Mixing
Coin Tossings probability
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I want to find the probability that in ten tossings a coin falls heads at least five times in succession. Is there any formula to compute this probability?
Answer provided is $frac72^6$
probability self-study combinatorics
asked 1 hour ago
Dhamnekar Winod
361210
add a comment |Â
up vote
1
down vote
favorite
I want to find the probability that in ten tossings a coin falls heads at least five times in succession. Is there any formula to compute this probability?
Answer provided is $frac72^6$
probability self-study combinatorics
asked 1 hour ago
Dhamnekar Winod
361210
Is it a homework?
– Tim♦
1 hour ago
@Tim You may say it is homework. Because I have taken this question from Williams Feller's book on Probability.
– Dhamnekar Winod
1 hour ago
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I want to find the probability that in ten tossings a coin falls heads at least five times in succession. Is there any formula to compute this probability?
Answer provided is $frac72^6$
probability self-study combinatorics
asked 1 hour ago
Dhamnekar Winod
361210
I want to find the probability that in ten tossings a coin falls heads at least five times in succession. Is there any formula to compute this probability?
Answer provided is $frac72^6$
probability self-study combinatorics
probability self-study combinatorics
asked 1 hour ago
Dhamnekar Winod
361210
asked 1 hour ago
Dhamnekar Winod
361210
edited 1 hour ago
asked 1 hour ago
Dhamnekar Winod
361210
asked 1 hour ago
Dhamnekar Winod
361210
asked 1 hour ago
Dhamnekar Winod
361210
361210
Is it a homework?
– Tim♦
1 hour ago
@Tim You may say it is homework. Because I have taken this question from Williams Feller's book on Probability.
– Dhamnekar Winod
1 hour ago
add a comment |Â
Is it a homework?
– Tim♦
1 hour ago
@Tim You may say it is homework. Because I have taken this question from Williams Feller's book on Probability.
– Dhamnekar Winod
1 hour ago
Is it a homework?
– Tim♦
1 hour ago
Is it a homework?
– Tim♦
1 hour ago
@Tim You may say it is homework. Because I have taken this question from Williams Feller's book on Probability.
– Dhamnekar Winod
1 hour ago
@Tim You may say it is homework. Because I have taken this question from Williams Feller's book on Probability.
– Dhamnekar Winod
1 hour ago
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
2
down vote
During writing this answer I saw that the question has already been answered but I will post my solution nonetheless.
The options are:
$(X,X,X,X,X,-,-,-,-,)$
$(-,X,X,X,X,X,-,-,-,-,)$
$(-,-,X,X,X,X,X,-,-,-,)$
$(-,-,-,X,X,X,X,X,-,-)$
$(-,-,-,-,X,X,X,X,X,-)$
$(-,-,-,-,-,X,X,X,X,X)$
where $X$ denotes heads and $-$ is a placeholder.
Line: there are $2^5$ options of obtaining at least those 5 heads at the first 5 tosses: $2^5$
Line: There are $2^4$ possibilities for the last 4, but only 1 for the first (which must be tails) because if the first was $X$ we would be in line 1: therefore $2^4$
Line: There are $2^3$ possibilities for the last three, but we can vary only the first one, the second one has to be tails as otherwise we are again in one of the upper cases: $2^3cdot 2^1$
Line: There are $2^2$ for the last two and $2^2$ for the first 3, as we can vary the first and the second one without falling into one of the upper cases (the fourth one being tails again)
Line: $2^1$ for the last one and $2^3$ for the first, the second and the third
Line: We can vary the first, second, third and fourth arbitrarily (as long as fifth is tails) without falling into the upper cases.
therefore we obtain $$frac2^5+5cdot2^42^10= frac72^6$$
answered 24 mins ago
0rangetree
551212
add a comment |Â
up vote
1
down vote
Corrected answer after Orangetree pointed out I forgot to take into account events were not mutually exclusive.
You need to think about how many different coin tossing sequences give at least $5$ consecutive heads, and how many coin tossing sequences there are in total, and then take the ratio of the two.
