想象你坐在一张 12 人的桌子边。4 只狼，8 个好人。但这一局没有预言家，没有女巫，也没有猎人。好人白天只能凭感觉投人，说白了就是乱投；狼每天晚上刀掉一个好人。

你猜狼赢的概率有多大？

我算了一下，是 105/128，大约 82%。狼只占三分之一，赢面却超过八成。

这个数后面藏着一个更有意思的问题：人多了以后，狼要占的比例还一样吗？12 个人要 4 只狼，那 1200 个人是不是就得 400 只？

差得远。2006 年，Mark Braverman、Omid Etesami 和 Elchanan Mossel 三个人专门为这个游戏写了一篇论文，后来发在概率论的老牌期刊《应用概率年刊》（The Annals of Applied Probability）上。下文用三人姓氏的首字母，叫它 BEM。论文致谢里说，Mossel 是 2004 年在一次学术年会上给一大群人主持 Mafia（狼人杀的前身）的时候，冒出这个念头的。

他们的结论是：好人完全没有信息时，狼的数量只要和总人数的平方根差不多，就能跟好人打平。一百万开根号是一千，按我后面的精确计算，674 只狼就够了。

这是“玩物有理”的第二篇，上一篇聊的是俄罗斯方块。这一篇想讲清楚三件事：为什么偏偏是平方根；为什么多一个预言家，局面就整个翻过来；还有，后来有人说这条定理有漏洞，他说得对不对。数学我尽量用大白话讲。想看证明的读者，灰框里有，不想看的直接跳过，不影响往下读。

数学家研究一个游戏，第一步总是把它削到只剩骨架。

论文里的狼人杀是这样的：一共 $R$ 个人，其中 $M$ 只狼。狼互相认识，好人谁也不认识谁。白天大家投票，票最多的人出局，平票就抽签；晚上狼刀一个好人。狼死光了好人赢，好人死光了狼赢。一天一夜下来，桌上少两个人。

没有神职，没有警长，也没有遗言。

好人该怎么投？答案有点反直觉：乱投就是最好的办法。

你想，好人手里一点信息都没有，所有人在他眼里都一样可疑。狼完全可以装成好人，你说什么它附和什么，跟着大家一起投。狼还没过半的时候，好人不管设计出多精巧的投票方法，狼都能混在里面照做，结果和乱投没有区别。等狼过了半数，票数由狼说了算，好人怎么投都没用。

所以好人唯一能做的，是保证“乱投”是真的乱。论文给了一个很漂亮的办法：每人同时报一个数，加起来除以人数取余数，余数是几就投几号。只要有一个好人报的是真随机数，狼就没法操纵结果。“同时报”这一步，现实里得靠法官收纸条。

当然，真实的狼人杀有发言、有逻辑、有人表情管理失败，好人是能读出信息的。模型故意把这些全拿掉，只问一个问题：完全没有信息的时候，人多到底有多大用。

白天投出去的人是狼，概率是多少？就是狼占的比例 $M/R$。晚上狼刀的一定是好人。这么一天天打下去，狼在少，好人也在少。有没有什么东西是不变的？

有。BEM 找到了这样一个数：

它不是每一局都纹丝不动，而是“平均来看”不变。打个比方：你和朋友抛硬币赌钱，正面赢一块，反面输一块。你兜里的钱会上下晃，但它的期望永远停在原地。数学上把这种“公平赌局里的筹码”叫作鞅。$X$ 就是狼人杀里的一个鞅。

这个式子有什么用？狼多的时候，$M(M-1)$ 和 $M^2$ 差不多，$X$ 就约等于 $(M/\sqrt{R}{)}^2$。翻译成人话：狼数除以根号人数，整局平均不变。

举个例子。一百万人的局，674 只狼，674 除以 1000 是 0.674。打到只剩一万人时，根号一万是 100，狼平均还剩 67 只左右，比例还是 0.674。人少了 99%，狼只少了 90%。狼被消耗得比好人慢得多。

