Probability of a particular finite combination of heads and tails in infinite tosses sequence
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I am asked to find the probability of getting HHHTHHTT or (111010011) combination in infinite coin tosses sequence where probability of getting heads is $p$.
So I thought that if it starts with the first toss it would be simply $p^3(p-1)p^2(p-1)^2=p^5(p-1)^3$. But it does not need to start with the first toss, so what then? How to take this into consideration?
probability probability-theory
edited Sep 10 at 14:14
Micah
28.9k1361102
asked Sep 10 at 13:03
Kacper
339112
add a comment |Â
up vote
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I am asked to find the probability of getting HHHTHHTT or (111010011) combination in infinite coin tosses sequence where probability of getting heads is $p$.
So I thought that if it starts with the first toss it would be simply $p^3(p-1)p^2(p-1)^2=p^5(p-1)^3$. But it does not need to start with the first toss, so what then? How to take this into consideration?
probability probability-theory
edited Sep 10 at 14:14
Micah
28.9k1361102
asked Sep 10 at 13:03
Kacper
339112
2
Well, the probability of finding any finite subsequence in an infinite sequence of tosses is 1. It's called Infinite-Monkey-Theorem :) The subsequence itself must be posible though, which ist true in your case.
– denklo
Sep 10 at 13:14
A more interesting question would be finding the expected tosses to get this particular sequence.
– karakfa
Sep 10 at 14:42
@karakfa That can be done by solving the following equations where $p+q=1$ and $mu$ denotes the expectation: $beginalignedmu & =1+pmu_H\ mu_H & =1+pmu_HH+qmu\ mu_HH & =1+pmu_HHH+qmu\ mu_HHH & =1+pmu_HHH+qmu_HHHT\ mu_HHHT & =1+pmu_HHHTH+qmu\ mu_HHHTH & =1+pmu_HHHTHH+qmu\ mu_HHHTHH & =1+pmu_HHH+qmu_HHHTHHT\ mu_HHHTHHT & =1+pmu_H+q endaligned $
– drhab
Sep 10 at 15:36
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I am asked to find the probability of getting HHHTHHTT or (111010011) combination in infinite coin tosses sequence where probability of getting heads is $p$.
So I thought that if it starts with the first toss it would be simply $p^3(p-1)p^2(p-1)^2=p^5(p-1)^3$. But it does not need to start with the first toss, so what then? How to take this into consideration?
probability probability-theory
edited Sep 10 at 14:14
Micah
28.9k1361102
asked Sep 10 at 13:03
Kacper
339112
I am asked to find the probability of getting HHHTHHTT or (111010011) combination in infinite coin tosses sequence where probability of getting heads is $p$.
So I thought that if it starts with the first toss it would be simply $p^3(p-1)p^2(p-1)^2=p^5(p-1)^3$. But it does not need to start with the first toss, so what then? How to take this into consideration?
probability probability-theory
probability probability-theory
edited Sep 10 at 14:14
Micah
28.9k1361102
asked Sep 10 at 13:03
Kacper
339112
edited Sep 10 at 14:14
Micah
28.9k1361102
asked Sep 10 at 13:03
Kacper
339112
edited Sep 10 at 14:14
Micah
28.9k1361102
edited Sep 10 at 14:14
Micah
28.9k1361102
edited Sep 10 at 14:14
Micah
28.9k1361102
28.9k1361102
asked Sep 10 at 13:03
Kacper
339112
asked Sep 10 at 13:03
Kacper
339112
asked Sep 10 at 13:03
Kacper
339112
339112
2
Well, the probability of finding any finite subsequence in an infinite sequence of tosses is 1. It's called Infinite-Monkey-Theorem :) The subsequence itself must be posible though, which ist true in your case.
– denklo
Sep 10 at 13:14
A more interesting question would be finding the expected tosses to get this particular sequence.
– karakfa
Sep 10 at 14:42
@karakfa That can be done by solving the following equations where $p+q=1$ and $mu$ denotes the expectation: $beginalignedmu & =1+pmu_H\ mu_H & =1+pmu_HH+qmu\ mu_HH & =1+pmu_HHH+qmu\ mu_HHH & =1+pmu_HHH+qmu_HHHT\ mu_HHHT & =1+pmu_HHHTH+qmu\ mu_HHHTH & =1+pmu_HHHTHH+qmu\ mu_HHHTHH & =1+pmu_HHH+qmu_HHHTHHT\ mu_HHHTHHT & =1+pmu_H+q endaligned $
– drhab
Sep 10 at 15:36
add a comment |Â
2
Well, the probability of finding any finite subsequence in an infinite sequence of tosses is 1. It's called Infinite-Monkey-Theorem :) The subsequence itself must be posible though, which ist true in your case.