Clearly there is $2^10$ coin tossing sequences in total, since each of $10$ coins have $2$ possible outcomes.
To see how many sequences have $5$ consecutive heads consider the following. In any sequence
$$X-X-X-X-X-X-X-X-X-X$$
We can put the $5$ consecutive heads in $5$ places. This can be seen by putting it at the start of thee sequence, and then moving it over step by step.
$$H-H-H-H-H-X-X-X-X-X$$
$$X-H-H-H-H-H-X-X-X-X$$
$$X-X-H-H-H-H-H-X-X-X$$
$$X-X-X-H-H-H-H-H-X-X$$
$$X-X-X-X-H-H-H-H-H-X$$
$$X-X-X-X-X-H-H-H-H-H$$
For the first scenario, we have $2^5$ possible sequences since the remaining coins can be either heads or tails without affecting the result.
For the second scenario, we can only take $2^4$ since we need to fix the first coin as tails to avoid counting sequences more than once.
Similarly, for the third scenario we can count $2^4$ un-counted sequences, as we fix the coin preceeding the sequence as tails.. and so on
This gives total number of ways
$$2^5 + 2^4 + 2^4 + 2^4 + 2^4 + 2^4 = 2^4(2+1+1+1+1+1) = 2^4times 7$$
Hence the probability should be
$$frac2^4 times 72^10 = frac72^6$$
To actually answer your question if a general formula exists, the reasoning above can be extended to give the probability of at least $m$ heads out of $n$ coin tosses as
$$frac2^m-1(2+(n-m))2^n = frac2+n-m2^n-m+1$$
answered 1 hour ago
Xiaomi
1135
New contributor
Xiaomi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Not "five times in succession". At least five times in succession.
– Alvaro Fuentes
1 hour ago
2
Why would that affect the result? The remaining 5 coins can also be heads. The above gives the probability of at least 5 heads, not 5 heads exactly.
– Xiaomi
1 hour ago
the events are not mutually exclusive so you cannot simply add their probabilities
– 0rangetree
58 mins ago
Oh I see your logic. But you're counting some results more than once, as @0rangetree points out.
– Alvaro Fuentes
56 mins ago
Oh I see now. Thanks @0rangetree
– Xiaomi
56 mins ago
 |Â
show 1 more comment
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
During writing this answer I saw that the question has already been answered but I will post my solution nonetheless.
The options are:
$(X,X,X,X,X,-,-,-,-,)$
$(-,X,X,X,X,X,-,-,-,-,)$
$(-,-,X,X,X,X,X,-,-,-,)$
$(-,-,-,X,X,X,X,X,-,-)$
$(-,-,-,-,X,X,X,X,X,-)$
$(-,-,-,-,-,X,X,X,X,X)$
where $X$ denotes heads and $-$ is a placeholder.
Line: there are $2^5$ options of obtaining at least those 5 heads at the first 5 tosses: $2^5$
Line: There are $2^4$ possibilities for the last 4, but only 1 for the first (which must be tails) because if the first was $X$ we would be in line 1: therefore $2^4$
Line: There are $2^3$ possibilities for the last three, but we can vary only the first one, the second one has to be tails as otherwise we are again in one of the upper cases: $2^3cdot 2^1$
Line: There are $2^2$ for the last two and $2^2$ for the first 3, as we can vary the first and the second one without falling into one of the upper cases (the fourth one being tails again)
Line: $2^1$ for the last one and $2^3$ for the first, the second and the third
Line: We can vary the first, second, third and fourth arbitrarily (as long as fifth is tails) without falling into the upper cases.
therefore we obtain $$frac2^5+5cdot2^42^10= frac72^6$$
answered 24 mins ago
0rangetree
551212
add a comment |Â
up vote
2
down vote
During writing this answer I saw that the question has already been answered but I will post my solution nonetheless.