图 3 是我模拟的三局。前面几十万天，三条线几乎贴在一起，平得像拿尺子画的；要到最后几十个人，才开始上蹿下跳。道理不难想：每天只出局一个人，对一百万人来说就是一滴水；人少了，投出去一个就能改写整个比例。

所以这种局真正的胜负，是在最后几十个人里定下来的。前面的几十万天，基本是在走流程。

有了这个不变的数，平方根就藏不住了。

假设狼数是根号人数的 $\eta$ 倍（$\eta$ 读作“艾塔”，就是个倍数）。那 $X$ 一开局就大约是 ${\eta }^2$，而且整局平均都停在这个水平。$\eta$ 很小，狼手里的“筹码”就很少，几乎翻不了盘；$\eta$ 很大，狼几乎稳赢。BEM 把这个直觉写成了一条定理。

关键在“只跟 $\eta$ 有关”，和总人数无关。100 个人的局放 10 只狼，一百万人的局放 1000 只狼，两边的 $\eta$ 都是 1。按精确计算，前者狼的胜率是 70.6%，后者是 68.3%，差不多都在七成上下。

另一半，狼多时几乎必胜，证明要麻烦得多。作者换了一个复杂得多的量，要证明它在狼过半之前只涨不跌，最后把一个九次多项式交给 Mathematica 展开，再一项一项比大小。这里就不展开了。

定理只说了量级，没说到底几只。具体数字可以硬算。记 $w(R,M)$ 为 $R$ 人、$M$ 狼时狼的胜率。白天投出好人，就变成少两个好人的局面；投出狼，就变成少一只狼、少一个好人的局面。写成式子：

狼过半时 $w=1$，没狼了 $w=0$。从小局一路往上推，电脑几十秒就能推到一百万人。结果画在图 4 里：总人数是偶数时，狼数达到 $0.674\sqrt{R}$ 就五五开，一百万人对应 674 只；总人数是奇数时更少，$0.494\sqrt{R}$ 就够了。一万、十万、一百万人三条曲线几乎完全叠在一起。

顺带一提，原论文用模拟估过这个常数，得到 0.643；Piotr Migdał 2010 年给出了胜率的精确公式，再用一个粗略近似估成 0.627。精确递推出来是 0.674。

图 4 里有个怪地方：红线（奇数人数）明显比灰线高。人数差一个，狼的处境能差这么多？

先看最简单的情况：桌上只有一只狼。

狼要赢，只需要每个白天都不被投中。还剩 $k$ 个人的那天，它躲过去的概率是 $(k-1)/k$，一路乘下去就是它的胜率。区别出在最后一天。

总人数是偶数的话，最后会剩一狼一好人，白天两人互投，只能抽签，狼只有一半机会。总人数是奇数的话，最后一个白天剩三个人，狼只要躲过这一票（概率 2/3），晚上把最后一个好人刀掉就赢了。

所以 5 个人里藏 1 只狼，狼的胜率是 8/15，已经过半；4 个人里藏 1 只狼，只有 3/8。

人越多，一只狼越难活到最后：它的胜率一直压在 $\sqrt{2/(R+1)}$ 以下，上一节的证明用的就是这个界。

回到那个不变的数。既然前期“狼数除以根号人数”几乎不动，那剩 $k$ 个人的时候，狼大概有 $\eta \sqrt{k}$ 只。

还能说得更细。每只狼能不能活到最后，基本是各凭运气，彼此不太相干。很多件“概率很小、互不相干”的事凑在一起，发生的次数服从泊松分布，就像一页书上的错别字个数。所以剩 $k$ 人时，狼的只数大致服从平均值为 $\eta \sqrt{k}$ 的泊松分布。

这给了另一种算法：随便取一个 $k$，从 $k$ 个人、泊松个狼开局，玩一局小游戏，算狼的胜率。$k$ 取 1600 时，结果和一百万人的精确值差不到千分之一（$\eta =1$ 时是 0.6825 对 0.6829）。这一步我没有写成严格证明，只算数值上的观察。

