Probability that THHT occurs in a sequence of 10 coin tossesFinding probability of recording head in a sequence of coin tossProbability that a coin lands on tails an odd number of times when it is tossed $100$ timesWhat's the expected number of coin tosses in order to get a sequence HHTTHH?Probability in a series of events, during the series.Chance on winning by throwing a head on first toss.probability of sequence of exactly 3 headsSequence of non-independent coin tossesis it possible to find out how many times coin will return head in n-sequence?A fair coin is tossed $text10 times$. What is the probability that ONLY the first two tosses will yield heads?What is the probability of player A winning this Coin Toss Game?
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Probability that THHT occurs in a sequence of 10 coin tosses
Finding probability of recording head in a sequence of coin tossProbability that a coin lands on tails an odd number of times when it is tossed $100$ timesWhat's the expected number of coin tosses in order to get a sequence HHTTHH?Probability in a series of events, during the series.Chance on winning by throwing a head on first toss.probability of sequence of exactly 3 headsSequence of non-independent coin tossesis it possible to find out how many times coin will return head in n-sequence?A fair coin is tossed $text10 times$. What is the probability that ONLY the first two tosses will yield heads?What is the probability of player A winning this Coin Toss Game?
$begingroup$
Assume we have a fair coin, and we throw the coin $10$ times in a row.
I want to calculate the probability that the sequence 'tail, head, head, tail' occurs.
So I think I can interpret this event as a binary number with $10$ digits. So $1$ means tail, $0$ means head. Therefore we have $2^10 = 1024$ different outcomes of the $10$ throws. The sequence 'tail, head, head, tail' can start at $7$ different positions and so there are $7cdot2^6 = 448$ different outcomes of the $10$ throws with the sequence 'tail, head, head, tail'. So the probability would be $frac4481024 = 0.4375$.
But I have a feeling there's something wrong?
probability combinatorics
edited Mar 22 at 20:02
Robert Howard
2,30331035
asked Mar 22 at 14:55
geirageira
713
$endgroup$
add a comment |
$begingroup$
Assume we have a fair coin, and we throw the coin $10$ times in a row.
I want to calculate the probability that the sequence 'tail, head, head, tail' occurs.
So I think I can interpret this event as a binary number with $10$ digits. So $1$ means tail, $0$ means head. Therefore we have $2^10 = 1024$ different outcomes of the $10$ throws. The sequence 'tail, head, head, tail' can start at $7$ different positions and so there are $7cdot2^6 = 448$ different outcomes of the $10$ throws with the sequence 'tail, head, head, tail'. So the probability would be $frac4481024 = 0.4375$.
But I have a feeling there's something wrong?
probability combinatorics
edited Mar 22 at 20:02
Robert Howard
2,30331035
asked Mar 22 at 14:55
geirageira
713
$endgroup$
2
$begingroup$
You are over-counting sequences like T H H T H H T .
$endgroup$
– kimchi lover
Mar 22 at 15:01
1
$begingroup$
@kimchilover ...and even THHTHHTHHT.
$endgroup$
– CiaPan
Mar 22 at 15:32
add a comment |
$begingroup$
Assume we have a fair coin, and we throw the coin $10$ times in a row.
I want to calculate the probability that the sequence 'tail, head, head, tail' occurs.
So I think I can interpret this event as a binary number with $10$ digits. So $1$ means tail, $0$ means head. Therefore we have $2^10 = 1024$ different outcomes of the $10$ throws. The sequence 'tail, head, head, tail' can start at $7$ different positions and so there are $7cdot2^6 = 448$ different outcomes of the $10$ throws with the sequence 'tail, head, head, tail'. So the probability would be $frac4481024 = 0.4375$.
But I have a feeling there's something wrong?
probability combinatorics
edited Mar 22 at 20:02
Robert Howard
2,30331035
asked Mar 22 at 14:55
geirageira
713
$endgroup$
Assume we have a fair coin, and we throw the coin $10$ times in a row.
I want to calculate the probability that the sequence 'tail, head, head, tail' occurs.
