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：存在一组n个不可分割的物品（或杂务）和一组n个参与者。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认为在最后一天得到了公平对待，而且在每一天之后都如此。我们提出了关于置换序列的两个“平衡”条件。第一个条件总是可以满足，但可能过于宽松；第二个条件较强，对于所有n≤11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——对于所有与物品排序一致的估值函数均成立。我们提出了第三个平衡条件，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个较弱的平衡条件，该条件能保证至多两件物品的序数比例公平性（PROP2）。此条件是否对所有n均可满足仍是一个开放问题。