positions of the switches: given that there is the unlikely posibility that the counter may be the first prisoner in the room and the left switch could be set in the down position, the counter must count (via the left switch being swithed down) everyone twice (minus one). Otherwise, he may mis-count a prisoner having been in the room.


Up until that point the guards can manipulate it so that someone hasn't been in the room.


...


I realize (now and after this concluded) that this wasn't exactly what was going on here but in case anyone was holding their breath I thought I'd provide the solution to the little riddle.
 
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