Finding the expected number of flips needed for a coin having probability $p$ of landing on heads
$begingroup$
A coin, having probability $p$ of landing on heads and probability of $q=1-p$ of landing on tails. It is continuously flipped until at least one head and one tail have been flipped.
a) Find the expected number of flips needed.
This is not part of a homework assignment. I am studying for a final and don't understand the professors solutions. Since this is clearly geometric, I would think the solution would be:
$$E(N)=sum_{i=0}^{infty}ip^{n-1}q+sum_{i=0}^{infty}iq^{n-1}p=frac{1}{q}+frac{1}{p}.$$
However, I am completely wrong. The answer is
$$E(N)=pleft(1+frac{1}{q}right)+qleft(1+frac{1}{p}right).$$
For example, consider we flip for heads first. Then we have
$$E(Nmid H)=p+psum_{i=0}^{infty}np^{n-1}q.$$
I am not sure why this makes sense. I am not entirely sure why we have an added $1$ and a factored $p$, $q$. Could someone carefully explain why it makes sense that this is the right answer?
probability expected-value
asked Apr 16 at 22:28
Mistah WhiteMistah White
62
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Mistah White is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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$endgroup$
add a comment |
$begingroup$
A coin, having probability $p$ of landing on heads and probability of $q=1-p$ of landing on tails. It is continuously flipped until at least one head and one tail have been flipped.
a) Find the expected number of flips needed.
This is not part of a homework assignment. I am studying for a final and don't understand the professors solutions. Since this is clearly geometric, I would think the solution would be:
$$E(N)=sum_{i=0}^{infty}ip^{n-1}q+sum_{i=0}^{infty}iq^{n-1}p=frac{1}{q}+frac{1}{p}.$$
However, I am completely wrong. The answer is
$$E(N)=pleft(1+frac{1}{q}right)+qleft(1+frac{1}{p}right).$$
For example, consider we flip for heads first. Then we have
$$E(Nmid H)=p+psum_{i=0}^{infty}np^{n-1}q.$$
I am not sure why this makes sense. I am not entirely sure why we have an added $1$ and a factored $p$, $q$. Could someone carefully explain why it makes sense that this is the right answer?
probability expected-value
asked Apr 16 at 22:28
Mistah WhiteMistah White
62
New contributor
Mistah White is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
2
$begingroup$
It's all a question of the first toss. If it is $H$ then you just get one more than the expected time to get a $T$, if it is $T$ then you just get one more than the expected time to get $H$. Your method is incorrect because the expected number of tosses needed to get one of the two is $1$.
$endgroup$
– lulu
Apr 16 at 22:30
3
$begingroup$
In both the title and first paragraph it appears there is $0$ chance of landing tails, so you will wait forever.
$endgroup$
– Ross Millikan
Apr 16 at 22:32
$begingroup$
Note: your sums are hard to follow. What's $n$? The upper limit of the sums should be $infty$, the exponent of the probability ought to be a simple function of $i$. Done correctly, your method ought to work (though it's easier to do it the other way).
$endgroup$
– lulu
Apr 16 at 22:58
add a comment |
$begingroup$
A coin, having probability $p$ of landing on heads and probability of $q=1-p$ of landing on tails. It is continuously flipped until at least one head and one tail have been flipped.
a) Find the expected number of flips needed.
This is not part of a homework assignment. I am studying for a final and don't understand the professors solutions. Since this is clearly geometric, I would think the solution would be:
$$E(N)=sum_{i=0}^{infty}ip^{n-1}q+sum_{i=0}^{infty}iq^{n-1}p=frac{1}{q}+frac{1}{p}.$$
However, I am completely wrong. The answer is
$$E(N)=pleft(1+frac{1}{q}right)+qleft(1+frac{1}{p}right).$$
For example, consider we flip for heads first. Then we have
$$E(Nmid H)=p+psum_{i=0}^{infty}np^{n-1}q.$$
I am not sure why this makes sense. I am not entirely sure why we have an added $1$ and a factored $p$, $q$. Could someone carefully explain why it makes sense that this is the right answer?
probability expected-value
asked Apr 16 at 22:28
Mistah WhiteMistah White
62
New contributor
Mistah White is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
A coin, having probability $p$ of landing on heads and probability of $q=1-p$ of landing on tails. It is continuously flipped until at least one head and one tail have been flipped.
a) Find the expected number of flips needed.
