为什么在ConcurrentHashMap中使用0.5作为泊松分布的参数？

Question

根据ConcurrentHashMap中的Java文档：

 * Ideally, the frequency of nodes in bins follows a Poisson distribution
 * (http://en.wikipedia.org/wiki/Poisson_distribution) with a
 * parameter of about 0.5 on average, given the resizing threshold
 * of 0.75, although with a large variance because of resizing
 * granularity. Ignoring variance, the expected occurrences of
 * list size k are (exp(-0.5) * pow(0.5, k) / factorial(k)). The
 * first values are:

问题是：0.5参数是怎么来的？

Answer 1

我认为我得到0.5的原因： 最大元素计数为0.75 * bin[].length，则将调整垃圾箱的大小。 因此，假设元素数为0.5 * bin[].length，则每个垃圾箱的平均元素数为0.5。