The options are:
$(X,X,X,X,X,-,-,-,-,)$
$(-,X,X,X,X,X,-,-,-,-,)$
$(-,-,X,X,X,X,X,-,-,-,)$
$(-,-,-,X,X,X,X,X,-,-)$
$(-,-,-,-,X,X,X,X,X,-)$
$(-,-,-,-,-,X,X,X,X,X)$
where $X$ denotes heads and $-$ is a placeholder.
Line: there are $2^5$ options of obtaining at least those 5 heads at the first 5 tosses: $2^5$
Line: There are $2^4$ possibilities for the last 4, but only 1 for the first (which must be tails) because if the first was $X$ we would be in line 1: therefore $2^4$
Line: There are $2^3$ possibilities for the last three, but we can vary only the first one, the second one has to be tails as otherwise we are again in one of the upper cases: $2^3cdot 2^1$
Line: There are $2^2$ for the last two and $2^2$ for the first 3, as we can vary the first and the second one without falling into one of the upper cases (the fourth one being tails again)
Line: $2^1$ for the last one and $2^3$ for the first, the second and the third
Line: We can vary the first, second, third and fourth arbitrarily (as long as fifth is tails) without falling into the upper cases.
therefore we obtain $$frac2^5+5cdot2^42^10= frac72^6$$
answered 24 mins ago
0rangetree
551212
add a comment |Â
up vote
2
down vote
up vote
2
down vote
During writing this answer I saw that the question has already been answered but I will post my solution nonetheless.
The options are:
$(X,X,X,X,X,-,-,-,-,)$
$(-,X,X,X,X,X,-,-,-,-,)$
$(-,-,X,X,X,X,X,-,-,-,)$
$(-,-,-,X,X,X,X,X,-,-)$
$(-,-,-,-,X,X,X,X,X,-)$
$(-,-,-,-,-,X,X,X,X,X)$
where $X$ denotes heads and $-$ is a placeholder.
Line: there are $2^5$ options of obtaining at least those 5 heads at the first 5 tosses: $2^5$
Line: There are $2^4$ possibilities for the last 4, but only 1 for the first (which must be tails) because if the first was $X$ we would be in line 1: therefore $2^4$
Line: There are $2^3$ possibilities for the last three, but we can vary only the first one, the second one has to be tails as otherwise we are again in one of the upper cases: $2^3cdot 2^1$
Line: There are $2^2$ for the last two and $2^2$ for the first 3, as we can vary the first and the second one without falling into one of the upper cases (the fourth one being tails again)
Line: $2^1$ for the last one and $2^3$ for the first, the second and the third
Line: We can vary the first, second, third and fourth arbitrarily (as long as fifth is tails) without falling into the upper cases.
therefore we obtain $$frac2^5+5cdot2^42^10= frac72^6$$
answered 24 mins ago
0rangetree
551212
During writing this answer I saw that the question has already been answered but I will post my solution nonetheless.
The options are:
$(X,X,X,X,X,-,-,-,-,)$
$(-,X,X,X,X,X,-,-,-,-,)$
$(-,-,X,X,X,X,X,-,-,-,)$
$(-,-,-,X,X,X,X,X,-,-)$
$(-,-,-,-,X,X,X,X,X,-)$
$(-,-,-,-,-,X,X,X,X,X)$
where $X$ denotes heads and $-$ is a placeholder.
Line: there are $2^5$ options of obtaining at least those 5 heads at the first 5 tosses: $2^5$
Line: There are $2^4$ possibilities for the last 4, but only 1 for the first (which must be tails) because if the first was $X$ we would be in line 1: therefore $2^4$
Line: There are $2^3$ possibilities for the last three, but we can vary only the first one, the second one has to be tails as otherwise we are again in one of the upper cases: $2^3cdot 2^1$
Line: There are $2^2$ for the last two and $2^2$ for the first 3, as we can vary the first and the second one without falling into one of the upper cases (the fourth one being tails again)
Line: $2^1$ for the last one and $2^3$ for the first, the second and the third
Line: We can vary the first, second, third and fourth arbitrarily (as long as fifth is tails) without falling into the upper cases.
therefore we obtain $$frac2^5+5cdot2^42^10= frac72^6$$
answered 24 mins ago
0rangetree
551212
answered 24 mins ago
0rangetree
551212
answered 24 mins ago
0rangetree
551212
answered 24 mins ago
0rangetree
551212
551212
add a comment |Â
add a comment |Â
up vote
1
down vote
Corrected answer after Orangetree pointed out I forgot to take into account events were not mutually exclusive.