– denklo
Sep 10 at 13:14
A more interesting question would be finding the expected tosses to get this particular sequence.
– karakfa
Sep 10 at 14:42
@karakfa That can be done by solving the following equations where $p+q=1$ and $mu$ denotes the expectation: $beginalignedmu & =1+pmu_H\ mu_H & =1+pmu_HH+qmu\ mu_HH & =1+pmu_HHH+qmu\ mu_HHH & =1+pmu_HHH+qmu_HHHT\ mu_HHHT & =1+pmu_HHHTH+qmu\ mu_HHHTH & =1+pmu_HHHTHH+qmu\ mu_HHHTHH & =1+pmu_HHH+qmu_HHHTHHT\ mu_HHHTHHT & =1+pmu_H+q endaligned $
– drhab
Sep 10 at 15:36
2
2
Well, the probability of finding any finite subsequence in an infinite sequence of tosses is 1. It's called Infinite-Monkey-Theorem :) The subsequence itself must be posible though, which ist true in your case.
– denklo
Sep 10 at 13:14
Well, the probability of finding any finite subsequence in an infinite sequence of tosses is 1. It's called Infinite-Monkey-Theorem :) The subsequence itself must be posible though, which ist true in your case.
– denklo
Sep 10 at 13:14
A more interesting question would be finding the expected tosses to get this particular sequence.
– karakfa
Sep 10 at 14:42
A more interesting question would be finding the expected tosses to get this particular sequence.
– karakfa
Sep 10 at 14:42
@karakfa That can be done by solving the following equations where $p+q=1$ and $mu$ denotes the expectation: $beginalignedmu & =1+pmu_H\ mu_H & =1+pmu_HH+qmu\ mu_HH & =1+pmu_HHH+qmu\ mu_HHH & =1+pmu_HHH+qmu_HHHT\ mu_HHHT & =1+pmu_HHHTH+qmu\ mu_HHHTH & =1+pmu_HHHTHH+qmu\ mu_HHHTHH & =1+pmu_HHH+qmu_HHHTHHT\ mu_HHHTHHT & =1+pmu_H+q endaligned $
– drhab
Sep 10 at 15:36
@karakfa That can be done by solving the following equations where $p+q=1$ and $mu$ denotes the expectation: $beginalignedmu & =1+pmu_H\ mu_H & =1+pmu_HH+qmu\ mu_HH & =1+pmu_HHH+qmu\ mu_HHH & =1+pmu_HHH+qmu_HHHT\ mu_HHHT & =1+pmu_HHHTH+qmu\ mu_HHHTH & =1+pmu_HHHTHH+qmu\ mu_HHHTHH & =1+pmu_HHH+qmu_HHHTHHT\ mu_HHHTHHT & =1+pmu_H+q endaligned $
– drhab
Sep 10 at 15:36
add a comment |Â
1 Answer
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Look at the string as concatenation of strings of length $8$ and call these strings the special strings.
The probability that the first special string is not HHHTHHTT is a number $r<1$.
Then the probability that none of the first $n$ special strings is HHHTHHTT is $r^n$.
If $s$ denote the probability that none of the special strings is HHHTHHTT then $sleq r^n$ for every $n$.
This for $n=1,2,3,dots$ and $r<1$ so this can only be true if $s=0$.
That means that the probability that at least one of the special strings is HHHTHHTT is $1$.
Then of course the same is true if we look at all strings of length $8$ (not only the special ones).
It can also be proved for every integer $n$ that the probability that not more than $n$ strings of length $8$ will be HHHTHHTT is $0$, and from this it follows that the probability that an infinite number of strings of length $8$ will be HHHTHHTT equals $1$.
answered Sep 10 at 14:10
drhab
89.4k541123
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
Look at the string as concatenation of strings of length $8$ and call these strings the special strings.
The probability that the first special string is not HHHTHHTT is a number $r<1$.
Then the probability that none of the first $n$ special strings is HHHTHHTT is $r^n$.
If $s$ denote the probability that none of the special strings is HHHTHHTT then $sleq r^n$ for every $n$.
This for $n=1,2,3,dots$ and $r<1$ so this can only be true if $s=0$.