但它说的事很直观：一百万人的大局，其实等价于一局几十人的小局，只不过开局时狼有几只，是摇骰子摇出来的。

玩过狼人杀的都知道预言家有多要命。论文的后半部分，就是把预言家加了进来：每晚私下查一个人是好是坏，结果只有她自己知道。

和上面对比一下就知道这有多夸张。没有神职时，根号量级的狼就能打平；多一个预言家，狼得占到总人数的一个固定比例才有戏。一百万人的局，674 只狼本来能五五开，现在碰上一个预言家，基本没机会。

那预言家该怎么用？直觉是让她去找狼。可是算一算就知道不够：她平均只能活大约 $R/4$ 天，这段时间能查出大约一半的狼。就算这一半全被投出去，剩下那一半还是远多于根号量级，好人照样输。

论文的办法很妙：让预言家专门攒好人。用玩家的话说，就是攒金水。

第一段，预言家装成普通好人，闷声查人，攒下一份金水名单，名单上的人要比狼多。第二段，她把名单私下发给名单上的每个人，然后请大家白天把自己投出去。她一出局，身份就公开了，名单上的人由此知道这份名单是真的。第三段，名单上的人互相认识了，像狼一样抱团：每天由名单上排第一的人在名单外挑一个，全体金水一起投他。

好人从此也有了自己的“狼队友”。金水只投名单外的人，每一票都比乱投准得多。

狼会不会冒充预言家，也发一份假名单？期刊版补了一条规则：凡是自称预言家的，按跳出来的先后顺序全部投出去。狼冒充预言家只会白送一条命，所以不会干。用玩家的话说，在这个模型里，悍跳没有收益。

这套打法有个前提：好人之间能私下传消息、能匿名投票，论文用法官或者密码学协议来保证。原文还给了一个不靠这些的版本：预言家查到一半以上的活人都是好人，就公开名单，再以死自证。代价是好人的胜率只能保证大于 0，根号量级的狼可以抢在她攒够名单之前，先把她刀了。

真实的狼人杀里，预言家跳身份、报金水和查杀、留警徽流，干的是同一件事：死之前，把信息交到好人手里。定理告诉你这件事值多少钱。它不是多抓一两只狼的问题，它把狼需要的数量，从根号一下子抬到了一个固定的比例。

2008 年 4 月，中国科学院计算技术研究所的 Erlin Yao 在 arXiv 上贴出一篇论文，说定理 2 在原文的条件下根本不成立，就算成立，也推不出“根号量级打平”，还给了反例。

我把两篇对着读了一遍，觉得问题出在一个很细的地方：量词的顺序。

先看一个生活里的例子。“每个人都能找到一双合脚的鞋”和“有一双鞋，每个人穿都合脚”，只是把“每个”和“有一个”换了个位置，意思天差地别。前一句显然对，后一句显然错。

定理 2 说的是：对每个 $\eta$，只要人数足够多，结论就成立。需要多少人，可以随 $\eta$ 变。$\eta$ 越小，要的人越多，至少得多到狼数 $\eta \sqrt{R}$ 不少于 1 只。

Yao 的第一条反驳，是先把人数固定住，再让 $\eta$ 越变越小。人数固定、$\eta$ 足够小，算出来的狼数就不到 1 只，可狼至少得有 1 只，胜率当然不会趋于 0。这等于拿“有一双鞋人人合脚”去反驳“每个人都有合脚的鞋”。原定理没说过前一句。

第二条是一个反例。他构造的胜率是 $w=M/(R+M)$，配了一个下界 $p(\eta )=\eta /(\eta +\sqrt{R})$。问题是，这个本该只和 $\eta$ 有关的下界，偷偷带上了人数 $R$，人越多它越接近 0。原定理要的，偏偏是一个不随人数缩水的下界，而这个反例在人多时胜率也趋于 0，正好被它排除了。

