x****r
发帖数: 129 | 1 如果2222old，别peng我
扔硬币，第一次得到连续三个相同结果（三个正，或三个反）时扔了N次。N期望值是？ |
p*****k
发帖数: 318 | |
h*****0
发帖数: 4889 | 3 能详细说一下为啥是half吗？
【在 p*****k 的大作中提到】
: some relevant discussion here:
: http://www.mitbbs.com/article_t/Quant/31218015.html
|
p*****k
发帖数: 318 | 4 one way to see this is:
by symmetry, equal prob. that "HHH" appears before/after "TTT".
note the expectation operator is linear:
E["TTT"]=E[N]+(1/2)*E["TTT" after "HHH"],
where both E["TTT"]=14 (by the Markov property). |
h*****0
发帖数: 4889 | 5 how do you get from 'linear':
"TTT" = N + (1/2) {"TTT" after "HHH"}
【在 p*****k 的大作中提到】
: one way to see this is:
: by symmetry, equal prob. that "HHH" appears before/after "TTT".
: note the expectation operator is linear:
: E["TTT"]=E[N]+(1/2)*E["TTT" after "HHH"],
: where both E["TTT"]=14 (by the Markov property).
|
p*****k
发帖数: 318 | 6 sorry if the "linearity" i mentioned caused confusion.
# of tosses to get "TTT"
= # of tosses to get "TTT" or "HHH", whichever appears first
+ # of tosses from either pattern to "TTT"
then take average on both sides.
the last term is 0 if "TTT" appears first; and is same as E["TTT"]
if "HHH" appears first. |
h*****0
发帖数: 4889 | 7 so the last term is actually I_{"HHH" appears first} * N_{"TTT"}, where I_{a
} is defined to be 1 if and only if a is true.
【在 p*****k 的大作中提到】
: sorry if the "linearity" i mentioned caused confusion.
: # of tosses to get "TTT"
: = # of tosses to get "TTT" or "HHH", whichever appears first
: + # of tosses from either pattern to "TTT"
: then take average on both sides.
: the last term is 0 if "TTT" appears first; and is same as E["TTT"]
: if "HHH" appears first.
|
B********h
发帖数: 59 | 8 solve this.
S=(1/2)(S+1) + (1/2) (3/2 + (1/2)(S+2))
=> S=7 |
