1. 为什么谈判很重要？
答：In life, you don't get what you deserve. Instead, you get what you negotiate.
2. 两个人合作的基础是什么？
答：合作的基础（pie）是来自于两人合作的利润-A单独做的利润-B单独做的利润。如果pie代表着节省（即负的）或者如果pie代表着创造了更多利润（即正的），那么就有可能合作的基础。
举一个增益的例子。例如A不合作可以创造1个单位，B不合作可以创造2个单位，如果他们合作就可以创造9个单位，此时pie是什么？应该怎么划分？pie=9-1-2=6，应该是针对6平分，即A拿到1+3=4，B拿到2+3=4。为什么需要对pie进行平分，因为双方都需要对方，离开了对方都不能拿到因为合作而带来的新收益。
再举个节省的例子，如果双方合作能省下1000元，则可以基于fairness（平等原则）来对节省的1000元进行平等划分。为什么能平分？原因是谁也离不开谁，离开了对方，就无法节省这1000元。
继续举个节省的例子，如果A搭车回家需要花150元，B搭地铁回家需要40元，此时如果A和B一起搭车，就可以节省40元。则pie为40元，但是如果B愿意花费最多100元来搭车，则pie可以提升为100元。心理价位和实际价位是不一样的。
对于实际生活中的例子，是有成本计算的，例如A和B一起合作，A单方面有收益100元，B单方面有收益200元，合作的成本是共同share一个50元成本的软件；如果不合作，则需要各自买50元的软件。此时该如何计算合作的时候50元的成本该如何平摊？我们应该先计算合作的利润和分开的利润，合作的利润为200+100-50=250，A单独做的话收益为100-50=50，B单独做的话收益为200-50=150，此时pie=250-50-150=50，如果平分pie的话，则A拿到的收益为25，其支出的成本为50-25=25，B拿到的收益为25，其支出的成本为50-25=25.
3. 现实生活中合作的难点是什么？
答：不知道pie是什么，或者双方对于pie是什么不能达成统一意见，或者单方面隐藏部分pie。除此之外，对于如何平分pie也存在着分歧。
4. 计算pie的基本原则是什么？
答：穷尽双方的option，合作和不合作带来的收益。
5. 古代法律中对于pie（其实是有争议的部分）的分配是如何的？（the Principle of Divided Cloth）
- 在古希腊法律中，如果A申明要100%的一块布，而B申明要50%的一块布。则无争议的是另外50%的布，属于A，有争议的50%则会被平分，所以A拿到75%，B拿到了25%。（这种方式的竞品是相对于直觉上应该按照比例来分，propotional的算法并不会把各自不合作而应得的部分算进去）
- 如果一个人欠了A公司100元，B公司50元，而欠款人身上只有100元，则按照pie的方式其中50元没有争议，只有另外的50元有争议。所以A拿到了75元，B拿到了25元。
6. 谈判的流程是什么？
- 先发表申明，为什么我要拿到这么多（风险平摊，时间，金钱，距离，舒适度，空间大小上的收益）
- 选择对自己有利的算法（proportional或者divided cloth），如果是占比例比较大的，则可以选择proportional来做，否则的就按照平分pie来做
- 再来分配pie
7. 如果是多人合作的话，有哪些算法可以平摊费用？
- 方法A的好处：容易理解，不需要知道每个人都花多少钱去做事情，很多时候也不知道对方需要单独花多少钱去做事情。可以按照哪些路段/物品是三人一起共用，则一起平摊那部分的钱，省下的部分如果还有两人一起共用，则再一起平摊那部分的钱；这种方式对于用的最少的人来说最有利，但是对于用的最多的人来说，不一定公平，例如时间上可能用的最多的人时间花的更多了。
- 方法B（Shapley Value）的好处：对于用的最多的人来说价格会优惠一些；还有很多时候并不是完全知道什么物品或者路线是会共用的；核心思想是大家的pie都要一样。可以按照方法A来算，然后对于detour的部分进行分摊。可以按照pie的算法来计算，但是多人的情况在平分pie的时候，简单的情况可以选择平分pie，对于用的最少的人而言还是最赚的，所以此时可以选择random ordering，先算出指定情况下合作的总费用是多少，然后再根据不同情况下的ordering来计算每次每方需要花费多少钱，最后再来计算平均值。这种算法有个好处，因为各方的加入会使得各方都具有贡献值，有些人的加入会贡献大，有些人的加入会贡献小一些，有些人的加入会产生负作用。
- 方法C（Nucleolus）的好处：对于用的最多的人来说价格会优惠一些；理解上也比方法B容易一些，核心思想是大家的gain（而不仅仅是pie）都要一样多，这样才公平。
I want to be careful that I don’t leave you with the impression that the Shapley Value is the only fair way to divide the pie. I think it has many desirable properties, but it isn’t the only option. Here I’ll explain one other approach, called the Nucleolus (like the largest structure inside the nucleus of a cell), which was developed by David Schmeidler in 1969.