You need to think about how many different coin tossing sequences give at least $5$ consecutive heads, and how many coin tossing sequences there are in total, and then take the ratio of the two.
Clearly there is $2^10$ coin tossing sequences in total, since each of $10$ coins have $2$ possible outcomes.
To see how many sequences have $5$ consecutive heads consider the following. In any sequence
$$X-X-X-X-X-X-X-X-X-X$$
We can put the $5$ consecutive heads in $5$ places. This can be seen by putting it at the start of thee sequence, and then moving it over step by step.
$$H-H-H-H-H-X-X-X-X-X$$
$$X-H-H-H-H-H-X-X-X-X$$
$$X-X-H-H-H-H-H-X-X-X$$
$$X-X-X-H-H-H-H-H-X-X$$
$$X-X-X-X-H-H-H-H-H-X$$
$$X-X-X-X-X-H-H-H-H-H$$
For the first scenario, we have $2^5$ possible sequences since the remaining coins can be either heads or tails without affecting the result.
For the second scenario, we can only take $2^4$ since we need to fix the first coin as tails to avoid counting sequences more than once.
Similarly, for the third scenario we can count $2^4$ un-counted sequences, as we fix the coin preceeding the sequence as tails.. and so on
This gives total number of ways
$$2^5 + 2^4 + 2^4 + 2^4 + 2^4 + 2^4 = 2^4(2+1+1+1+1+1) = 2^4times 7$$
Hence the probability should be
$$frac2^4 times 72^10 = frac72^6$$
To actually answer your question if a general formula exists, the reasoning above can be extended to give the probability of at least $m$ heads out of $n$ coin tosses as
$$frac2^m-1(2+(n-m))2^n = frac2+n-m2^n-m+1$$
answered 1 hour ago
Xiaomi
1135
New contributor
Xiaomi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Not "five times in succession". At least five times in succession.
– Alvaro Fuentes
1 hour ago
2
Why would that affect the result? The remaining 5 coins can also be heads. The above gives the probability of at least 5 heads, not 5 heads exactly.
– Xiaomi
1 hour ago
the events are not mutually exclusive so you cannot simply add their probabilities
– 0rangetree
58 mins ago
Oh I see your logic. But you're counting some results more than once, as @0rangetree points out.
– Alvaro Fuentes
56 mins ago
Oh I see now. Thanks @0rangetree
– Xiaomi
56 mins ago
 |Â
show 1 more comment
up vote
1
down vote
Corrected answer after Orangetree pointed out I forgot to take into account events were not mutually exclusive.
You need to think about how many different coin tossing sequences give at least $5$ consecutive heads, and how many coin tossing sequences there are in total, and then take the ratio of the two.
Clearly there is $2^10$ coin tossing sequences in total, since each of $10$ coins have $2$ possible outcomes.
To see how many sequences have $5$ consecutive heads consider the following. In any sequence
$$X-X-X-X-X-X-X-X-X-X$$
We can put the $5$ consecutive heads in $5$ places. This can be seen by putting it at the start of thee sequence, and then moving it over step by step.
$$H-H-H-H-H-X-X-X-X-X$$
$$X-H-H-H-H-H-X-X-X-X$$
$$X-X-H-H-H-H-H-X-X-X$$
$$X-X-X-H-H-H-H-H-X-X$$
$$X-X-X-X-H-H-H-H-H-X$$
$$X-X-X-X-X-H-H-H-H-H$$
For the first scenario, we have $2^5$ possible sequences since the remaining coins can be either heads or tails without affecting the result.
For the second scenario, we can only take $2^4$ since we need to fix the first coin as tails to avoid counting sequences more than once.
