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Probability of 'i' occurrences K consecutive '1's in an n-bit stream
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I am interested in finding the probability of $i$ occurrences of $k$ consecutive $1$s in an $n$-bit stream.
For example, if the bit sequence is
$$10bf 1110001010bf 111010100110bf 111001010bf 111010,$$
then $i = 4$, $k = 3$ and $n = 40$.
probability
edited Jul 27 at 21:32
Antoine
2,485925
asked Jul 26 at 21:21
qzyDO
11
 |Â
show 2 more comments
up vote
-1
down vote
favorite
I am interested in finding the probability of $i$ occurrences of $k$ consecutive $1$s in an $n$-bit stream.
For example, if the bit sequence is
$$10bf 1110001010bf 111010100110bf 111001010bf 111010,$$
then $i = 4$, $k = 3$ and $n = 40$.
probability
edited Jul 27 at 21:32
Antoine
2,485925
asked Jul 26 at 21:21
qzyDO
11
@Clayton forming part of something :)
– Antoine
Jul 26 at 21:29
qzyDO: Have you tried to count, for example, the number of $n$-bit streams with $i = 0$ occurrences of $k = 2$ consecutive $1$s?
– Antoine
Jul 26 at 21:32
1
qzyDO: For clarification, how many occurrences of two consecutive ones are in the stream $111$?
– Antoine
Jul 26 at 21:34
Sorry, its consequitive. Not constitutive
– qzyDO
Jul 27 at 9:45
1
Of course there is need to do so. You are interested in probability $P(i, k, n)$ of finding $i$ occ. of $k$ cons. $1$s in $n$-bit streams. This probability equals (under the assumption that all streams are equiprobable) $textnumber of streams with $i$ occ. of $k$ cons. $1$s in $n$-bit streams / textnumber of all $n$-bit streams$. The number of all $n$-bit streams is $2^n$. The other number is trickier. But we cannot help you unless you define your problem well and tell us how exactly consecutive groups are counted.
– Antoine
Jul 27 at 21:27
 |Â
show 2 more comments
up vote
-1
down vote
favorite
up vote
-1
down vote
favorite
I am interested in finding the probability of $i$ occurrences of $k$ consecutive $1$s in an $n$-bit stream.
For example, if the bit sequence is
$$10bf 1110001010bf 111010100110bf 111001010bf 111010,$$
then $i = 4$, $k = 3$ and $n = 40$.
probability
edited Jul 27 at 21:32
Antoine
2,485925
asked Jul 26 at 21:21
qzyDO
11
I am interested in finding the probability of $i$ occurrences of $k$ consecutive $1$s in an $n$-bit stream.
For example, if the bit sequence is
$$10bf 1110001010bf 111010100110bf 111001010bf 111010,$$
then $i = 4$, $k = 3$ and $n = 40$.
probability
edited Jul 27 at 21:32
Antoine
2,485925
asked Jul 26 at 21:21
qzyDO
11
edited Jul 27 at 21:32
Antoine
2,485925
edited Jul 27 at 21:32
Antoine
2,485925
edited Jul 27 at 21:32
Antoine
2,485925
2,485925
asked Jul 26 at 21:21
qzyDO
11
asked Jul 26 at 21:21
qzyDO
11
asked Jul 26 at 21:21
qzyDO
11
11
@Clayton forming part of something :)
– Antoine
Jul 26 at 21:29
qzyDO: Have you tried to count, for example, the number of $n$-bit streams with $i = 0$ occurrences of $k = 2$ consecutive $1$s?
– Antoine
Jul 26 at 21:32
1
qzyDO: For clarification, how many occurrences of two consecutive ones are in the stream $111$?
– Antoine
Jul 26 at 21:34
Sorry, its consequitive. Not constitutive
– qzyDO
Jul 27 at 9:45
1
Of course there is need to do so. You are interested in probability $P(i, k, n)$ of finding $i$ occ. of $k$ cons. $1$s in $n$-bit streams. This probability equals (under the assumption that all streams are equiprobable) $textnumber of streams with $i$ occ. of $k$ cons. $1$s in $n$-bit streams / textnumber of all $n$-bit streams$. The number of all $n$-bit streams is $2^n$. The other number is trickier. But we cannot help you unless you define your problem well and tell us how exactly consecutive groups are counted.