To explain how the Nucleolus works, let’s take a slightly different version of the Runway Problem. We’ll have three airlines as before.
A needs a runway of length 12
B needs a runway of length 18
C needs a runway of length 18
Under the Shapley Value approach, we’d say the three airlines would split the cost of the first length three ways and then B and C, as the only users of the next half length, would share that cost equally. Thus,
A pays 4
B pays 4 + 3 = 7
C pays 4 + 3 = 7
Before turning to the Nucleolus, we can determine a few properties that any fair solution should obey.
For starters, since A is using less of the runway than B or C, whatever amount A pays, it should be no more than what B or C pays.
From this, it follows that B and C should pay the exact same amount, as B uses no more than C and C uses no more than B.
Combining these two properties, we can conclude the most A should ever pay is 6. Since B and C must pay at least as much as A, once A pays 6 so must B and C, and together that covers the full cost of the runway.
The most A should pay is 6. What is the least? Well, it wouldn’t be fair for A to pay less than 4, as that is its equal share of the cost of the runway it uses. Therefore, A should pay something between 4 and 6. The Shapley Value has A paying 4. The Nucleolus makes the argument for why A should pay 6. It has to do with how much each side gains when one individual (or group) joins another.
Imagine B and C have already formed a partnership. In that case, the two of them have already saved 18 by coming together. If A joins them, the three-way partnership will create another 12 of savings. Under the Shapley Value, A pays 4 and thus saves 8, while the (B, C) partnership only saves 4. Thus Airline A gains double what the (B, C) partnership gets. This seems unfair. A needs the (B, C) group just as much as the (B, C) pair needs A. The Nucleolus proposes this gain be split evenly, namely 6 and 6, which requires A to pay 6.
You might be wondering why I picked the combination of A joining (B, C). Why not look at B joining an (A, C) partnership? Indeed, the Nucleolus looks at all the possible combinations. When B joins (A, C), there is a gain of 18. Using the Shapley Value approach, B would pay 7 and save 11, while (A, C) would only save 7. This isn’t equal, either. To make this equal, we’d have to have B pay 9. By symmetry, C would have to pay 9. But that creates a bigger issue as then A would pay nothing. We’ve made the asymmetry even worse than in the case where A joins (B, C). If A pays nothing, then all of the gain goes to A and none to (B, C).
What the Nucleolus does is finds the division that maximizes the smallest gain. And once that is done, it maximizes the next smallest gain subject to not lowering the smallest gain. It isn’t always possible to make all the gains equal and when it isn’t, the Nucleolus comes as close as possible.
Here’s what I mean. Recall that when A joins (B, C), there is only 12 to go around. If we split this evenly, A saves 6 which implies it pays 6. And since B and C must not pay less than A, they pay 6 as well. So, we have a cost division where A, B, and C all pay 6.
