l*******l
发帖数: 248 | 1 You throw a fair coin one million times. What is the probability of getting
six heads after six tails some time during this experiment | e******0
发帖数: 211 | 2 Posion分布近似？
6H6T 时间 A
Ai 与 Aj 独立,
总共1M-12个事件.
P(A)的概率(1/2)^12.
二项式的Poisson 近似
getting
【在 l*******l 的大作中提到】
: You throw a fair coin one million times. What is the probability of getting
: six heads after six tails some time during this experiment
| m*******t
发帖数: 6 | 3 说真的，您能不能买本红宝书和绿宝书，也没几个钱，以后找到工作全回来了。这是红
宝书上的原题，
答案我都还记得，（1000,000-11）/2^12.
getting
【在 l*******l 的大作中提到】
: You throw a fair coin one million times. What is the probability of getting
: six heads after six tails some time during this experiment
| m******e
发帖数: 45 | 4 LZ is asking for probability of at least one occurrance, your answer sounds
like expected number of occurance.
【在 m*******t 的大作中提到】
: 说真的，您能不能买本红宝书和绿宝书，也没几个钱，以后找到工作全回来了。这是红
: 宝书上的原题，
: 答案我都还记得，（1000,000-11）/2^12.
:
: getting
| p*****k
发帖数: 318 | 5 latingirl, i guess it's 6 consecutive T's followed by 6 consecutive H's, i.e.,
the probability of the pattern "TTTTTTHHHHHH" ever occurring in 10^6 tosses?
if so, you could consider the dual probability, i.e., the pattern never
appears within total of n tosses: p(n), for which the recurrence is
p(n+12)=p(n+11)-p(n)/2^12
with p(0)=p(1)=...=p(11)=1.
standard technique applies. some details and extension could be found here:
http://www.wilmott.com/messageview.cfm?catid=26&threadid=74043
all you need is 1-p(10^6), which would be extremely close to 1 | l*******l
发帖数: 248 | 6 re
sounds
【在 m******e 的大作中提到】
: LZ is asking for probability of at least one occurrance, your answer sounds
: like expected number of occurance.
| l*******l
发帖数: 248 | 7 谢谢你！
when n is sufficiently large, one just needs the root with the largest norm,
so the probability is ~1.4*(0.92)^n.
这个1.4和0.92是怎么来的？我没看懂。
.e.,
tosses?
here:
【在 p*****k 的大作中提到】
: latingirl, i guess it's 6 consecutive T's followed by 6 consecutive H's, i.e.,
: the probability of the pattern "TTTTTTHHHHHH" ever occurring in 10^6 tosses?
: if so, you could consider the dual probability, i.e., the pattern never
: appears within total of n tosses: p(n), for which the recurrence is
: p(n+12)=p(n+11)-p(n)/2^12
: with p(0)=p(1)=...=p(11)=1.
: standard technique applies. some details and extension could be found here:
: http://www.wilmott.com/messageview.cfm?catid=26&threadid=74043
: all you need is 1-p(10^6), which would be extremely close to 1
| c**********e
发帖数: 2007 | | c**********e
发帖数: 2007 | 9 Suppose the solutions of the moment generating function are t_1, t_2,...,t_4
(for the Wilmott one, the order is 4). As 1/2 is one of its root, and its 2
roots are not real, it is easy to numerically get t_1=0.92...
As indicated in Wilmott, p(n) can thus be written as a1*(x1^n)+a2*(x2^n)+a3*
(x3^n)+a4*(x4^n), where a's are determined by the initial conditions. one
can get a_i=1+(3/4)/(4*x_i^3-1) by inverting a Vandermonde matrix.
Here we have a order 12 moment generating function. But the approach is
similar.
norm,
【在 l*******l 的大作中提到】
: 谢谢你！
: when n is sufficiently large, one just needs the root with the largest norm,
: so the probability is ~1.4*(0.92)^n.
: 这个1.4和0.92是怎么来的？我没看懂。
:
: .e.,
: tosses?
: here:
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