并不是真正的完整答案(也不是有用的参考),而只是一个扩展的注释。对于任何给定的垃圾箱,在垃圾箱中恰好有球的概率将由p B = ( mB。我们可以使用因桑多((b+1)apB=(mB)(1n)B(n−1n)m−B,得到pB<((r+1)r+1((b+1)aa)<((b+1)b+1bb)a,其中r=mpB<((r+1)r+1rr)B(1n)B(n−1n)m−B。请注意,由于((b+1)ar=mB−1。((b+1)aa)>14ab((b+1)b+1bb)a
因此,我们有。现在,既然你有兴趣在寻找的概率乙中我们可以考虑一个垃圾桶或更多的球p ≥ 乙 = Σ 米b = 乙 p b <pB<eB(r+1)ln(r+1)−Brlnr−mlnn+(m−B)ln(n−1)Bp≥B=∑mb=Bpb<∑mb=Beb(r+1)ln(r+1)−brlnr−mlnn+(m−b)ln(n−1). Rearranging the terms, we get
p≥B<e−mlnnn−1×eB(r+1)ln(r+1)−Brlnr−Bln(n−1)∑b=0m−Beb(r+1)ln(r+1)−brlnr−bln(n−1).
Note the summation above is merely a geometric series, so we can simplify this to give
p≥B<e−mlnnn−1×eB(r+1)ln(r+1)−Brlnr−Bln(n−1)×1−((r+1)r+1rr(n−1))m−B+11−((r+1)r+1rr(n−1)).
If we rewrite
(r+1)r+1rr(n−1) terms using exponentials, we get
p≥B<e−mlnnn−1×eB(r+1)ln(r+1)−Brlnr−Bln(n−1)×1−(e(r+1)ln(r+1)−rlnr−ln(n−1))m−B+11−e(r+1)ln(r+1)−rlnr−ln(n−1),
which then becomes
p≥B<e−mlnnn−1×(eB((r+1)ln(r+1)−rlnr−ln(n−1))−e(m+1)((r+1)ln(r+1)−rlnr−ln(n−1)))1−e(r+1)ln(r+1)−rlnr−ln(n−1).
Now, I take it you care about finding some B such that p≥B<Cn for some constant C, since this gives the total probability of any bin having B or more balls as bounded from above by C. This criteria is satisfied by taking
e−mlnnn−1×(eB((r+1)ln(r+1)−rlnr−ln(n−1))−e(m+1)((r+1)ln(r+1)−rlnr−ln(n−1)))1−e(r+1)ln(r+1)−rlnr−ln(n−1)=Cn,
which can be rewritten as
B=ln(Cnemlnnn−1(1−e(r+1)ln(r+1)−rlnr−ln(n−1))+e(m+1)((r+1)ln(r+1)−rlnr−ln(n−1)))(r+1)ln(r+1)−rlnr−ln(n−1).
I'm not entirely sure how useful this comment will be to you (it's entirely possible I've made a mistake somewhere), but hopefully it can be of some use.