Yao 自己也证明了一个界：狼数不超过 $k$ 时，

他说这验证了根号结论。可一旦狼数是根号量级，$k$ 也得取那么大，左边的下界就缩成大约 $\sqrt{2\eta }{R}^{-1/4}$，人越多越接近 0。最需要它的地方，它偏偏没用了。

学过数学分析的读者会觉得眼熟：“对每个 $x$，$n$ 够大以后误差都很小”和“$n$ 够大以后，对所有 $x$ 误差都很小”，一个是逐点收敛，一个是一致收敛。定理 2 是前一种，而前一种已经足够说明根号是门槛：随便定一个 $\eta$，人一多，胜率就被夹在两个和人数无关的常数之间。

再补两个背景。BEM 的论文 2007 年 5 月投稿、6 月修订，期刊版里定理 2 的写法和 2006 年底的 arXiv 版一字不差。Yao 的论文在 arXiv 上注明投给了同一份期刊，我没有查到它正式发表。

以上是我的读法，欢迎指正。

说回真实的游戏。中文狼人杀最常见的 12 人局（预女猎白）是 4 狼、4 民、4 神，通常还用屠边规则：狼刀光神职或者刀光平民就算赢，比刀光所有好人容易。按这个模型，12 个人里 3 只狼就已经过了五五开（胜率 21/32），4 只狼在没有神职时胜率 82%。

狼占三分之一，是一个固定比例。按定理 4，这么多狼，好人只有靠神职才能拉回来。预言家、女巫、猎人、白痴，给好人的不是信息，就是额外的行动。

血染钟楼走得更远。它的配置大致是 5 到 9 人有 2 个邪恶玩家，10 到 12 人有 3 个，13 到 15 人有 4 个（一个恶魔加几个爪牙）。好人几乎人人有技能，很多是每晚拿信息的角色，相当于定理 4 里的“很多个预言家”。照定理的说法，信息位这么多，好人应该稳赢才对。

设计者的办法是让信息变得靠不住。官方规则写得很清楚：醉酒或中毒的玩家其实没有能力，说书人却会假装他有，照样给他信息，信息可以是假的，他自己还不知道。再加上处决要过半数票、死人照样能说话还留一张票、只剩两个人时邪恶获胜，这几条都在把好人从“很多个靠谱的预言家”往回拽。

定理给了两个端点：没有信息，狼只要根号那么多；信息又多又可靠，狼接近一半也白搭。设计师做的事，是在这两头之间找一个让人玩得下去的位置。

论文结尾还留了一个猜想。现在的模型是一天一夜，好人两轮里只有一轮能投票。节奏换一换，门槛的指数大概也会跟着变。

一天一夜就是 $d=1$、$r=2$，对应根号。Yao 的论文说证明了这个猜想，但证明只有梗概，又建立在上一节那个会失效的下界上，所以我还是把它当猜想看。

算完这些数，我最大的感受是：在这个模型里，人数不怎么值钱，信息才值钱。一百万个互不信任的好人，只能和 674 只互相认识的狼打个平手；一个会发金水的预言家，就能把门槛从根号抬到一个固定比例。

论文开头把这类问题比作公司里的员工和股东、学校里的学生和管理者、城市里的市民和黑帮。数学只回答了最简化的版本，真实的桌上，每个人都在发言、观察、推理。

所以下次你在桌上第一晚就被刀了，也别太郁闷。至少你知道，狼手里那点筹码，从开局到现在平均一分都没少过。

一百万个互不信任的人，
只能和 674 只狼打平。

Imagine you are sitting at a table of 12. Four wolves, eight villagers. But this game has no Seer, no Witch and no Hunter. By day the villagers can only vote on gut feeling, which is to say at random; every night the wolves kill one villager.

How likely do you think the wolves are to win?

I worked it out: 105/128, about 82%. The wolves are only a third of the table, yet they win more than eight games in ten.