Let’s see how the gains are split in each combination of an individual joining a pair. We know that the gains from A joining (B, C) are split evenly. When B joins (A, C), since B pays 6, it gains 12 while (A, C) gains 6. And the same is true for C when it joins (A, B). We might like to increase how much (A, C) or (A, B) gains as the divisions are lopsided toward the other party, but to do so would require that B pays more and, by symmetry, C pays more. If B and C were to each pay more than 6, then the (B, C) pair would gain less than 6 when A joins them. Thus we can’t make the result for when B joins (A, C) more fair without making the A joins (B, C) scenario even less fair.
John Rawls argued that society should work to maximize the welfare of its worst-off members. In a similar vein, the Nucleolus looks to find the cost division that maximizes the gain to the group that is getting the least from coming together. And it keeps on doing that to the extent possible.
Let me provide a few more examples.
A needs a runway of length 12
B needs a runway of length 24
C needs a runway of length 24
Under the Shapley Value approach, we’d say that all three airlines would split the cost of the first length three ways and then B and C, as the only users of the next length, would share that cost equally. Therefore,
A pays 4
B pays 4 + 6 = 10
C pays 4 + 6 = 10
But under this cost division, when A joins (B, C), A gains 8 and (B, C) only gains 4. To equalize this, we should have A pay 6. Thus the Nucleolus solution is:
A pays 6
B pays 9
C pays 9
Unlike our first example, it is not the case here that all three parties split the cost evenly. Indeed, as the runway needed by Airlines B and C increases, they pay all of the additional costs. A never pays more than half the cost of the first length.
And to the extent that the runway needs of B and C decrease (from their starting point of 18), the three airlines will continue to split the full cost three ways. For example, with the numbers below, A, B, and C would each pay 5 under the Nucleolus.
A needs a runway of length 12
B needs a runway of length 15
C needs a runway of length 15
Still wondering why it’s fair for A to be paying an equal share of the full runway cost? The reason is B and C can together create a large amount of savings without A. Basically, A is very lucky to be joining (B, C) and should be happy to split the surplus created when it joins this pair.
This becomes even clearer if we add additional airlines that need the longer runway. Consider this case:
A needs a runway of length 12
B needs a runway of length 16
C needs a runway of length 16
D needs a runway of length 16
Under the Shapley Value, A pays 3, while under the Nucleolus, A pays 4. Either way, A gets a great deal in joining the (B, C, D) trio. A gains either 9 or 8 while the trio gains 3 or 4. To make the gains as equal as possible, A should pay 4. (A can’t pay more than 4 as that would mean it pays more than B, C, and D.)
For our final example, we’ll forgo airport runways for estate settlements. The Babylonian Talmud provides an illustration for how to divide up an estate in the face of competing claims. Surprisingly (for a 2,000 year old book), the unusual solution proposed is the same as the Nucleolus.
Here's the set up. A person dies owing money to three creditors. They are owed 100, 200, and 300 respectively. These debts are larger than the assets of the estate, so they can't all be paid off. The table below shows how the assets should be allocated, according to the Talmud. In the event the estate only has 100 to pay out, each of the three creditors is paid 33 1/3. If the estate has 200, A is paid 50, while B and C receive 75. If the estate has 300, A is paid 50, B is paid 100, and C is paid 150.
It seems like there are three different approaches being taken. The first row suggests equal treatment, the third row suggests proportional treatment, and the middle row just looks weird.
| Estate | Creditor A (100) | Creditor B (200) | Creditor C (300) |
| 100 | 33 1/3 | 33 1/3 | 33 1/3 |
| 200 | 50 | 75 | 75 |
| 300 | 50 | 100 | 150 |
As it turns out, all three rows are consistent with the Nucleolus.
