I believe this boils down to the probability that the first bucket holds H-1 balls (because your probability is really the probability that the bucket you pick to drop a ball into has H-1 balls. I'm guessing this should be solvable with combinatorics, then, but that is not my strong point.
As a side note: this is not a statistics problem, but a probability problem.
if you could afford to store for each brigade the number n[h] of units with h hits for each possible h, then the problem becomes straighforward: with probability n[h]/N you select a unit with h hits, and then increment n[h+1] and decrement n[h], or if you've selected h=max-1 you decrement n[h] and N.
If you can't afford the extra memory, a reasonable and tractable choice would be the maximum entropy distribution, see here for example
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我相信这可以归结为第一个桶装有 H-1 个球的概率（因为你的概率实际上是你选择放入球的桶有 H-1 个球的概率。
我猜这应该可以用组合学解决，但这不是我的强项。
附带说明：这不是统计问题，而是概率问题。
I believe this boils down to the probability that the first bucket holds H-1 balls (because your probability is really the probability that the bucket you pick to drop a ball into has H-1 balls.
I'm guessing this should be solvable with combinatorics, then, but that is not my strong point.
As a side note: this is not a statistics problem, but a probability problem.
如果您有能力为每个旅存储每个可能的 h 命中次数为 h 的单位数量 n[h]，那么问题就变得很简单：以 n[h]/N 的概率选择一个命中次数为 h 的单位，然后递增n[h+1] 并递减 n[h]，或者如果您选择了 h=max-1，则递减 n[h] 和 N。
如果您负担不起额外的内存，一个合理且易于处理的选择是最大熵分布，请参阅此处例如
if you could afford to store for each brigade the number n[h] of units with h hits for each possible h, then the problem becomes straighforward: with probability n[h]/N you select a unit with h hits, and then increment n[h+1] and decrement n[h], or if you've selected h=max-1 you decrement n[h] and N.
If you can't afford the extra memory, a reasonable and tractable choice would be the maximum entropy distribution, see here for example