Similarly, for the third scenario we can count $2^4$ un-counted sequences, as we fix the coin preceeding the sequence as tails.. and so on
This gives total number of ways
$$2^5 + 2^4 + 2^4 + 2^4 + 2^4 + 2^4 = 2^4(2+1+1+1+1+1) = 2^4times 7$$
Hence the probability should be
$$frac2^4 times 72^10 = frac72^6$$
To actually answer your question if a general formula exists, the reasoning above can be extended to give the probability of at least $m$ heads out of $n$ coin tosses as
$$frac2^m-1(2+(n-m))2^n = frac2+n-m2^n-m+1$$
answered 1 hour ago
Xiaomi
1135
New contributor
Xiaomi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Not "five times in succession". At least five times in succession.
– Alvaro Fuentes
1 hour ago
2
Why would that affect the result? The remaining 5 coins can also be heads. The above gives the probability of at least 5 heads, not 5 heads exactly.
– Xiaomi
1 hour ago
the events are not mutually exclusive so you cannot simply add their probabilities
– 0rangetree
58 mins ago
Oh I see your logic. But you're counting some results more than once, as @0rangetree points out.
– Alvaro Fuentes
56 mins ago
Oh I see now. Thanks @0rangetree
– Xiaomi
56 mins ago
 |Â
show 1 more comment
up vote
1
down vote
up vote
1
down vote
Corrected answer after Orangetree pointed out I forgot to take into account events were not mutually exclusive.
You need to think about how many different coin tossing sequences give at least $5$ consecutive heads, and how many coin tossing sequences there are in total, and then take the ratio of the two.
Clearly there is $2^10$ coin tossing sequences in total, since each of $10$ coins have $2$ possible outcomes.
To see how many sequences have $5$ consecutive heads consider the following. In any sequence
$$X-X-X-X-X-X-X-X-X-X$$
We can put the $5$ consecutive heads in $5$ places. This can be seen by putting it at the start of thee sequence, and then moving it over step by step.
$$H-H-H-H-H-X-X-X-X-X$$
$$X-H-H-H-H-H-X-X-X-X$$
$$X-X-H-H-H-H-H-X-X-X$$
$$X-X-X-H-H-H-H-H-X-X$$
$$X-X-X-X-H-H-H-H-H-X$$
$$X-X-X-X-X-H-H-H-H-H$$
For the first scenario, we have $2^5$ possible sequences since the remaining coins can be either heads or tails without affecting the result.
For the second scenario, we can only take $2^4$ since we need to fix the first coin as tails to avoid counting sequences more than once.
Similarly, for the third scenario we can count $2^4$ un-counted sequences, as we fix the coin preceeding the sequence as tails.. and so on
This gives total number of ways
$$2^5 + 2^4 + 2^4 + 2^4 + 2^4 + 2^4 = 2^4(2+1+1+1+1+1) = 2^4times 7$$
Hence the probability should be
$$frac2^4 times 72^10 = frac72^6$$
To actually answer your question if a general formula exists, the reasoning above can be extended to give the probability of at least $m$ heads out of $n$ coin tosses as
$$frac2^m-1(2+(n-m))2^n = frac2+n-m2^n-m+1$$
answered 1 hour ago
Xiaomi
1135
New contributor
Xiaomi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Corrected answer after Orangetree pointed out I forgot to take into account events were not mutually exclusive.
You need to think about how many different coin tossing sequences give at least $5$ consecutive heads, and how many coin tossing sequences there are in total, and then take the ratio of the two.
Clearly there is $2^10$ coin tossing sequences in total, since each of $10$ coins have $2$ possible outcomes.
To see how many sequences have $5$ consecutive heads consider the following. In any sequence
$$X-X-X-X-X-X-X-X-X-X$$
We can put the $5$ consecutive heads in $5$ places. This can be seen by putting it at the start of thee sequence, and then moving it over step by step.
$$H-H-H-H-H-X-X-X-X-X$$
$$X-H-H-H-H-H-X-X-X-X$$
$$X-X-H-H-H-H-H-X-X-X$$
$$X-X-X-H-H-H-H-H-X-X$$
$$X-X-X-X-H-H-H-H-H-X$$
$$X-X-X-X-X-H-H-H-H-H$$
For the first scenario, we have $2^5$ possible sequences since the remaining coins can be either heads or tails without affecting the result.