This is not part of a homework assignment. I am studying for a final and don't understand the professors solutions. Since this is clearly geometric, I would think the solution would be:
$$E(N)=sum_{i=0}^{infty}ip^{n-1}q+sum_{i=0}^{infty}iq^{n-1}p=frac{1}{q}+frac{1}{p}.$$
However, I am completely wrong. The answer is
$$E(N)=pleft(1+frac{1}{q}right)+qleft(1+frac{1}{p}right).$$
For example, consider we flip for heads first. Then we have
$$E(Nmid H)=p+psum_{i=0}^{infty}np^{n-1}q.$$
I am not sure why this makes sense. I am not entirely sure why we have an added $1$ and a factored $p$, $q$. Could someone carefully explain why it makes sense that this is the right answer?
probability expected-value
probability expected-value
asked Apr 16 at 22:28
Mistah WhiteMistah White
62
New contributor
Mistah White is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Apr 16 at 22:28
Mistah WhiteMistah White
62
New contributor
Mistah White is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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edited Apr 17 at 16:36
Mistah White
asked Apr 16 at 22:28
Mistah WhiteMistah White
62
New contributor
Mistah White is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Apr 16 at 22:28
Mistah WhiteMistah White
62
asked Apr 16 at 22:28
Mistah WhiteMistah White
62
62
New contributor
Mistah White is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Mistah White is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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Check out our Code of Conduct.
2
$begingroup$
It's all a question of the first toss. If it is $H$ then you just get one more than the expected time to get a $T$, if it is $T$ then you just get one more than the expected time to get $H$. Your method is incorrect because the expected number of tosses needed to get one of the two is $1$.
$endgroup$
– lulu
Apr 16 at 22:30
3
$begingroup$
In both the title and first paragraph it appears there is $0$ chance of landing tails, so you will wait forever.
$endgroup$
– Ross Millikan
Apr 16 at 22:32
$begingroup$
Note: your sums are hard to follow. What's $n$? The upper limit of the sums should be $infty$, the exponent of the probability ought to be a simple function of $i$. Done correctly, your method ought to work (though it's easier to do it the other way).
$endgroup$
– lulu
Apr 16 at 22:58
add a comment |
2
$begingroup$
It's all a question of the first toss. If it is $H$ then you just get one more than the expected time to get a $T$, if it is $T$ then you just get one more than the expected time to get $H$. Your method is incorrect because the expected number of tosses needed to get one of the two is $1$.
$endgroup$
– lulu
Apr 16 at 22:30
3
$begingroup$
In both the title and first paragraph it appears there is $0$ chance of landing tails, so you will wait forever.
$endgroup$
– Ross Millikan
Apr 16 at 22:32
$begingroup$
Note: your sums are hard to follow. What's $n$? The upper limit of the sums should be $infty$, the exponent of the probability ought to be a simple function of $i$. Done correctly, your method ought to work (though it's easier to do it the other way).
$endgroup$
– lulu
Apr 16 at 22:58
2
2
$begingroup$
It's all a question of the first toss. If it is $H$ then you just get one more than the expected time to get a $T$, if it is $T$ then you just get one more than the expected time to get $H$. Your method is incorrect because the expected number of tosses needed to get one of the two is $1$.
$endgroup$
– lulu
Apr 16 at 22:30
$begingroup$
It's all a question of the first toss. If it is $H$ then you just get one more than the expected time to get a $T$, if it is $T$ then you just get one more than the expected time to get $H$. Your method is incorrect because the expected number of tosses needed to get one of the two is $1$.
$endgroup$
– lulu
Apr 16 at 22:30
3
3
$begingroup$
In both the title and first paragraph it appears there is $0$ chance of landing tails, so you will wait forever.
$endgroup$
– Ross Millikan
Apr 16 at 22:32
$begingroup$
In both the title and first paragraph it appears there is $0$ chance of landing tails, so you will wait forever.
$endgroup$
– Ross Millikan
Apr 16 at 22:32
$begingroup$
Note: your sums are hard to follow. What's $n$? The upper limit of the sums should be $infty$, the exponent of the probability ought to be a simple function of $i$. Done correctly, your method ought to work (though it's easier to do it the other way).
$endgroup$
– lulu
Apr 16 at 22:58
$begingroup$
Note: your sums are hard to follow. What's $n$? The upper limit of the sums should be $infty$, the exponent of the probability ought to be a simple function of $i$. Done correctly, your method ought to work (though it's easier to do it the other way).
