Y******u
发帖数: 1912 | 1 In a set of cereals, there are 4 kinds of coupons. A child wants to collect
all the 4 kinds of coupons. How many boxes in average should he buy?
n kinds of coupons? |
s********7
发帖数: 52 | |
p*b
发帖数: 30 | 3 Let k be the number of boxes the child buys.
Let Ai be the event that coupon i is not collected. Then P(Ai)=(1-1/n)^k
The probability that the child collects all coupons is
P((UAi)^c) = 1 -P(UAi) = 1-sum P(Ai)+ sum P(Ai,Aj) -....
= 1 - n(1-1/n)^k + n(n-1)/2 (1-2/n)^k - ...
To understand the problem, we can simplify n(n-1)...(n-i+1)/i! (1-i/n)^k by
n^i/i! exp(-ki/n)
Choose k=n(logn+c), then P((UAi)^c) can be approximated by 1-exp(-c)+exp(-2c
)/2! - ... = exp(-exp(-c))
collect
【在 Y******u 的大作中提到】
: In a set of cereals, there are 4 kinds of coupons. A child wants to collect
: all the 4 kinds of coupons. How many boxes in average should he buy?
: n kinds of coupons?
|
w*********i
发帖数: 77 | 4 4\times (\frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \frac{1}{4})
collect
【在 Y******u 的大作中提到】
: In a set of cereals, there are 4 kinds of coupons. A child wants to collect
: all the 4 kinds of coupons. How many boxes in average should he buy?
: n kinds of coupons?
|
B****n
发帖数: 11290 | 5 This is based on the assumption that the numbers of the four kinds of
coupons are the same.
【在 w*********i 的大作中提到】
: 4\times (\frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \frac{1}{4})
:
: collect
|
w*******x
发帖数: 489 | 6 1+4/3+4/2+4/1=8.3333
就是拿到一个不一样得长度为1
拿到第二个不一样的平均长度为4/3，etc。。
collect
【在 Y******u 的大作中提到】
: In a set of cereals, there are 4 kinds of coupons. A child wants to collect
: all the 4 kinds of coupons. How many boxes in average should he buy?
: n kinds of coupons?
|
s*****t
发帖数: 987 | 7
by
2c
re this
好像某本概率书上面有
【在 p*b 的大作中提到】
: Let k be the number of boxes the child buys.
: Let Ai be the event that coupon i is not collected. Then P(Ai)=(1-1/n)^k
: The probability that the child collects all coupons is
: P((UAi)^c) = 1 -P(UAi) = 1-sum P(Ai)+ sum P(Ai,Aj) -....
: = 1 - n(1-1/n)^k + n(n-1)/2 (1-2/n)^k - ...
: To understand the problem, we can simplify n(n-1)...(n-i+1)/i! (1-i/n)^k by
: n^i/i! exp(-ki/n)
: Choose k=n(logn+c), then P((UAi)^c) can be approximated by 1-exp(-c)+exp(-2c
: )/2! - ... = exp(-exp(-c))
:
|
B****n
发帖数: 11290 | 8 我覺得這問題最有意思的倒不是數學怎麼算
而是大家一開始就假定每種coupons的數目一樣 機率都是1/4
從簡化數學的角度這樣也沒錯
不過我從小到大就發現每次要收集什麼東西 幾乎總是有一兩個收集不到 我總是懷疑數
目真的是一樣的嗎
collect
【在 Y******u 的大作中提到】
: In a set of cereals, there are 4 kinds of coupons. A child wants to collect
: all the 4 kinds of coupons. How many boxes in average should he buy?
: n kinds of coupons?
|
s****i
发帖数: 216 | 9 X_i the length until the ith coupon,
then X_i-X_(i-1) are independent of each other, i guess |