For the second scenario, we can only take $2^4$ since we need to fix the first coin as tails to avoid counting sequences more than once.
Similarly, for the third scenario we can count $2^4$ un-counted sequences, as we fix the coin preceeding the sequence as tails.. and so on
This gives total number of ways
$$2^5 + 2^4 + 2^4 + 2^4 + 2^4 + 2^4 = 2^4(2+1+1+1+1+1) = 2^4times 7$$
Hence the probability should be
$$frac2^4 times 72^10 = frac72^6$$
To actually answer your question if a general formula exists, the reasoning above can be extended to give the probability of at least $m$ heads out of $n$ coin tosses as
$$frac2^m-1(2+(n-m))2^n = frac2+n-m2^n-m+1$$
answered 1 hour ago
Xiaomi
1135
New contributor
Xiaomi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 22 mins ago
answered 1 hour ago
Xiaomi
1135
New contributor
Xiaomi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 1 hour ago
Xiaomi
1135
answered 1 hour ago
Xiaomi
1135
1135
New contributor
Xiaomi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Xiaomi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Xiaomi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Not "five times in succession". At least five times in succession.
– Alvaro Fuentes
1 hour ago
2
Why would that affect the result? The remaining 5 coins can also be heads. The above gives the probability of at least 5 heads, not 5 heads exactly.
– Xiaomi
1 hour ago
the events are not mutually exclusive so you cannot simply add their probabilities
– 0rangetree
58 mins ago
Oh I see your logic. But you're counting some results more than once, as @0rangetree points out.
– Alvaro Fuentes
56 mins ago
Oh I see now. Thanks @0rangetree
– Xiaomi
56 mins ago
 |Â
show 1 more comment
Not "five times in succession". At least five times in succession.
– Alvaro Fuentes
1 hour ago
2
Why would that affect the result? The remaining 5 coins can also be heads. The above gives the probability of at least 5 heads, not 5 heads exactly.
– Xiaomi
1 hour ago
the events are not mutually exclusive so you cannot simply add their probabilities
– 0rangetree
58 mins ago
Oh I see your logic. But you're counting some results more than once, as @0rangetree points out.
– Alvaro Fuentes
56 mins ago
Oh I see now. Thanks @0rangetree
– Xiaomi
56 mins ago
Not "five times in succession". At least five times in succession.
– Alvaro Fuentes
1 hour ago
Not "five times in succession". At least five times in succession.
– Alvaro Fuentes
1 hour ago
2
2
Why would that affect the result? The remaining 5 coins can also be heads. The above gives the probability of at least 5 heads, not 5 heads exactly.
– Xiaomi
1 hour ago
Why would that affect the result? The remaining 5 coins can also be heads. The above gives the probability of at least 5 heads, not 5 heads exactly.
– Xiaomi
1 hour ago
the events are not mutually exclusive so you cannot simply add their probabilities
– 0rangetree
58 mins ago
the events are not mutually exclusive so you cannot simply add their probabilities
– 0rangetree
58 mins ago
Oh I see your logic. But you're counting some results more than once, as @0rangetree points out.
– Alvaro Fuentes
56 mins ago
Oh I see your logic. But you're counting some results more than once, as @0rangetree points out.
– Alvaro Fuentes
56 mins ago
Oh I see now. Thanks @0rangetree
– Xiaomi
56 mins ago
Oh I see now. Thanks @0rangetree
– Xiaomi
56 mins ago
 |Â
show 1 more comment
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Is it a homework?
– Tim♦
1 hour ago
@Tim You may say it is homework. Because I have taken this question from Williams Feller's book on Probability.
– Dhamnekar Winod
1 hour ago