Behind this number lies a more interesting question: as the table grows, do the wolves need the same share? If 12 players need 4 wolves, do 1,200 players need 400?

Not even close. In 2006 Mark Braverman, Omid Etesami and Elchanan Mossel wrote a paper about exactly this game, later published in a venerable probability journal, The Annals of Applied Probability. I will call it BEM, after the authors' initials. According to the paper's acknowledgements, the idea came to Mossel in 2004, while he was moderating a game of Mafia (the forerunner of Werewolf) for a large crowd at an annual academic meeting.

Their conclusion: when the villagers have no information at all, wolves numbering roughly the square root of the total number of players can hold the villagers to a draw. The square root of a million is a thousand, and by the exact calculation I give later, 674 wolves are enough.

This is the second essay in “The Logic of Play”; the first was about Tetris. This one sets out to explain three things: why it is the square root, of all things; why one extra Seer flips the whole picture; and whether the person who later said the theorem had a hole in it was right. I will keep the mathematics in plain language as far as I can. Readers who want the proofs will find them in the gray boxes; anyone else can skip them without losing the thread.

The first thing a mathematician does with a game is strip it down to the bone.

In the paper, the game goes like this. There are $R$ players, $M$ of them wolves. The wolves know one another; the villagers know nobody. By day everyone votes, and the player with the most votes is exiled, with ties settled by lot; by night the wolves kill a villager. The villagers win when every wolf is dead, the wolves when every villager is dead. Each day and night, the table loses two players.

No special roles, no sheriff, no last words.

How should the villagers vote? The answer is a little counterintuitive: voting at random is the best they can do.

Think about it. The villagers have no information at all, so to each of them everyone looks equally suspicious. A wolf can simply pose as a villager, agree with whatever you say, and vote along with everyone else. As long as the wolves are short of half, however clever a voting scheme the villagers devise, the wolves can blend in and follow it, and the result is no different from voting at random. Once the wolves are past half, they decide every vote, and nothing the villagers do makes any difference.

So the one thing the villagers can do is make sure that “random” really is random. The paper has a lovely way to do it: everyone calls out a number at the same moment, the numbers are added up and divided by the number of players, and everyone votes for the seat whose number is the remainder. As long as one villager calls out a truly random number, the wolves cannot steer the result. In real life, the “at the same moment” part takes a moderator collecting slips of paper.

Real Werewolf, of course, has speeches, logic, and players whose poker faces slip, so villagers can pick up information. The model strips all of this away on purpose and asks a single question: with no information at all, how much are numbers really worth?

What is the chance that the player exiled by day is a wolf? It is just the wolves' share, $M/R$. The player killed at night is always a villager. As the game goes on, day after day, the wolves dwindle, and so do the villagers. Does anything stay the same?

Something does. BEM found this number:

It does not sit perfectly still in every game; it stays the same “on average.” Here is an analogy. You and a friend bet on coin tosses, winning a dollar on heads and losing one on tails. The money in your pocket goes up and down, but its expected value never moves. Mathematicians call this kind of “stack of chips in a fair game” a martingale. $X$ is a martingale hiding inside Werewolf.

What is this formula good for? When there are many wolves, $M(M-1)$ and $M^2$ are nearly equal, so $X$ is roughly $(M/\sqrt{R}{)}^2$. In plain words: the number of wolves divided by the square root of the number of players stays the same, on average, all game.

An example. In a million-player game with 674 wolves, 674 divided by 1,000 is 0.674. By the time only 10,000 players are left, the square root of 10,000 is 100, and on average about 67 wolves remain: the ratio is still 0.674. The players have fallen by 99%, the wolves by only 90%. The wolves are used up far more slowly than the villagers.

Figure 3 shows three games I simulated. For the first few hundred thousand days the three lines lie almost on top of one another, as flat as if drawn with a ruler; only in the last few dozen players do they start leaping up and down. The reason is not hard to see: one player voted out a day is a drop in the ocean for a million players, but once few are left, a single vote can rewrite the whole ratio.