That means that the probability that at least one of the special strings is HHHTHHTT is $1$.
Then of course the same is true if we look at all strings of length $8$ (not only the special ones).
It can also be proved for every integer $n$ that the probability that not more than $n$ strings of length $8$ will be HHHTHHTT is $0$, and from this it follows that the probability that an infinite number of strings of length $8$ will be HHHTHHTT equals $1$.
answered Sep 10 at 14:10
drhab
89.4k541123
add a comment |Â
up vote
4
down vote
accepted
Look at the string as concatenation of strings of length $8$ and call these strings the special strings.
The probability that the first special string is not HHHTHHTT is a number $r<1$.
Then the probability that none of the first $n$ special strings is HHHTHHTT is $r^n$.
If $s$ denote the probability that none of the special strings is HHHTHHTT then $sleq r^n$ for every $n$.
This for $n=1,2,3,dots$ and $r<1$ so this can only be true if $s=0$.
That means that the probability that at least one of the special strings is HHHTHHTT is $1$.
Then of course the same is true if we look at all strings of length $8$ (not only the special ones).
It can also be proved for every integer $n$ that the probability that not more than $n$ strings of length $8$ will be HHHTHHTT is $0$, and from this it follows that the probability that an infinite number of strings of length $8$ will be HHHTHHTT equals $1$.
answered Sep 10 at 14:10
drhab
89.4k541123
add a comment |Â
up vote
4
down vote
accepted
up vote
4
down vote
accepted
Look at the string as concatenation of strings of length $8$ and call these strings the special strings.
The probability that the first special string is not HHHTHHTT is a number $r<1$.
Then the probability that none of the first $n$ special strings is HHHTHHTT is $r^n$.
If $s$ denote the probability that none of the special strings is HHHTHHTT then $sleq r^n$ for every $n$.
This for $n=1,2,3,dots$ and $r<1$ so this can only be true if $s=0$.
That means that the probability that at least one of the special strings is HHHTHHTT is $1$.
Then of course the same is true if we look at all strings of length $8$ (not only the special ones).
It can also be proved for every integer $n$ that the probability that not more than $n$ strings of length $8$ will be HHHTHHTT is $0$, and from this it follows that the probability that an infinite number of strings of length $8$ will be HHHTHHTT equals $1$.
answered Sep 10 at 14:10
drhab
89.4k541123
Look at the string as concatenation of strings of length $8$ and call these strings the special strings.
The probability that the first special string is not HHHTHHTT is a number $r<1$.
Then the probability that none of the first $n$ special strings is HHHTHHTT is $r^n$.
If $s$ denote the probability that none of the special strings is HHHTHHTT then $sleq r^n$ for every $n$.
This for $n=1,2,3,dots$ and $r<1$ so this can only be true if $s=0$.
That means that the probability that at least one of the special strings is HHHTHHTT is $1$.
Then of course the same is true if we look at all strings of length $8$ (not only the special ones).
It can also be proved for every integer $n$ that the probability that not more than $n$ strings of length $8$ will be HHHTHHTT is $0$, and from this it follows that the probability that an infinite number of strings of length $8$ will be HHHTHHTT equals $1$.
answered Sep 10 at 14:10
drhab
89.4k541123
edited Sep 10 at 14:56
answered Sep 10 at 14:10
drhab
89.4k541123
answered Sep 10 at 14:10
drhab
89.4k541123
answered Sep 10 at 14:10
drhab
89.4k541123
89.4k541123
add a comment |Â
add a comment |Â
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2
Well, the probability of finding any finite subsequence in an infinite sequence of tosses is 1. It's called Infinite-Monkey-Theorem :) The subsequence itself must be posible though, which ist true in your case.
– denklo
Sep 10 at 13:14
A more interesting question would be finding the expected tosses to get this particular sequence.
– karakfa
Sep 10 at 14:42
@karakfa That can be done by solving the following equations where $p+q=1$ and $mu$ denotes the expectation: $beginalignedmu & =1+pmu_H\ mu_H & =1+pmu_HH+qmu\ mu_HH & =1+pmu_HHH+qmu\ mu_HHH & =1+pmu_HHH+qmu_HHHT\ mu_HHHT & =1+pmu_HHHTH+qmu\ mu_HHHTH & =1+pmu_HHHTHH+qmu\ mu_HHHTHH & =1+pmu_HHH+qmu_HHHTHHT\ mu_HHHTHHT & =1+pmu_H+q endaligned $
– drhab
Sep 10 at 15:36