So I think I can interpret this event as a binary number with $10$ digits. So $1$ means tail, $0$ means head. Therefore we have $2^10 = 1024$ different outcomes of the $10$ throws. The sequence 'tail, head, head, tail' can start at $7$ different positions and so there are $7cdot2^6 = 448$ different outcomes of the $10$ throws with the sequence 'tail, head, head, tail'. So the probability would be $frac4481024 = 0.4375$.
But I have a feeling there's something wrong?
probability combinatorics
probability combinatorics
edited Mar 22 at 20:02
Robert Howard
2,30331035
asked Mar 22 at 14:55
geirageira
713
edited Mar 22 at 20:02
Robert Howard
2,30331035
asked Mar 22 at 14:55
geirageira
713
edited Mar 22 at 20:02
Robert Howard
2,30331035
edited Mar 22 at 20:02
Robert Howard
2,30331035
edited Mar 22 at 20:02
Robert Howard
2,30331035
2,30331035
asked Mar 22 at 14:55
geirageira
713
asked Mar 22 at 14:55
geirageira
713
asked Mar 22 at 14:55
geirageira
713
713
2
$begingroup$
You are over-counting sequences like T H H T H H T .
$endgroup$
– kimchi lover
Mar 22 at 15:01
1
$begingroup$
@kimchilover ...and even THHTHHTHHT.
$endgroup$
– CiaPan
Mar 22 at 15:32
add a comment |
2
$begingroup$
You are over-counting sequences like T H H T H H T .
$endgroup$
– kimchi lover
Mar 22 at 15:01
1
$begingroup$
@kimchilover ...and even THHTHHTHHT.
$endgroup$
– CiaPan
Mar 22 at 15:32
2
2
$begingroup$
You are over-counting sequences like T H H T H H T .
$endgroup$
– kimchi lover
Mar 22 at 15:01
$begingroup$
You are over-counting sequences like T H H T H H T .
$endgroup$
– kimchi lover
Mar 22 at 15:01
1
1
$begingroup$
@kimchilover ...and even THHTHHTHHT.
$endgroup$
– CiaPan
Mar 22 at 15:32
$begingroup$
@kimchilover ...and even THHTHHTHHT.
$endgroup$
– CiaPan
Mar 22 at 15:32
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
You can use the Inclusion-exclusion principle to solve this. When I do this I get $$7cdot 2^6 - 6cdot 2^2 - 4cdot 2^3+1$$ where the first term is what you get, where the second and third terms count the number of sequences with two non-overlapping instances of T H H T and of sequences with one overlap, like T H H T H H T, and finally the number of sequences with a triple overlap, T H H T H H T H H T.
Confession: I has earlier got $45cdot2^2$ for the second term, by a mental blunder, as AnnaSaabel pointed out. There are 2 "gaps" to separate the two instances of THHT, which can occur before, between, or after the 2 instances; they can be distributed in any of the 6 ways 200,020,002,110,101, or 011.
Added: if the number of coin tosses were $n=100$ (say), and the pattern sought was still THHT, this method becomes clumsy. A different method is to construct a Markov chain with states representing how far a string matching algorithm has progressed in matching the given pattern. If $M$ is the transition matrix for this chain, the desired answer is the entry in the matrix $M^n$ corresponing to the pair $(textstart state, textaccepting state)$.
answered Mar 22 at 15:14
kimchi loverkimchi lover
12.2k31229
$endgroup$
1
$begingroup$
How did you calculate 45 for the second term?
$endgroup$
– Anna Saabel
Mar 22 at 16:27
$begingroup$
@AnnaSaabel By a mental blunder. Thanks for the catch!
$endgroup$
– kimchi lover
Mar 22 at 17:00
1
$begingroup$
This is correct, as this Python 3 code snippet gives the same number: $tag*$print(sum('0110' in s for s in [bin(n)[-10:] for n in range(1<<10,1<<11)]))
$endgroup$
– Mike Earnest
Mar 22 at 17:08
$begingroup$
Your transition matrix is wrong: a match failure in state 1 or 2 will result in state 1, not state 0. In general, the transition matrix is basically the same thing as the array produced in the preprocessing step of KMP algorithm.
$endgroup$
– infmagic2047
Mar 22 at 17:52
$begingroup$
@infmagic2047 You are right; I've hopefully corrected this.