So the real outcome of a game like this is decided among the last few dozen players. The hundreds of thousands of days before that are mostly a formality.

With this steady number in hand, the square root has nowhere left to hide.

Suppose the number of wolves is $\eta$ times the square root of the number of players ($\eta$, read “eta,” is just a multiplier). Then $X$ starts out at about ${\eta }^2$, and on average it stays at that level all game. If $\eta$ is small, the wolves hold very few “chips” and can hardly ever turn the game around; if $\eta$ is large, the wolves all but win. BEM turned this intuition into a theorem.

The key is “depend only on $\eta$”: the total number of players does not come into it. A 100-player game with 10 wolves and a million-player game with 1,000 wolves both have $\eta$ equal to 1. By exact calculation, the wolves win 70.6% of the time in the first and 68.3% in the second: both around 70%.

The other half, that many wolves almost always win, is much harder to prove. The authors switch to a far more complicated quantity and have to show that it only rises, never falls, until the wolves are past half; in the end they hand a polynomial of degree nine to Mathematica, expand it, and compare it term by term. I will not go into it here.

The theorem gives only the order of magnitude, not how many wolves exactly. The exact numbers can be computed by brute force. Let $w(R,M)$ be the wolves' chance of winning with $R$ players and $M$ wolves. If a villager is exiled by day, we move to a position with two villagers fewer; if a wolf is exiled, to one with a wolf and a villager fewer. As a formula:

When the wolves are past half, $w=1$; when no wolves are left, $w=0$. Working up from small games, a computer gets all the way to a million players in a few dozen seconds. The results are plotted in Figure 4. With an even number of players, the game is 50–50 once the wolves reach $0.674\sqrt{R}$, which for a million players means 674; with an odd number it takes fewer, just $0.494\sqrt{R}$. The curves for ten thousand, a hundred thousand and a million players lie almost exactly on top of one another.

Incidentally, the original paper estimated this constant by simulation and got 0.643; in 2010 Piotr Migdał gave an exact formula for the winning probability, then used a rough approximation to estimate the constant at 0.627. The exact recursion gives 0.674.

There is something strange in Figure 4: the red line (an odd number of players) sits clearly above the gray ones. Can one player more or less really make that much difference to the wolves?

Start with the simplest case: a single wolf at the table.

To win, the wolf only has to survive every daytime vote. On the day when $k$ players are left, it escapes with probability $(k-1)/k$, and multiplying these all the way down gives its chance of winning. The difference comes on the last day.

If the total is even, the game ends with one wolf and one villager. By day the two vote for each other, so lots are drawn, and the wolf has only a 50–50 chance. If the total is odd, three players are left on the last day; the wolf just has to dodge that vote (probability 2/3) and kill the last villager that night, and it has won.

So with 1 wolf hidden among 5 players, the wolf wins with probability 8/15, already better than half; with 1 wolf among 4 players, only 3/8.

The more players there are, the harder it is for a single wolf to last to the end: its chance of winning always stays below $\sqrt{2/(R+1)}$, and this is the bound the proof in the previous section used.

Back to the number that stays put. Since “wolves divided by the square root of the players” barely moves early on, when $k$ players are left there are about $\eta \sqrt{k}$ wolves.

We can say more. Whether each wolf survives to the end is basically down to its own luck, largely independent of the others. When many unlikely, unrelated events are lumped together, the number that happen follows a Poisson distribution, like the number of typos on a page of a book. So when $k$ players are left, the number of wolves roughly follows a Poisson distribution with mean $\eta \sqrt{k}$.

This gives another way to compute: pick any $k$, start a small game with $k$ players and a Poisson number of wolves, and work out the wolves' chance of winning. With $k$ set to 1600, the result differs from the exact value for a million players by less than one part in a thousand (0.6825 against 0.6829 at $\eta =1$). I have not turned this step into a rigorous proof; treat it as a numerical observation.

