bbsriver |
2007-07-21 14:59 |
看了一下“杀手人渣”论坛上面的原文:
The Exact Game Mechanics
The first phase is Day 0.25, in which all twelve players may speak and vote for one player whom they wish to revive. This is done with lynching mechanics (simple majority). Once a decision is reached the game progresses to Day 0.5, in which all players must again revive a player, but obviously not the same player. Repeat on Day 0.75.
After this, 3 players will be alive and 9 dead. If at this point the mafia outnumber the town then the mafia win. If not, the most likely outcome, the game moves into Night 1. Any and all night choices are carried out by only the players who are alive, and they may only target other living players unless their role explicitly says that they may target limbo players as well, except that no killing will be allowed on Night 1.
In the morning, Day 1 begins. All players may again speak, and again the aim is to revive a player, but now only living players may vote. At the end of the day a player will be made living and Night 2 begins.
This continues until either side's win condition is fulfilled, and they are:
Mafia: More living mafia than living town players. Town: Either: all of the pro-town players revived (it is not required that all pro-town players survive after being revived though), half of the total pro-town players alive, or all mafia dead.
Also, to ensure that you cannot determine who is mafia based on when nightkills start to happen, I have assigned each member of the mafia a randomly generated value that determines whether they may or may not kill on the first night after they are revived.
At day you may use any methods to get players revived that you would to get them lynched or saved from lynch in a normal mafia game. This includes claiming, if you so desire.
这个规则实际上再玩的只有复活过来的这几个人。死者虽然可以选择让某人复活,但一天只能复活一人,通常对局面只有间接影响。而复活的人数有限,因此这是一个微型杀人游戏。微型得可以写出每一步可能性。
我按小蝶的修改版把每一步可能性都写出来推算一下,很短:
第一夜杀手不杀人:只有两种可能性:平民:杀手为 4:0或3:1 (2:2/1:3都是杀手胜)
第二夜杀手不杀人:只有一种可能性:3:2 (5:0是平民胜,4:1可能导致平民第三天获胜)
杀手为了避免让平民判断出唯一可能性的存在,第二夜杀人的可能性居多。(如果不杀人,第三天将是3:3或4:2。3:3杀手下一夜获胜;4:2杀手下一夜必须杀人,局面又变成3:2。)
第二夜杀手杀人的话,第三天有两种可能性:3:2/4:1 则前5人中一定有1-2个杀手,平民优先从这5个人之中寻找杀手。
最好的情况:平民白天处死1杀,复活1平民,杀手夜间处死1平民,再有1-2天游戏结束,平民胜 最差的情况:平民白天处死1平民,复活1杀,杀手夜间处死1平民,3:2时再有1天游戏结束,平民胜;4:1时一天后变成2:2进入白天。 居中的情况:平民白天处死1平民,复活1平民;或白天处死1杀,复活1杀,杀手夜间处死1平民,3:2-->2:2;4:1-->3:1
所以第四天游戏还没有结束的话,又只有两种可能性:2:2/3:1 2:2的情况下平民白天无法处决杀手,杀手夜间杀1名平民,变成1:2,杀手胜。 3:1的情况下,杀手如果在夜间杀1人,2:1进入第五天,范围太小不利于隐藏,所以杀手优先考虑不杀人。
所以第五天游戏还没有结束的话,第四天只能是3:1,前4人中一定有1杀手,平民优先从这4个人之中寻找杀手。
最好的情况:平民白天处死1杀,复活1平民,4:0进入下一夜,下一天5:0平民胜,或4:1进入下一夜,杀手杀1平民,3:1进入再下一天。 最差的情况:平民白天处死1平民,复活1杀,杀手夜间处死1平民,变成2:2进入白天,杀手胜。 居中的情况:平民白天处死1平民,复活1平民;或白天处死1杀,复活1杀,仍然是3:1。
所以游戏如果能一直玩下去,最后就是在3:1这个比例上循环……这个时候任何一方犯一个错误就会失败。
这样游戏的变化略显单调~~ |
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