# (Un)fair Division by Consensus

There are many methods of dividing fairly in some sense, say, loot between robbers or inheritance between heirs to a fortune (e.g., Method of Lone Divider, Method of Markers, Method of Sealed Bids.) Here's one related problem whose solution may come as a surprise.

A gang of robbers are set to divide their loot. The main step in the division is that one of the robbers takes on himself to apportion the loot between all the gang members alive. After that the robbers vote whether to accept or reject the division. The division only passes by the majority vote. In case of rejection, the responsible fellow gets nothing and does not participate in further attempts to divide the loot. (In some versions of the problem the fellow is physically eliminated to satisfy the non participation condition.) An additional condition is that the robbers within the gang are subject to a strict hierarchy, and the apportionment is always carried on by the highest ranking still participating member.

The question is "How the process of the division will be carried on and what will be the final partition?"

For definiteness sake, let the loot consist of 100 coins of equal worth and let there be seven robbers in the decreasing order of importance: A, B, C, D, E, F, G.

### References

- Haim Shapira,
*Conversations on Game Theory*, Kinneret, Zmora-Bitan, Dvir - Publishing House Ltd., 2008 (in Hebrew)

|Contact| |Front page| |Contents| |Algebra|

Copyright © 1996-2018 Alexander Bogomolny

### Solution

It's easier to start with smaller number of robbers. If there is just one, he walks away with the whole loot. That the problem is not quite trivial becomes apparent alreay when there are just two robbers: one divides, but the other's view on the partition is crucial because he may block any division that it is not to his taste. For, if the division fails, he takes it all. For this reason the only passable division for the ranking robber is to give the other guy the whole of it and take nothing to himself (still he stays alive.)

If there are three of them: G, F, E, with E dividing, there is no way E may satisfy G because the latter's negative vote is a certainty unless he gets the ehole lot. F, on the other hand, would be satisfied with anything, however little, because, if the division fails he will be forced to get nothing. It follows that E can safely divide the loot as follows G(0), F(1), E(99).

If that is D's turn to try his math skills, he knows there is no way to satisfy E. To get a concensus he should provide G and F with enough impetus for a "yea" vote. The following division satisfies the minimum condition to ensure just that: G(1), F(2), E(0), D(97).

Based on this strategy, the further possibilities are, depending on the number of robbers:

G(2), F(0), E(1), D(0), C(97).

G(0), F(1), E(2), D(1), C(0), B(96).

G(1), F(0), E(0), D(2), C(1), B(0), A(95), where A randomly chooses between D and F. It may be thought that A should prefer another division, e.g., G(1), F(1), E(0), D(1), C(0), B(0), A(96). In this case, though, D and F votes could not be taken for granted as each may be expecting to get paid for the support of A's division.

|Contact| |Front page| |Contents| |Algebra|

Copyright © 1996-2018 Alexander Bogomolny67005393