But what it says is intuitive: a giant game of a million players is really the same as a small game of a few dozen, except that the number of wolves at the start is decided by a roll of the dice.

Anyone who has played Werewolf knows how deadly the Seer is. The second half of the paper brings her in: each night she privately checks whether one player is good or bad, and only she knows the result.

Compare this with what came before to see how dramatic it is. With no special roles, wolves on the order of the square root are enough for a draw; add one Seer, and the wolves need a fixed fraction of the players just to stand a chance. In a million-player game, 674 wolves could have held the villagers to 50–50; facing one Seer, they have essentially no chance.

So how should the Seer be used? The obvious idea is to send her hunting wolves. But a little arithmetic shows that is not enough: on average she lives only about $R/4$ days, and in that time she can find about half the wolves. Even if that half were all voted out, the other half would still be far more than the square-root scale, and the villagers would lose anyway.

The paper's idea is clever: have the Seer collect villagers instead. In players' terms, she builds up a list of cleared players.

In stage one, the Seer poses as an ordinary villager and quietly checks players, building up a cleared list that must be longer than the number of wolves. In stage two, she privately sends the list to everyone on it, then asks the table to vote her out by day. Once she is out, her role is revealed, and so the people on the list know the list is genuine. In stage three, the people on the list know one another and band together like wolves: each day the first person on the list picks someone off the list, and all the cleared players vote for that player.

From then on, the villagers have “wolf teammates” of their own. The cleared players vote only for people off the list, and every vote is far more accurate than a random one.

Could a wolf pose as the Seer and send out a fake list? The journal version adds a rule: everyone who claims to be the Seer is voted out, in the order they came forward. A wolf impersonating the Seer would only be throwing away a life, so no wolf will do it. In players' terms: in this model, a fake claim never pays.

The plan has a precondition: the villagers must be able to pass messages privately and vote anonymously, which the paper guarantees with a moderator or a cryptographic protocol. The paper also gives a version that needs neither: once the Seer has found more than half of the living players to be villagers, she makes the list public and then dies to prove it. The price is that the villagers' chance of winning is only guaranteed to be above 0: wolves on the square-root scale can kill her before she has built up a long enough list.

In real Werewolf, when the Seer reveals her role, announces whom she has cleared and whom she has caught as a wolf, and arranges where her sheriff's badge goes if she dies, she is doing the same thing: putting her information in the villagers' hands before she dies. The theorem tells you what that is worth. It is not a matter of catching one or two more wolves; it lifts the number of wolves needed from the square root all the way up to a fixed fraction.

In April 2008 Erlin Yao, of the Institute of Computing Technology at the Chinese Academy of Sciences, posted a paper on arXiv arguing that Theorem 2 simply does not hold under the original paper's conditions, and that even if it did, it would not imply a draw at the square-root scale. He also gave counterexamples.

I read the two papers side by side, and I think the trouble lies in a very fine point: the order of quantifiers.

Start with an everyday example. “Everyone can find a pair of shoes that fits” and “There is one pair of shoes that fits everyone” differ only in where “every” and “there is one” sit, yet they mean completely different things. The first is obviously true; the second is obviously false.

Theorem 2 says: for every $\eta$, once there are enough players, the conclusion holds. How many players are enough is allowed to depend on $\eta$. The smaller $\eta$ is, the more players it takes; at the very least, enough that the number of wolves $\eta \sqrt{R}$ is at least 1.

Yao's first objection fixes the number of players first and then lets $\eta$ get smaller and smaller. With the number of players fixed and $\eta$ small enough, the computed number of wolves drops below 1; but there has to be at least one wolf, so of course the chance of winning does not tend to 0. That amounts to taking “one pair of shoes fits everyone” and using it to rebut “everyone has a pair of shoes that fits.” The theorem never said the first.