$endgroup$
– kimchi lover
Mar 22 at 18:02
add a comment |
$begingroup$
As @kimchilover states in the comments, you are counting some 10-digit binary numbers more than once in the number $7cdot 2^6$. To make this more obvious, consider a different problem: to find the probability that the sequence 'heads' appears. By your counting logic, there are 10 places for it to begin, so there are $10cdot 2^9$ different outcomes of the 10 throws with the sequence 'heads', so the probability would be $frac10cdot 2^92^10 = 5$. That can't be good. It's very clear now that the issue is overcounting -- you have five times as many sequences with 'heads' in them as the number of sequences total! The problem is that we have counted sequences with multiple heads many times. For example, the sequence of all heads is counted $10$ times, once for each of the places where the sequence 'heads' begins within it.
As I write this, I see that @kimchilover also just posted an answer to the question which directs you to the inclusion-exclusion principle, so I'll stop here with an answer which could just help you to try generalizing arguments which feel fishy to see where they go wrong. Good job detecting the fishiness!
answered Mar 22 at 15:19
cspruncsprun
2,839211
$endgroup$
add a comment |
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2 Answers
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2 Answers
2
active
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$begingroup$
You can use the Inclusion-exclusion principle to solve this. When I do this I get $$7cdot 2^6 - 6cdot 2^2 - 4cdot 2^3+1$$ where the first term is what you get, where the second and third terms count the number of sequences with two non-overlapping instances of T H H T and of sequences with one overlap, like T H H T H H T, and finally the number of sequences with a triple overlap, T H H T H H T H H T.
Confession: I has earlier got $45cdot2^2$ for the second term, by a mental blunder, as AnnaSaabel pointed out. There are 2 "gaps" to separate the two instances of THHT, which can occur before, between, or after the 2 instances; they can be distributed in any of the 6 ways 200,020,002,110,101, or 011.
Added: if the number of coin tosses were $n=100$ (say), and the pattern sought was still THHT, this method becomes clumsy. A different method is to construct a Markov chain with states representing how far a string matching algorithm has progressed in matching the given pattern. If $M$ is the transition matrix for this chain, the desired answer is the entry in the matrix $M^n$ corresponing to the pair $(textstart state, textaccepting state)$.
answered Mar 22 at 15:14
kimchi loverkimchi lover
12.2k31229
$endgroup$
1
$begingroup$
How did you calculate 45 for the second term?
$endgroup$
– Anna Saabel
Mar 22 at 16:27
$begingroup$
@AnnaSaabel By a mental blunder. Thanks for the catch!
$endgroup$
– kimchi lover
Mar 22 at 17:00
1
$begingroup$
This is correct, as this Python 3 code snippet gives the same number: $tag*$print(sum('0110' in s for s in [bin(n)[-10:] for n in range(1<<10,1<<11)]))
$endgroup$
– Mike Earnest
Mar 22 at 17:08
$begingroup$
Your transition matrix is wrong: a match failure in state 1 or 2 will result in state 1, not state 0. In general, the transition matrix is basically the same thing as the array produced in the preprocessing step of KMP algorithm.
$endgroup$
– infmagic2047
Mar 22 at 17:52
$begingroup$
@infmagic2047 You are right; I've hopefully corrected this.
$endgroup$
– kimchi lover
Mar 22 at 18:02
add a comment |
$begingroup$
You can use the Inclusion-exclusion principle to solve this. When I do this I get $$7cdot 2^6 - 6cdot 2^2 - 4cdot 2^3+1$$ where the first term is what you get, where the second and third terms count the number of sequences with two non-overlapping instances of T H H T and of sequences with one overlap, like T H H T H H T, and finally the number of sequences with a triple overlap, T H H T H H T H H T.
Confession: I has earlier got $45cdot2^2$ for the second term, by a mental blunder, as AnnaSaabel pointed out. There are 2 "gaps" to separate the two instances of THHT, which can occur before, between, or after the 2 instances; they can be distributed in any of the 6 ways 200,020,002,110,101, or 011.
