d******g
发帖数: 34 | 1 Given the fact that boys and girls are born with equal probability and you
want a girl, you will stop making babies until you have a girl. What are the
expected number of born boys and girls? | r*******s
发帖数: 303 | 2 1,1
跟那个扔硬币是一回事。扔到出现一个head的要平均2次，两个head同时出现是6次。..
.. | m*******s
发帖数: 758 | 3 如果连续生两个男的的停止呢？
突然想起可以构造一个题目及其拓展
throw fair coins, T is the first time to have Head,
what is E[1/T]?
..
【在 r*******s 的大作中提到】
: 1,1
: 跟那个扔硬币是一回事。扔到出现一个head的要平均2次，两个head同时出现是6次。..
: ..
| p*****k
发帖数: 318 | 4
one H is easy, the sum is exactly the taylor expansion of
-(p/q)*log(1-q), where p (or q) is the prob of getting H (or T).
for a fair coin, E[1/T] = log(2) ~ 0.69
case of two H's is a little messy. for a fair coin, i got:
E[1/T] = log(2) + log([sqrt(5)-1]/[sqrt(5)+1])/sqrt(5) ~ 0.26
anyone interested could help checking with monte-carlo
【在 m*******s 的大作中提到】
: 如果连续生两个男的的停止呢？
: 突然想起可以构造一个题目及其拓展
: throw fair coins, T is the first time to have Head,
: what is E[1/T]?
:
: ..
| m*******s
发帖数: 758 | 5 从哪搞出来的 log那个式子
我想问题的背景是， 生男=1，生女=0；每次只生一个，
生一个男就停止，停时生了T个娃，那么停时的"平均"性别=1/T。
【在 p*****k 的大作中提到】
:
: one H is easy, the sum is exactly the taylor expansion of
: -(p/q)*log(1-q), where p (or q) is the prob of getting H (or T).
: for a fair coin, E[1/T] = log(2) ~ 0.69
: case of two H's is a little messy. for a fair coin, i got:
: E[1/T] = log(2) + log([sqrt(5)-1]/[sqrt(5)+1])/sqrt(5) ~ 0.26
: anyone interested could help checking with monte-carlo
| p*****k
发帖数: 318 | 6 hmm, this is misleading. mathematically the expected "ratio" is a
well-defined question, but it does not really have a physical meaning.
let's say there are N (->infty) families, would you measure the ratio
of boy vs. girl for each family, then average? no. the sensible thing
to do is to get a weighted average of the ratio by the # of children
each family has. this is exactly what was asked in the original
problem: the averaged # of boys (and girls). | v********l
发帖数: 7 | 7 Suppose you will have 2n babies in the end, when n->infinity, you will have
n boys because the probability for boy is 1/2, name them b1,b2,...bn
Let Xi=1 when Bi is born before you have a girl, 0 otherwise.
So E[Xi]=1/(n+1) (because have a boy's probability before any girl is to
choose 1 from 1 boy and n girls)
and X is the total boys were born(before you got a girl)
so X=X1+X2+...Xn
E[X]=n*E[X1]=n/(1+n)
when n goes to infinity. E[x]=1.
So the expected boy is 1. |
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