$endgroup$
– lulu
Apr 16 at 22:58
add a comment |
2 Answers
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$begingroup$
If you get a head with probability $p$ then the expected number of throws is $1+E(X)$ where $X$ is a geometric distribution requiring a tail to be thrown with probability $q$ so $1+E(X)=1+frac1q$. Similarly if you throw a tail with probability $q$ then the expected number of throws is $1+E(Y)$ where $Y$ is a geometric distribution requiring a head to be thrown with probability $p$ so $1+E(Y)=1+frac1p$. This means that the overall expected number of throws is
$$pleft(1+frac1qright)+qleft(1+frac1pright)$$
because there is a probability $p$ that the expected number of throws is given by $1+E(X)$ and probability $q$ that it is given by $1+E(Y)$.
answered Apr 16 at 22:42
Peter ForemanPeter Foreman
8,5171321
$endgroup$
add a comment |
$begingroup$
Let $X$ be the time of the first head, and $Y$ the time of the first tail, and $W$ the first time when a head and a tail has been flipped.
You are right in assuming that $E[X]=frac{1}{p}$ and $E[Y]=frac{1}{q}$, But you are wrong in assuming that $W=X+Y$, that's simply not true, actually $W=max(X,Y)$.
A possible approach. Let $A$ be the indicator variable of the event: "first coin was a head" (hence $X=1$).
Then use $$E[W]=E[E[W | A ]] = P(A=1) E[W|A=1]+P(A=0) E[W|A=0]=\=p(E[Y]+1)+q(E[X]+1)$$
answered Apr 16 at 22:43
leonbloyleonbloy
42.6k647108
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2 Answers
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$begingroup$
If you get a head with probability $p$ then the expected number of throws is $1+E(X)$ where $X$ is a geometric distribution requiring a tail to be thrown with probability $q$ so $1+E(X)=1+frac1q$. Similarly if you throw a tail with probability $q$ then the expected number of throws is $1+E(Y)$ where $Y$ is a geometric distribution requiring a head to be thrown with probability $p$ so $1+E(Y)=1+frac1p$. This means that the overall expected number of throws is
$$pleft(1+frac1qright)+qleft(1+frac1pright)$$
because there is a probability $p$ that the expected number of throws is given by $1+E(X)$ and probability $q$ that it is given by $1+E(Y)$.
answered Apr 16 at 22:42
Peter ForemanPeter Foreman
8,5171321
$endgroup$
add a comment |
$begingroup$
If you get a head with probability $p$ then the expected number of throws is $1+E(X)$ where $X$ is a geometric distribution requiring a tail to be thrown with probability $q$ so $1+E(X)=1+frac1q$. Similarly if you throw a tail with probability $q$ then the expected number of throws is $1+E(Y)$ where $Y$ is a geometric distribution requiring a head to be thrown with probability $p$ so $1+E(Y)=1+frac1p$. This means that the overall expected number of throws is
$$pleft(1+frac1qright)+qleft(1+frac1pright)$$
because there is a probability $p$ that the expected number of throws is given by $1+E(X)$ and probability $q$ that it is given by $1+E(Y)$.
answered Apr 16 at 22:42
Peter ForemanPeter Foreman
8,5171321
$endgroup$
add a comment |
$begingroup$
If you get a head with probability $p$ then the expected number of throws is $1+E(X)$ where $X$ is a geometric distribution requiring a tail to be thrown with probability $q$ so $1+E(X)=1+frac1q$. Similarly if you throw a tail with probability $q$ then the expected number of throws is $1+E(Y)$ where $Y$ is a geometric distribution requiring a head to be thrown with probability $p$ so $1+E(Y)=1+frac1p$. This means that the overall expected number of throws is
$$pleft(1+frac1qright)+qleft(1+frac1pright)$$
because there is a probability $p$ that the expected number of throws is given by $1+E(X)$ and probability $q$ that it is given by $1+E(Y)$.
answered Apr 16 at 22:42
Peter ForemanPeter Foreman
8,5171321
$endgroup$
If you get a head with probability $p$ then the expected number of throws is $1+E(X)$ where $X$ is a geometric distribution requiring a tail to be thrown with probability $q$ so $1+E(X)=1+frac1q$. Similarly if you throw a tail with probability $q$ then the expected number of throws is $1+E(Y)$ where $Y$ is a geometric distribution requiring a head to be thrown with probability $p$ so $1+E(Y)=1+frac1p$. This means that the overall expected number of throws is
$$pleft(1+frac1qright)+qleft(1+frac1pright)$$
because there is a probability $p$ that the expected number of throws is given by $1+E(X)$ and probability $q$ that it is given by $1+E(Y)$.