Added: if the number of coin tosses were $n=100$ (say), and the pattern sought was still THHT, this method becomes clumsy. A different method is to construct a Markov chain with states representing how far a string matching algorithm has progressed in matching the given pattern. If $M$ is the transition matrix for this chain, the desired answer is the entry in the matrix $M^n$ corresponing to the pair $(textstart state, textaccepting state)$.
answered Mar 22 at 15:14
kimchi loverkimchi lover
12.2k31229
$endgroup$
1
$begingroup$
How did you calculate 45 for the second term?
$endgroup$
– Anna Saabel
Mar 22 at 16:27
$begingroup$
@AnnaSaabel By a mental blunder. Thanks for the catch!
$endgroup$
– kimchi lover
Mar 22 at 17:00
1
$begingroup$
This is correct, as this Python 3 code snippet gives the same number: $tag*$print(sum('0110' in s for s in [bin(n)[-10:] for n in range(1<<10,1<<11)]))
$endgroup$
– Mike Earnest
Mar 22 at 17:08
$begingroup$
Your transition matrix is wrong: a match failure in state 1 or 2 will result in state 1, not state 0. In general, the transition matrix is basically the same thing as the array produced in the preprocessing step of KMP algorithm.
$endgroup$
– infmagic2047
Mar 22 at 17:52
$begingroup$
@infmagic2047 You are right; I've hopefully corrected this.
$endgroup$
– kimchi lover
Mar 22 at 18:02
add a comment |
$begingroup$
You can use the Inclusion-exclusion principle to solve this. When I do this I get $$7cdot 2^6 - 6cdot 2^2 - 4cdot 2^3+1$$ where the first term is what you get, where the second and third terms count the number of sequences with two non-overlapping instances of T H H T and of sequences with one overlap, like T H H T H H T, and finally the number of sequences with a triple overlap, T H H T H H T H H T.
Confession: I has earlier got $45cdot2^2$ for the second term, by a mental blunder, as AnnaSaabel pointed out. There are 2 "gaps" to separate the two instances of THHT, which can occur before, between, or after the 2 instances; they can be distributed in any of the 6 ways 200,020,002,110,101, or 011.
Added: if the number of coin tosses were $n=100$ (say), and the pattern sought was still THHT, this method becomes clumsy. A different method is to construct a Markov chain with states representing how far a string matching algorithm has progressed in matching the given pattern. If $M$ is the transition matrix for this chain, the desired answer is the entry in the matrix $M^n$ corresponing to the pair $(textstart state, textaccepting state)$.
answered Mar 22 at 15:14
kimchi loverkimchi lover
12.2k31229
$endgroup$
You can use the Inclusion-exclusion principle to solve this. When I do this I get $$7cdot 2^6 - 6cdot 2^2 - 4cdot 2^3+1$$ where the first term is what you get, where the second and third terms count the number of sequences with two non-overlapping instances of T H H T and of sequences with one overlap, like T H H T H H T, and finally the number of sequences with a triple overlap, T H H T H H T H H T.
Confession: I has earlier got $45cdot2^2$ for the second term, by a mental blunder, as AnnaSaabel pointed out. There are 2 "gaps" to separate the two instances of THHT, which can occur before, between, or after the 2 instances; they can be distributed in any of the 6 ways 200,020,002,110,101, or 011.
Added: if the number of coin tosses were $n=100$ (say), and the pattern sought was still THHT, this method becomes clumsy. A different method is to construct a Markov chain with states representing how far a string matching algorithm has progressed in matching the given pattern. If $M$ is the transition matrix for this chain, the desired answer is the entry in the matrix $M^n$ corresponing to the pair $(textstart state, textaccepting state)$.
answered Mar 22 at 15:14
kimchi loverkimchi lover
12.2k31229
edited Mar 22 at 17:59
answered Mar 22 at 15:14
kimchi loverkimchi lover
12.2k31229
answered Mar 22 at 15:14
kimchi loverkimchi lover
12.2k31229
answered Mar 22 at 15:14
kimchi loverkimchi lover
12.2k31229
12.2k31229
1
$begingroup$
How did you calculate 45 for the second term?
$endgroup$
– Anna Saabel
Mar 22 at 16:27
$begingroup$
@AnnaSaabel By a mental blunder. Thanks for the catch!