First, look at the case where there is 300 to divide. If (B, C) get together, they can divide up 200 without any agreement from A. That is because (B, C) could write A a check for 100 and A would have no more claim. Thus (B, C) can get 200 on their own, but they need to bring A onboard to divide up that last 100. If they divide it evenly, then A’s gain from joining the group will be 50, the same as the gain to (B, C). You can check that this is the smallest amount anyone or pair gets in forming a threesome. For example, (A, B) on their own can get nothing as if they pay C off, that leaves nothing for themselves. So if (A, B) want to get anything, they need to bring C into the group. Similarly, C can get nothing on its own as paying off A and B would use up all the assets. Thus bringing C together with (A, B) creates 300 to share. In the proposed solution, C gets 150 and (A, B) get 150, both of which are much bigger than 50.
The middle row looks more confusing. Here (B, C) can get 100 of the 200 without an agreement from A (again by writing a check to A). Thus, just as in the last row, the addition of A is worth 100. If A gets 50 and (B, C) gets 50, then the two gains from joining are equal and they are also tied for the lowest amount any party gets for joining another. For example, (A, C) can get nothing on its own, so there is then 200 to split up when B joins (A,C). Here B gets 75 and (A, C) gets 125. To equalize this, B would have to get more and either A or C get less. But A can't get less as it is only getting 50 when joining (B, C) and that is already the minimum. If C were to get more than 75, then when C joins (A, B) that means B would have to get less than 75 (as A is getting 50)—but then B would be getting less for joining (A, C) than C would be getting.
If we go to the first row, no pair can get anything on their own. Even (B, C) without A gets nothing as absent an agreement there is no ability to pay A his claim of 100 and have anything left over. Thus if the payouts were anything but equal, then whoever got the lowest payout would get less for joining the other pair. Only when the payouts are equalized do we maximize the smallest payout.
If you are wondering how the Shapley Value would work in the Talmud case, just imagine that the three creditors get in line in a random order. Each is paid off until the money runs out. Thus if the order is (A, B, C) and the amount is 200, then A gets 100 for being first in line. There's still 100 left and B, as next in line, gets all of it leaving nothing for C. If the order is (C, A, B) then C gets 200 while A and B each get nothing. And so on. The Shapley Value is the average of all these possibilities.
There's one other way of explaining the Nucleolus solution. Recall our discussion of the Principle of the Divided Cloth. The Nucleolus has the attractive feature that if we take away one of the players and the amount that person gets then the other two players divide what's left according to the Principle of the Divided Cloth. Consider, for example, the middle row. If we take away A who is paid 50, then B and C have 150 to divide up. since they are claiming 200 and 300 respectively, each is claiming the entire cloth and so each is given half or 75. If we remove B with his 75 payment, that leaves 125 for A and C to divide. A is only claiming 100 which concedes 25 to C; meanwhile C claims 300 conceding nothing to A. Thus 100 is in dispute, and this is split 50/50. So A gets 50 and C gets 50 plus the conceded 25 for 75 in total.
I don’t want to pretend this is a rigorous explanation of the Nucleolus, but I hope I've given you a brief taste of why it is a very reasonable alternative to the Shapley Value. For more information about the Nucleolus, have a look at the original article by David Schmeidler and an elegant application of the Nucleolus to our airport cost-sharing problem by S.C. Littlechild. The application of bargaining theory to problems from the Talmud starts with Barry O'Neill in his 1982 paper; the connection to the Nucleolus was developed by Nobel Prize winner Robert Aumann and Michael Maschler.
Further Reading
Littlechild, S.C. “A Simple Expression for the Nucleolus in a Special Case.” *Int. Journal of Game Theory *3 (1974): 21–29. [LINK = ]
Schmeidler, David. “The Nucleolus of a Characteristic Function Game.” SIAM Journal on Applied Mathematics 17 (1969): 1163–1170.
O'Neill, Barry. “A Problem of Rights Arbitration from the Talmud.” Mathematical Social Sciences 2 (1982): 345-371. [LINK = ]
Aumann, Robert J. and Michael Maschler. “Game Theoretic Analysis of a Bankruptcy Problem from the Talmud.” Journal of Economic Theory 36 (1985): 195-213.
