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Robjection

让我们来学习怎么玩 福尔摩斯13 ！

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本教程的存在归功于：Robjection0000
Robjection作者柯南道尔的小说世界里，一场罪恶已被犯下，但谁是罪魁祸首呢？ 是莫利亚提教授是主谋，还是雷斯垂德探长、华生医生可能是主谋，或者甚至是夏洛克福尔摩斯他自己是？ 请试着在其他人搞清楚之前发现真相。 [green][tip] 这个教程最好在电脑上观看。[/green]1maintitlebar_content000
Robjection在一局3-4人游戏的开始，这局游戏[blue]13张卡片中的12张[/blue]会被平均地分发给各位玩家，并且剩下的一张卡片会背面朝上被放置在桌子上。 我之后会解释一场双人游戏该如何设置。 每张卡片都对应着一位源自于夏洛克福尔摩斯系列书籍中的人物。并且[red]这张背面朝上的卡片对应的人物[/red] 就是你要尝试去指认的罪魁祸首。2maintitlebar_content000innocent_criminal_wrap 1 playertablecard_94894832 2 playertablecard_85015709 2 playertablecard_89918668 2
Robjection我们已经搞定了[green]迈克洛夫特·福尔摩斯，雷斯垂德探长。 雷斯垂德，詹姆斯·莫里亚蒂以及塞巴斯蒂安·莫兰在我们手上，因此我们得知，他们之中没有人会是罪犯。 每张卡片底部有2或3个不同的标记。 [red]这张便捷线索表格里[/red]有每一个角色的姓名和每个人的标志物。 举个例子, [blue] 哈德森太太拥有一个烟斗和一条项链。[/blue]3playertable_94894832000note_10_icon 2 note_text_10_name 2 note_13_text 16 card_front_10 3 card_front_4 3 card_front_8 3 card_front_9 3
Robjection我们通过询问对手手中拿着什么符号来获取线索，并将结果[red]记录在我们的线索表上。[/red] [tip] 注意，同一张卡片上不会出现重复的标志物，所以举个例子，有2个拳头= 有2张带拳头标志的卡片。 那么，我们要怎样询问对手关于他们手中线索的事情？ 那么，有两种方式。4clue_89918668_name000clue_85015709_4 16
Robjection第一个是选择一个符号，然后问我们所有的对手*"你有这个符号吗?*"然后他们都会说*"是"*或*"不是"*. 让我们现在就试试吧！ 首先，点击线索表顶部的[red]书[/red]，然后点击[blue]“询问其他玩家”[/blue]按钮。5clue_header_4010btnBroadInvestigation 2 clue_header_4 1
Robjection干的漂亮！ 现在我们知道[blue]蓝色玩家至少有一张带有书本符号的卡牌，[/blue]但[red]红色玩家没有。[/red] 现在蓝色玩家将通过第二种方式获得线索:选择一个符号并问其中一个玩家*“你有多少个这个符号?”*1clue_89918668_466086f5d5264100clue_85015709_4 2 clue_89918668_4 1
Robjection[green]他们问我们有多少书,我们说2。[/green] 但这为什么要写在线索纸上呢? 我们已经知道我们有两本书了，所以这对我们来说不是什么新信息! 还有，为什么线索表上的数字有些是黑色的，有些是灰色的? And what's with the parentheses around some of the grey ones?1clue_94894832_466086f73bf95800clue_94894832_4 3
Robjection表格顶部[red]每个符号下方的灰色数字[/red]告诉你游戏中有多少张牌带有该符号。 你可以自己检查一下线索表下半部分的表格。 例如，[blue]雷斯垂德，格雷森，麦克罗夫特和玛丽·摩斯坦[/blue]各有一本书，所以总共有4本书。2clue_header_766086f73bf95800clue_header_0 1 clue_header_1 1 clue_header_2 1 clue_header_3 1 clue_header_4 1 clue_header_5 1 clue_header_6 1 clue_header_7 1 note_2_icon 2 note_2_name 2 note_9_icon 2 note_9_name 2 note_3_icon 2 note_3_name 2 note_11_icon 2 note_11_name 2