$endgroup$
– kimchi lover
Mar 22 at 17:00
1
$begingroup$
This is correct, as this Python 3 code snippet gives the same number: $tag*$print(sum('0110' in s for s in [bin(n)[-10:] for n in range(1<<10,1<<11)]))
$endgroup$
– Mike Earnest
Mar 22 at 17:08
$begingroup$
Your transition matrix is wrong: a match failure in state 1 or 2 will result in state 1, not state 0. In general, the transition matrix is basically the same thing as the array produced in the preprocessing step of KMP algorithm.
$endgroup$
– infmagic2047
Mar 22 at 17:52
$begingroup$
@infmagic2047 You are right; I've hopefully corrected this.
$endgroup$
– kimchi lover
Mar 22 at 18:02
add a comment |
1
$begingroup$
How did you calculate 45 for the second term?
$endgroup$
– Anna Saabel
Mar 22 at 16:27
$begingroup$
@AnnaSaabel By a mental blunder. Thanks for the catch!
$endgroup$
– kimchi lover
Mar 22 at 17:00
1
$begingroup$
This is correct, as this Python 3 code snippet gives the same number: $tag*$print(sum('0110' in s for s in [bin(n)[-10:] for n in range(1<<10,1<<11)]))
$endgroup$
– Mike Earnest
Mar 22 at 17:08
$begingroup$
Your transition matrix is wrong: a match failure in state 1 or 2 will result in state 1, not state 0. In general, the transition matrix is basically the same thing as the array produced in the preprocessing step of KMP algorithm.
$endgroup$
– infmagic2047
Mar 22 at 17:52
$begingroup$
@infmagic2047 You are right; I've hopefully corrected this.
$endgroup$
– kimchi lover
Mar 22 at 18:02
1
1
$begingroup$
How did you calculate 45 for the second term?
$endgroup$
– Anna Saabel
Mar 22 at 16:27
$begingroup$
How did you calculate 45 for the second term?
$endgroup$
– Anna Saabel
Mar 22 at 16:27
$begingroup$
@AnnaSaabel By a mental blunder. Thanks for the catch!
$endgroup$
– kimchi lover
Mar 22 at 17:00
$begingroup$
@AnnaSaabel By a mental blunder. Thanks for the catch!
$endgroup$
– kimchi lover
Mar 22 at 17:00
1
1
$begingroup$
This is correct, as this Python 3 code snippet gives the same number: $tag*$
print(sum('0110' in s for s in [bin(n)[-10:] for n in range(1<<10,1<<11)]))$endgroup$
– Mike Earnest
Mar 22 at 17:08
$begingroup$
This is correct, as this Python 3 code snippet gives the same number: $tag*$
print(sum('0110' in s for s in [bin(n)[-10:] for n in range(1<<10,1<<11)]))$endgroup$
– Mike Earnest
Mar 22 at 17:08
$begingroup$
Your transition matrix is wrong: a match failure in state 1 or 2 will result in state 1, not state 0. In general, the transition matrix is basically the same thing as the array produced in the preprocessing step of KMP algorithm.
$endgroup$
– infmagic2047
Mar 22 at 17:52
$begingroup$
Your transition matrix is wrong: a match failure in state 1 or 2 will result in state 1, not state 0. In general, the transition matrix is basically the same thing as the array produced in the preprocessing step of KMP algorithm.
$endgroup$
– infmagic2047
Mar 22 at 17:52
$begingroup$
@infmagic2047 You are right; I've hopefully corrected this.
$endgroup$
– kimchi lover
Mar 22 at 18:02
$begingroup$
@infmagic2047 You are right; I've hopefully corrected this.
$endgroup$
– kimchi lover
Mar 22 at 18:02
add a comment |
$begingroup$
As @kimchilover states in the comments, you are counting some 10-digit binary numbers more than once in the number $7cdot 2^6$. To make this more obvious, consider a different problem: to find the probability that the sequence 'heads' appears. By your counting logic, there are 10 places for it to begin, so there are $10cdot 2^9$ different outcomes of the 10 throws with the sequence 'heads', so the probability would be $frac10cdot 2^92^10 = 5$. That can't be good. It's very clear now that the issue is overcounting -- you have five times as many sequences with 'heads' in them as the number of sequences total! The problem is that we have counted sequences with multiple heads many times. For example, the sequence of all heads is counted $10$ times, once for each of the places where the sequence 'heads' begins within it.