– Antoine
Jul 27 at 21:27
 |Â
show 2 more comments
@Clayton forming part of something :)
– Antoine
Jul 26 at 21:29
qzyDO: Have you tried to count, for example, the number of $n$-bit streams with $i = 0$ occurrences of $k = 2$ consecutive $1$s?
– Antoine
Jul 26 at 21:32
1
qzyDO: For clarification, how many occurrences of two consecutive ones are in the stream $111$?
– Antoine
Jul 26 at 21:34
Sorry, its consequitive. Not constitutive
– qzyDO
Jul 27 at 9:45
1
Of course there is need to do so. You are interested in probability $P(i, k, n)$ of finding $i$ occ. of $k$ cons. $1$s in $n$-bit streams. This probability equals (under the assumption that all streams are equiprobable) $textnumber of streams with $i$ occ. of $k$ cons. $1$s in $n$-bit streams / textnumber of all $n$-bit streams$. The number of all $n$-bit streams is $2^n$. The other number is trickier. But we cannot help you unless you define your problem well and tell us how exactly consecutive groups are counted.
– Antoine
Jul 27 at 21:27
@Clayton forming part of something :)
– Antoine
Jul 26 at 21:29
@Clayton forming part of something :)
– Antoine
Jul 26 at 21:29
qzyDO: Have you tried to count, for example, the number of $n$-bit streams with $i = 0$ occurrences of $k = 2$ consecutive $1$s?
– Antoine
Jul 26 at 21:32
qzyDO: Have you tried to count, for example, the number of $n$-bit streams with $i = 0$ occurrences of $k = 2$ consecutive $1$s?
– Antoine
Jul 26 at 21:32
1
1
qzyDO: For clarification, how many occurrences of two consecutive ones are in the stream $111$?
– Antoine
Jul 26 at 21:34
qzyDO: For clarification, how many occurrences of two consecutive ones are in the stream $111$?
– Antoine
Jul 26 at 21:34
Sorry, its consequitive. Not constitutive
– qzyDO
Jul 27 at 9:45
Sorry, its consequitive. Not constitutive
– qzyDO
Jul 27 at 9:45
1
1
Of course there is need to do so. You are interested in probability $P(i, k, n)$ of finding $i$ occ. of $k$ cons. $1$s in $n$-bit streams. This probability equals (under the assumption that all streams are equiprobable) $textnumber of streams with $i$ occ. of $k$ cons. $1$s in $n$-bit streams / textnumber of all $n$-bit streams$. The number of all $n$-bit streams is $2^n$. The other number is trickier. But we cannot help you unless you define your problem well and tell us how exactly consecutive groups are counted.
– Antoine
Jul 27 at 21:27
Of course there is need to do so. You are interested in probability $P(i, k, n)$ of finding $i$ occ. of $k$ cons. $1$s in $n$-bit streams. This probability equals (under the assumption that all streams are equiprobable) $textnumber of streams with $i$ occ. of $k$ cons. $1$s in $n$-bit streams / textnumber of all $n$-bit streams$. The number of all $n$-bit streams is $2^n$. The other number is trickier. But we cannot help you unless you define your problem well and tell us how exactly consecutive groups are counted.
– Antoine
Jul 27 at 21:27
 |Â
show 2 more comments
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@Clayton forming part of something :)
– Antoine
Jul 26 at 21:29
qzyDO: Have you tried to count, for example, the number of $n$-bit streams with $i = 0$ occurrences of $k = 2$ consecutive $1$s?
– Antoine
Jul 26 at 21:32
1
qzyDO: For clarification, how many occurrences of two consecutive ones are in the stream $111$?
– Antoine
Jul 26 at 21:34
Sorry, its consequitive. Not constitutive
– qzyDO
Jul 27 at 9:45
1
Of course there is need to do so. You are interested in probability $P(i, k, n)$ of finding $i$ occ. of $k$ cons. $1$s in $n$-bit streams. This probability equals (under the assumption that all streams are equiprobable) $textnumber of streams with $i$ occ. of $k$ cons. $1$s in $n$-bit streams / textnumber of all $n$-bit streams$. The number of all $n$-bit streams is $2^n$. The other number is trickier. But we cannot help you unless you define your problem well and tell us how exactly consecutive groups are counted.
– Antoine
Jul 27 at 21:27