The second is a counterexample. The win probability he constructs is $w=M/(R+M)$, together with a lower bound $p(\eta )=\eta /(\eta +\sqrt{R})$. The trouble is that this lower bound, which ought to depend only on $\eta$, quietly carries the number of players $R$, and gets closer to 0 the more players there are. What the theorem asks for is precisely a lower bound that does not shrink as the number of players grows; and since the counterexample's win probability also tends to 0 when there are many players, the theorem rules it out.

Yao also proved a bound of his own: when there are at most $k$ wolves,

He says this confirms the square-root result. But once the number of wolves is on the square-root scale, $k$ has to be that large as well, and the lower bound on the left shrinks to about $\sqrt{2\eta }{R}^{-1/4}$, getting closer to 0 the more players there are. Just where it is needed most, it stops being of any use.

Readers who have studied real analysis will find this familiar: “for every $x$, the error is small once $n$ is large enough” and “once $n$ is large enough, the error is small for all $x$” are pointwise and uniform convergence respectively. Theorem 2 is of the first kind, and the first kind is already enough to make the square root the threshold: fix any $\eta$, and once there are many players, the chance of winning is squeezed between two constants that do not depend on the number of players.

Two more pieces of background. The BEM paper was submitted in May 2007 and revised in June, and the journal version states Theorem 2 word for word as the arXiv version of late 2006 does. Yao's paper notes on arXiv that it was submitted to the same journal; I could not find that it was ever published.

That is my reading; corrections are welcome.

Now for the real games. The most common setup in Chinese Werewolf is the 12-player game with Seer, Witch, Hunter and Idiot: 4 wolves, 4 plain villagers and 4 special roles. It usually adds the “side-kill” rule as well: the wolves win once they have killed all the special roles or all the plain villagers, which is easier than killing every villager. Under this model, 3 wolves among 12 players are already past 50–50 (they win with probability 21/32), and 4 wolves with no special roles to face win 82% of the time.

Wolves making up a third of the table is a fixed proportion. By Theorem 4, against that many wolves the villagers can claw their way back only through special roles. The Seer, the Witch, the Hunter and the Idiot each give the villagers either information or an extra action.

Blood on the Clocktower goes further. Its setups run roughly like this: 2 evil players in games of 5 to 9, 3 in games of 10 to 12, and 4 in games of 13 to 15 (one Demon plus a few Minions). Almost every good player has an ability, and many have roles that bring in information every night: in effect, the “many Seers” of Theorem 4. By the theorem, with that many sources of information, the good team ought to win easily.

The designers' answer is to make information unreliable. The official rules say so plainly: a drunk or poisoned player actually has no ability, but the Storyteller pretends they do and keeps giving them information, which can be false, without their knowing it. Add that an execution needs a majority vote, that the dead can still talk and keep one vote, and that evil wins once only two players are left, and every one of these rules drags the good team back from being “many reliable Seers.”

The theorems mark the two ends: with no information, the wolves need only about the square root; with plenty of reliable information, wolves close to half the table still lose. What a designer does is find a point between the two where the game stays worth playing.

The paper ends with a conjecture. In the current model a day is followed by a night, so the villagers vote in only one round out of two. Change the rhythm, and the exponent in the threshold will probably change with it.

One day and one night means $d=1$ and $r=2$, which gives the square root. Yao's paper claims a proof of this conjecture, but the proof is only a sketch, and it rests on the lower bound from the previous section, the one that breaks down, so I still treat it as a conjecture.

Having worked out all these numbers, what strikes me most is this: in this model, numbers are not worth much; information is. A million villagers who trust no one can only manage a draw against 674 wolves who know each other, while one Seer who can hand out a cleared list lifts the threshold from the square root to a fixed fraction.

At the start of the paper, the authors compare problems of this kind to employees and shareholders in a company, students and administrators in a school, citizens and gangs in a city. The mathematics answers only the most stripped-down version; at a real table, everyone is talking, watching and reasoning.

So next time you are killed on the first night, don't feel too bad. At least you know that the wolves' little stack of chips has not lost a single cent, on average, from the start of the game until now.

A million people who trust no one
can only draw with 674 wolves.