As I write this, I see that @kimchilover also just posted an answer to the question which directs you to the inclusion-exclusion principle, so I'll stop here with an answer which could just help you to try generalizing arguments which feel fishy to see where they go wrong. Good job detecting the fishiness!
answered Mar 22 at 15:19
cspruncsprun
2,839211
$endgroup$
add a comment |
$begingroup$
As @kimchilover states in the comments, you are counting some 10-digit binary numbers more than once in the number $7cdot 2^6$. To make this more obvious, consider a different problem: to find the probability that the sequence 'heads' appears. By your counting logic, there are 10 places for it to begin, so there are $10cdot 2^9$ different outcomes of the 10 throws with the sequence 'heads', so the probability would be $frac10cdot 2^92^10 = 5$. That can't be good. It's very clear now that the issue is overcounting -- you have five times as many sequences with 'heads' in them as the number of sequences total! The problem is that we have counted sequences with multiple heads many times. For example, the sequence of all heads is counted $10$ times, once for each of the places where the sequence 'heads' begins within it.
As I write this, I see that @kimchilover also just posted an answer to the question which directs you to the inclusion-exclusion principle, so I'll stop here with an answer which could just help you to try generalizing arguments which feel fishy to see where they go wrong. Good job detecting the fishiness!
answered Mar 22 at 15:19
cspruncsprun
2,839211
$endgroup$
add a comment |
$begingroup$
As @kimchilover states in the comments, you are counting some 10-digit binary numbers more than once in the number $7cdot 2^6$. To make this more obvious, consider a different problem: to find the probability that the sequence 'heads' appears. By your counting logic, there are 10 places for it to begin, so there are $10cdot 2^9$ different outcomes of the 10 throws with the sequence 'heads', so the probability would be $frac10cdot 2^92^10 = 5$. That can't be good. It's very clear now that the issue is overcounting -- you have five times as many sequences with 'heads' in them as the number of sequences total! The problem is that we have counted sequences with multiple heads many times. For example, the sequence of all heads is counted $10$ times, once for each of the places where the sequence 'heads' begins within it.
As I write this, I see that @kimchilover also just posted an answer to the question which directs you to the inclusion-exclusion principle, so I'll stop here with an answer which could just help you to try generalizing arguments which feel fishy to see where they go wrong. Good job detecting the fishiness!
answered Mar 22 at 15:19
cspruncsprun
2,839211
$endgroup$
As @kimchilover states in the comments, you are counting some 10-digit binary numbers more than once in the number $7cdot 2^6$. To make this more obvious, consider a different problem: to find the probability that the sequence 'heads' appears. By your counting logic, there are 10 places for it to begin, so there are $10cdot 2^9$ different outcomes of the 10 throws with the sequence 'heads', so the probability would be $frac10cdot 2^92^10 = 5$. That can't be good. It's very clear now that the issue is overcounting -- you have five times as many sequences with 'heads' in them as the number of sequences total! The problem is that we have counted sequences with multiple heads many times. For example, the sequence of all heads is counted $10$ times, once for each of the places where the sequence 'heads' begins within it.
As I write this, I see that @kimchilover also just posted an answer to the question which directs you to the inclusion-exclusion principle, so I'll stop here with an answer which could just help you to try generalizing arguments which feel fishy to see where they go wrong. Good job detecting the fishiness!
answered Mar 22 at 15:19
cspruncsprun
2,839211
edited Mar 22 at 15:26
answered Mar 22 at 15:19
cspruncsprun
2,839211
answered Mar 22 at 15:19
cspruncsprun
2,839211
answered Mar 22 at 15:19
cspruncsprun
2,839211
2,839211
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$begingroup$
You are over-counting sequences like T H H T H H T .
$endgroup$
– kimchi lover
Mar 22 at 15:01
1
$begingroup$
@kimchilover ...and even THHTHHTHHT.
$endgroup$
– CiaPan
Mar 22 at 15:32