m*i
2 楼
前院后院落叶铺了厚厚的一层,树太多了!怎么打扫收走落叶方便快捷呢?
thx!
thx!
m*n
3 楼
You can do repeated expectation:
let the expectation you want to be x;
fix the first person;
conditioning on the 2nd person has the same bday as the 1st person, you got
2;
conditioning on the 2nd person has a different bday as the 1st person, you
can condition on the 3rd person; if the 3rd person has the same bday as the
previous two people, you got 3;
otherwise keep going...
so x = 1/365*2+364/365*2/365*3+363/365*3/365*4+...
then x = 1/365* sum_{n = 1}^{364} n*(n+1)*(365-n+1)/365 and it is a power
summation
let the expectation you want to be x;
fix the first person;
conditioning on the 2nd person has the same bday as the 1st person, you got
2;
conditioning on the 2nd person has a different bday as the 1st person, you
can condition on the 3rd person; if the 3rd person has the same bday as the
previous two people, you got 3;
otherwise keep going...
so x = 1/365*2+364/365*2/365*3+363/365*3/365*4+...
then x = 1/365* sum_{n = 1}^{364} n*(n+1)*(365-n+1)/365 and it is a power
summation
z*p
5 楼
got
the
The solution seems to have a problem:
Look at the 4th person. The condition for the 4th person to have an
opportunity is that "the 2nd person failed and the 3rd person failed as well
". The probability for this to happen is
(364/365)*(363/365) instead of 363/365
【在 m***n 的大作中提到】
![](/moin_static193/solenoid/img/up.png)
: You can do repeated expectation:
: let the expectation you want to be x;
: fix the first person;
: conditioning on the 2nd person has the same bday as the 1st person, you got
: 2;
: conditioning on the 2nd person has a different bday as the 1st person, you
: can condition on the 3rd person; if the 3rd person has the same bday as the
: previous two people, you got 3;
: otherwise keep going...
: so x = 1/365*2+364/365*2/365*3+363/365*3/365*4+...
d*n
7 楼
Every person is Bernoulli distribution with various success probabilities.
Denote the i-th person has success probability Pi. And I assume we do not
count the last repeated b-day.
1st: P1 = 1
2nd: P2 = p(1st and 2nd have different b-days) = 1*(364/365)
3rd: P3 = p(1st, 2nd, and 3rd have different b-days) = p(3rd person
different b-day|1st and 2nd people different b-days)*p(1st and 2nd people
different b-days) = (363/365) *[1*(364/365)]
4th: P4 = p(4th person different b-day|1st, 2nd, and 3rd people different b-
days)*p(1st, 2nd, and 3rd people different b-days) = (362/365) *[1*(364/365)
*(363/365)]
...
Result: 1+1*(364/365)+1*(364/365)*(363/365)+1*(364/365)*(363/365)*(362/365)
+...
Denote the i-th person has success probability Pi. And I assume we do not
count the last repeated b-day.
1st: P1 = 1
2nd: P2 = p(1st and 2nd have different b-days) = 1*(364/365)
3rd: P3 = p(1st, 2nd, and 3rd have different b-days) = p(3rd person
different b-day|1st and 2nd people different b-days)*p(1st and 2nd people
different b-days) = (363/365) *[1*(364/365)]
4th: P4 = p(4th person different b-day|1st, 2nd, and 3rd people different b-
days)*p(1st, 2nd, and 3rd people different b-days) = (362/365) *[1*(364/365)
*(363/365)]
...
Result: 1+1*(364/365)+1*(364/365)*(363/365)+1*(364/365)*(363/365)*(362/365)
+...
s*n
9 楼
根据抽屉原理,不考虑闰年的话,最多有365个人没有相同的生日,到366的时候至少有
一对相同的生日。所以你列出的期望 应该是从第二个人到第366人 N×P(N)的和。
具体的计算好像可以用泰勒级数近似,但是这个Wiki里给出的结果只是 N个人里没有两
个人有相同生日的 CDF, 感觉上和计算从第几个人开始出现相同生日的期望不一定一
样。
http://en.wikipedia.org/wiki/Birthday_problem
b-
365)
【在 d*****n 的大作中提到】![](/moin_static193/solenoid/img/up.png)
: Every person is Bernoulli distribution with various success probabilities.
