s******e
发帖数: 128 | 1 发现自己数学能力大大退化了。
向大家请教一个概率题：
有24个连续的信道
每个信道空闲的概率是0.5
在这24个信道里至少找到4个连续空闲信道的概率是多大？
例如信道排列为1，2，3，...， 24
如果3，4，5，6信道是空闲，就是一组4个连续空闲信道
多谢指教！ | s**e
发帖数: 1834 | 2 Let f(n) be the probability that there are 4 consecutive empty spots in n
consecutive spots.
So
(1) f(1)=f(2)=f(3)=0, and f(4) = 1/16.
(2) f(n+4)=1/2*f(n+3)+1/4*f(n+2)+1/8*f(n+1)+1/16*f(n)+1/16
You can compute f(24) by computing f(5),f(6),..., one by one.
【在 s******e 的大作中提到】
: 发现自己数学能力大大退化了。
: 向大家请教一个概率题：
: 有24个连续的信道
: 每个信道空闲的概率是0.5
: 在这24个信道里至少找到4个连续空闲信道的概率是多大？
: 例如信道排列为1，2，3，...， 24
: 如果3，4，5，6信道是空闲，就是一组4个连续空闲信道
: 多谢指教！
| s******e
发帖数: 128 | 3 Thanks!
You are so brilliant!
n
【在 s**e 的大作中提到】
: Let f(n) be the probability that there are 4 consecutive empty spots in n
: consecutive spots.
: So
: (1) f(1)=f(2)=f(3)=0, and f(4) = 1/16.
: (2) f(n+4)=1/2*f(n+3)+1/4*f(n+2)+1/8*f(n+1)+1/16*f(n)+1/16
: You can compute f(24) by computing f(5),f(6),..., one by one.
| s******e
发帖数: 128 | 4 An extension of the problem:
What is the probability of 2 set of 2 continuous channels (e.g. 3-4 and
11-12) in 24 continuous channels, assuming the channel idle probability is
1/2?
Can the following solution be extended to the above extension?
Thanks again!
n
【在 s**e 的大作中提到】
: Let f(n) be the probability that there are 4 consecutive empty spots in n
: consecutive spots.
: So
: (1) f(1)=f(2)=f(3)=0, and f(4) = 1/16.
: (2) f(n+4)=1/2*f(n+3)+1/4*f(n+2)+1/8*f(n+1)+1/16*f(n)+1/16
: You can compute f(24) by computing f(5),f(6),..., one by one.
| s**e
发帖数: 1834 | 5 It can be done with some modification. Assume in your question, the 2 sets
of 2 continuous channels are not overlapping (e.g. 3-4-5 doesn't count).
Let f_{i,j}(n) be the probability of i set of j continuous empty channels in
n consecutive channels.
So we have:
(1) f_{1,2}(1)=0, f_{1,2}(2)=1/4;
(2) f_{1,2}(n+2)=1/2*f_{1,2}(n+1) + 1/4*f_{1,2}(n) + 1/4;
(3) f_{2,2}(1)=f_{2,2}(2)=f_{2,2}(3)=0, f_{2,2}(4)=1/16;
(4) f_{2,2}(n+2)=1/2*f_{2,2}(n+1) + 1/4*f_{2,2}(n) ...
+ 1/4*f_{1,2}(n)
So you first com
【在 s******e 的大作中提到】
: An extension of the problem:
: What is the probability of 2 set of 2 continuous channels (e.g. 3-4 and
: 11-12) in 24 continuous channels, assuming the channel idle probability is
: 1/2?
: Can the following solution be extended to the above extension?
: Thanks again!
:
: n
| s******e
发帖数: 128 | 6 Thanks again for the quick solution.
Not sure how (4) is derived.
Can you elaborate it a little bit?
sets
channels in
【在 s**e 的大作中提到】
: It can be done with some modification. Assume in your question, the 2 sets
: of 2 continuous channels are not overlapping (e.g. 3-4-5 doesn't count).
: Let f_{i,j}(n) be the probability of i set of j continuous empty channels in
: n consecutive channels.
: So we have:
: (1) f_{1,2}(1)=0, f_{1,2}(2)=1/4;
: (2) f_{1,2}(n+2)=1/2*f_{1,2}(n+1) + 1/4*f_{1,2}(n) + 1/4;
: (3) f_{2,2}(1)=f_{2,2}(2)=f_{2,2}(3)=0, f_{2,2}(4)=1/16;
: (4) f_{2,2}(n+2)=1/2*f_{2,2}(n+1) + 1/4*f_{2,2}(n) ...
: + 1/4*f_{1,2}(n)
| s**e
发帖数: 1834 | 7 (4) is derived the same way that (2) is derived.
To computer f_{2,2}(n+2) for n+2 channels, if the first channel is 1 (means
non-empty, 0 means empty), then it contribute 1/2*f_{2,2}(n+1); if the first
2 channels are 0 and 1, then it contribute 1/4*f_{2,2}(n); finally, if the
first 2 channels are 0 and 0, then it contributes 1/4*f_{1,2}(n) (remember,
we assumes 2 sets of 2 consecutive empty channels can not overlap).
【在 s******e 的大作中提到】
: Thanks again for the quick solution.
: Not sure how (4) is derived.
: Can you elaborate it a little bit?
:
: sets
: channels in
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