RobjectionThese [red]grey numbers in parentheses[/red] are just a reminder of how many of each symbol we have in our own hand. 在游戏开始时，除了我们没有人知道这些数字。3clue_94894832_166086f73bf95800clue_94894832_0 1 clue_94894832_1 1 clue_94894832_2 1 clue_94894832_3 1 clue_94894832_4 1 clue_94894832_6 1 clue_94894832_7 1
Robjection[red]黑色数字和勾号[/red]是游戏过程中添加的注释。 每个问题和答案都是公开的，所以我们的对手会在他们的桌子上有相同的黑色数字和勾号。 这意味着现在每个人都知道我们有2本书，而紫色的玩家没有，但只有蓝色的玩家知道他们有多少本书。 我们只知道它至少有一个。4clue_89918668_466086f73bf95800clue_94894832_4 1 clue_85015709_4 1 clue_89918668_4 1
Robjection不管怎样，让我们看看紫色的玩家会问什么。5maintitlebar_content66086f73bf95800
Robjection哇，这是怎么回事? 这些[red]项链，眼睛和头骨列[/red]早些时候已经变灰了! 这是因为，在第一轮，没有人可以问这三个符号。 因为他们出现的频率比其他的要低，他们的线索会让我们在游戏一开始就很容易缩小嫌疑人的范围。1clue_89918668_566086f7e0690300clue_header_5 1 clue_header_6 1 clue_header_7 1 clue_94894832_5 1 clue_94894832_6 1 clue_94894832_7 1 clue_85015709_5 1 clue_85015709_6 1 clue_85015709_7 1 clue_89918668_5 1 clue_89918668_6 1 clue_89918668_7 1
Robjection想象一下，如果在第一轮，我们问其他玩家是否有头骨，他们都说没有。 既然我们手上有[green]莫里亚蒂[/green] 和[blue]塞巴斯蒂安·莫兰[/blue]，那凶手一定是[red]艾琳·阿德勒[/red]，但我们的对手无法及时发现。2note_1_icon66086f7e0690300note_1_icon 1 note_text_1_name 1 card_front_8 3 card_front_9 2
Robjection总之，紫色玩家问谁有灯泡，我们和蓝色玩家都说有。 让我们来看看蓝色玩家到底有多少个灯泡! 首先，点击表格中的[red]单元格[/red]，然后点击顶部的[blue]“询问单个玩家”[/blue]按钮。3clue_85015709_166086f7e0690310btnNarrowInvestigation 2 clue_85015709_1 1
Robjection好的，[blue]蓝色玩家有两个灯泡。[/blue]我们不知道最后一个灯泡是否在紫色玩家的手中。 同时，蓝色玩家对灯泡的分布一无所知，只知道他们有2个，我们至少有1个。 让我们继续往下看。1clue_85015709_166086f99846bc00clue_85015709_1 2
Robjection啊，这是什么? 结果是[blue]蓝色的玩家有两本书。[/blue]这说明他们手里一定有[green]格雷森探长和玛丽·摩斯坦[/green]，所以他们俩都不是罪犯。1maintitlebar_content66086fdbc7d0d00clue_85015709_4 2 note_3_icon 3 note_3_name 3 note_11_icon 3 note_11_name 3
Robjection教程不会让你这么做，但在真正的游戏中，你可以点击 [green]角色名称旁边的空格[/green]来循环: - *无辜*(非罪犯) -对手的名字(角色在该玩家手中) -[red]*罪犯*[/red](罪犯) 做这些笔记没有机械的效果，但它可以帮助你收集你的想法。2maintitlebar_content66086fdbc7d0d00note_7_text 3 note_1_text 3 note_8_text 3 note_3_text 3 note_10_text 3 note_4_text 3 note_11_text 3 note_5_text 3 note_6_text 3
Robjection不管怎样，还记得我之前说的吗，如果蓝色玩家和紫色玩家都没有头骨，那么我们就知道艾琳·阿德勒一定是罪犯? 好吧，让我们看看她是不是! 点击表格顶部的[red]骷髅[/red]，然后点击[blue]“询问其他玩家”[/blue]按钮。3clue_header_766086fdbc7d0d10btnBroadInvestigation 2 clue_header_7 1