: Denote the i-th person has success probability Pi. And I assume we do not
: count the last repeated b-day.
: 1st: P1 = 1
: 2nd: P2 = p(1st and 2nd have different b-days) = 1*(364/365)
: 3rd: P3 = p(1st, 2nd, and 3rd have different b-days) = p(3rd person
: different b-day|1st and 2nd people different b-days)*p(1st and 2nd people
: different b-days) = (363/365) *[1*(364/365)]
: 4th: P4 = p(4th person different b-day|1st, 2nd, and 3rd people different b-
: days)*p(1st, 2nd, and 3rd people different b-days) = (362/365) *[1*(364/365)
一对相同的生日。所以你列出的期望 应该是从第二个人到第366人 N×P(N)的和。
具体的计算好像可以用泰勒级数近似,但是这个Wiki里给出的结果只是 N个人里没有两
个人有相同生日的 CDF, 感觉上和计算从第几个人开始出现相同生日的期望不一定一
样。
http://en.wikipedia.org/wiki/Birthday_problem
b-
365)
【在 d*****n 的大作中提到】
![](/moin_static193/solenoid/img/up.png)
: Every person is Bernoulli distribution with various success probabilities.
: Denote the i-th person has success probability Pi. And I assume we do not
: count the last repeated b-day.
: 1st: P1 = 1
: 2nd: P2 = p(1st and 2nd have different b-days) = 1*(364/365)
: 3rd: P3 = p(1st, 2nd, and 3rd have different b-days) = p(3rd person
: different b-day|1st and 2nd people different b-days)*p(1st and 2nd people
: different b-days) = (363/365) *[1*(364/365)]
: 4th: P4 = p(4th person different b-day|1st, 2nd, and 3rd people different b-
: days)*p(1st, 2nd, and 3rd people different b-days) = (362/365) *[1*(364/365)
a*a
15 楼
那个电动blower应该很管用
吹到一起,然后反向转,把树叶都打碎,然后直接放后院变成肥料
吹到一起,然后反向转,把树叶都打碎,然后直接放后院变成肥料
u*q
16 楼
Use a wide rake.
包。。。子。。。
包。。。子。。。
b*i
17 楼
见邻居用过lawn vacuum. 基本上大叶子全吸掉了。我自己没有,没用过。
c*o
19 楼
多的话,blow + hand job
少的话,直接割草机毛器掉
少的话,直接割草机毛器掉
d*e
22 楼
买了 个背着的吹叶子的 200多 还算挺好用
这个再加块布 差不多了
这个再加块布 差不多了
v*r
32 楼
我家有吹树叶的,吹出来的风也不小,可是不怎么吹得动啊,树叶老被草卡住。比起耙
子来也不轻松。我现在就大致耙一遍,再割下草就行了,可是很快就恢复原样,好累。
子来也不轻松。我现在就大致耙一遍,再割下草就行了,可是很快就恢复原样,好累。
d*a
34 楼
1】要看是什么叶子,如果是松树类的,必须用耙子先扒,没什么好办法。
2】如果是阔叶,可以用下图的blower,本地工具店,在11月中旬降雪前,或者city收
集树叶前,租一次也就几十块。
2】如果是阔叶,可以用下图的blower,本地工具店,在11月中旬降雪前,或者city收
集树叶前,租一次也就几十块。
d*a
35 楼
从一边,吹起,往路边吹。city 应该可以来收的。
相关阅读
已经提交485 (H1B) 可以申请homestead exemption吗?请推荐holmdel nj 一带的买房代理,谢谢!COSTCO换桃的极品中国人想起了小时候的事[出售] Southwest Airlines award机票2张,$120 for one, $230 (转载)请问Lowes的冬天种菜的塑料薄膜叫啥名字?谢谢!foreclosure疑问空调FILTER的选择我在想,如果能有会编程序的计算机吗?家里有咬人的小虫子急问:房子外面的woodtrim到底是stain还是paint?底楼地毯霉味很重,打算换地板了出租整个房子有没有可能加个条款:如果不付房租,自己同意放弃一(转载)恭喜欢哥, 居然没在这里发过包子...买Model Home求教Costco.com 有不少东东减价值不值得换热水器?EMF到底有没有安全标标准?inspection time欠操的节奏