There is a set of n indivisible items (or chores), and a set of n players. Each day, a single item should be assigned to each player. We want to ensure that all players feel that they have been treated fairly, not only after the last day, but after every single day. We present two 'balance' conditions on sequences of permutations. One condition can always be satisfied, but is arguably too weak; a second condition is strong, and can be satisfied for all n <= 11, but cannot be satisfied for some larger values of n, including all n>61. We then relate the 'balance' condition to the requirement that the cumulative assignment is proportional up to one item (PROP1), where proportionality holds in a strong ordinal sense -- for every valuations that are consistent with the item ranking. We present a third balance condition that implies ordinal PROP1. We show that a sequence guaranteeing this balance condition exists for all n <= 12, but might not exist when n=6k for any k >= 19. Finally, we present a fourth, weaker balance condition on a sequence, that guarantees ordinal proportionality up to two items (PROP2). Whether or not this condition can be satisfied for all n remains an open question.

翻译：存在一个包含n个不可分割物品（或杂务）的集合，以及一个包含n个参与者的集合。每天，每个参与者应被分配一个物品。我们希望确保所有参与者不仅在整个过程结束后，而且在每一天结束后都感到自己受到了公平对待。我们提出了关于置换序列的两种“平衡”条件。第一种条件总是可以满足，但可能被认为过于宽松；第二种条件较强，对于所有n≤11的情况均可满足，但对于某些更大的n值（包括所有n>61）则无法满足。接着，我们将“平衡”条件与累积分配满足“至多一件物品差异的比例公平性”（PROP1）的要求联系起来，其中比例公平性在强序数意义上成立——适用于所有与物品排序一致的估值函数。我们提出了第三种平衡条件，该条件蕴含序数PROP1。我们证明对于所有n≤12，存在保证此平衡条件的序列，但当n=6k（k≥19）时可能不存在。最后，我们提出了关于序列的第四种较弱平衡条件，该条件保证序数意义上的“至多两件物品差异的比例公平性”（PROP2）。该条件是否对所有n都能满足仍是一个开放问题。