Robjection[red]这次运气不好，[/red]但至少我们可以把她排除在外。 希望我们对手的问题能给我们提供更多信息让我们进一步缩小嫌疑人的范围。1clue_89918668_766086feeded8200clue_89918668_7 1
Robjection我们的对手都问对方有多少条项链。 因此，现在所有人都知道所有[red]项链符号[/red]和所有[blue]书符号[/blue]都在玩家手中，所以罪犯不能拥有这两个。 另外，我们(只有我们)知道所有的[green]骷髅符号[/green]都在玩家手中，所以罪犯也不能拥有这些符号。1clue_89918668_56608700aa3d0c00clue_header_4 2 clue_header_5 1 clue_header_7 3 clue_94894832_4 2 clue_94894832_5 1 clue_94894832_7 3 clue_85015709_4 2 clue_85015709_5 1 clue_85015709_7 3 clue_89918668_4 2 clue_89918668_5 1 clue_89918668_7 3
Robjection此外，由于她是唯一一个没有头骨的角色，我们知道紫色玩家一定有[red]艾琳阿德勒[/red]，她有[red]一个灯泡[/red]，这意味着紫色玩家手里有一个灯泡。 如果他们有1个灯泡，[green]我们有2个[/green]，[blue]蓝色玩家有2个[/blue]，这就是5个灯泡，所以罪犯一个都没有。2clue_85015709_16608700aa3d0c00clue_94894832_1 3 clue_85015709_1 2 note_1_icon 1 note_text_1_name 1
Robjection所以罪犯只能有烟斗、拳头、警徽和眼睛符号。 目前唯一可能的嫌疑人是[red]布拉德斯特里特探长[/red]，[blue]霍普金斯探长[/blue]和[green]约翰·华生[/green]。3maintitlebar_content6608700aa3d0c00note_8_icon 3 note_text_8_name 3 note_5_icon 1 note_text_5_name 1 note_6_icon 2 note_text_6_name 2
Robjection最后，一旦你有了足够的线索来找出谁是罪犯，你就可以通过[blue]在线索表[/blue]上选择他们的名字，[red]点击“指控一个角色”按钮来说出他们是谁。[/red] 如果你猜对了，你就赢了!4maintitlebar_content6608700aa3d0c00btnGuessCriminal 1 note_8_name 2 note_text_5_name 2 note_6_name 2
Robjection!!! 但是要小心! 如果你错了，你就出局了，你就会*输掉*游戏！ 你仍然要回答对手提出的任何问题，但你自己不能提出任何问题，也不能提出任何指控。 如果除了一个玩家外，所有玩家都猜错了罪犯，那么没有提出指控的玩家就默认获胜。5maintitlebar_content6608700aa3d0c00
Robjection这就是玩《福尔摩斯13》标准的3-4人游戏所需要知道的一切! 有一种专家版本，当你被问及你有哪些符号时，你不告诉任何人你的卡片上的符号。 你手里的某张牌会被忽略，就像这样，每一轮都会发生变化。6maintitlebar_content6608700aa3d0c00
Robjection在双人游戏中，两名玩家都有5张牌，剩下的3张牌面朝下放在桌上。 中间的那个是罪犯。 因为“询问其他玩家”的选项现在完全多余了，相反，你可以从你的手牌中交换一张在罪犯旁边的正面朝下的牌。 你丢弃的卡牌面朝上放在了罪犯的旁边，无法被再次拿起。7maintitlebar_content6608700aa3d0c00
Robjection《福尔摩斯13》教程到此结束。 非常感谢*Hope S。 本游戏由Hwang*设计，由*BoardM Factory*发布，由BGA用户*ufm*在本网站实现了它。 如果你觉得本教程有帮助，请务必给它5颗星。 调查愉快!8maintitlebar_content6608700aa3d0c00
Robjection

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福尔摩斯13 规则：互动式教程
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Sherlock 13

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