answered Apr 16 at 22:42
Peter ForemanPeter Foreman
8,5171321
edited Apr 16 at 22:53
answered Apr 16 at 22:42
Peter ForemanPeter Foreman
8,5171321
answered Apr 16 at 22:42
Peter ForemanPeter Foreman
8,5171321
answered Apr 16 at 22:42
Peter ForemanPeter Foreman
8,5171321
8,5171321
add a comment |
add a comment |
$begingroup$
Let $X$ be the time of the first head, and $Y$ the time of the first tail, and $W$ the first time when a head and a tail has been flipped.
You are right in assuming that $E[X]=frac{1}{p}$ and $E[Y]=frac{1}{q}$, But you are wrong in assuming that $W=X+Y$, that's simply not true, actually $W=max(X,Y)$.
A possible approach. Let $A$ be the indicator variable of the event: "first coin was a head" (hence $X=1$).
Then use $$E[W]=E[E[W | A ]] = P(A=1) E[W|A=1]+P(A=0) E[W|A=0]=\=p(E[Y]+1)+q(E[X]+1)$$
answered Apr 16 at 22:43
leonbloyleonbloy
42.6k647108
$endgroup$
add a comment |
$begingroup$
Let $X$ be the time of the first head, and $Y$ the time of the first tail, and $W$ the first time when a head and a tail has been flipped.
You are right in assuming that $E[X]=frac{1}{p}$ and $E[Y]=frac{1}{q}$, But you are wrong in assuming that $W=X+Y$, that's simply not true, actually $W=max(X,Y)$.
A possible approach. Let $A$ be the indicator variable of the event: "first coin was a head" (hence $X=1$).
Then use $$E[W]=E[E[W | A ]] = P(A=1) E[W|A=1]+P(A=0) E[W|A=0]=\=p(E[Y]+1)+q(E[X]+1)$$
answered Apr 16 at 22:43
leonbloyleonbloy
42.6k647108
$endgroup$
add a comment |
$begingroup$
Let $X$ be the time of the first head, and $Y$ the time of the first tail, and $W$ the first time when a head and a tail has been flipped.
You are right in assuming that $E[X]=frac{1}{p}$ and $E[Y]=frac{1}{q}$, But you are wrong in assuming that $W=X+Y$, that's simply not true, actually $W=max(X,Y)$.
A possible approach. Let $A$ be the indicator variable of the event: "first coin was a head" (hence $X=1$).
Then use $$E[W]=E[E[W | A ]] = P(A=1) E[W|A=1]+P(A=0) E[W|A=0]=\=p(E[Y]+1)+q(E[X]+1)$$
answered Apr 16 at 22:43
leonbloyleonbloy
42.6k647108
$endgroup$
Let $X$ be the time of the first head, and $Y$ the time of the first tail, and $W$ the first time when a head and a tail has been flipped.
You are right in assuming that $E[X]=frac{1}{p}$ and $E[Y]=frac{1}{q}$, But you are wrong in assuming that $W=X+Y$, that's simply not true, actually $W=max(X,Y)$.
A possible approach. Let $A$ be the indicator variable of the event: "first coin was a head" (hence $X=1$).
Then use $$E[W]=E[E[W | A ]] = P(A=1) E[W|A=1]+P(A=0) E[W|A=0]=\=p(E[Y]+1)+q(E[X]+1)$$
answered Apr 16 at 22:43
leonbloyleonbloy
42.6k647108
edited Apr 16 at 22:50
answered Apr 16 at 22:43
leonbloyleonbloy
42.6k647108
answered Apr 16 at 22:43
leonbloyleonbloy
42.6k647108
answered Apr 16 at 22:43
leonbloyleonbloy
42.6k647108
42.6k647108
add a comment |
add a comment |
Mistah White is a new contributor. Be nice, and check out our Code of Conduct.
Mistah White is a new contributor. Be nice, and check out our Code of Conduct.
Mistah White is a new contributor. Be nice, and check out our Code of Conduct.
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$begingroup$
It's all a question of the first toss. If it is $H$ then you just get one more than the expected time to get a $T$, if it is $T$ then you just get one more than the expected time to get $H$. Your method is incorrect because the expected number of tosses needed to get one of the two is $1$.
$endgroup$
– lulu
Apr 16 at 22:30
3
$begingroup$
In both the title and first paragraph it appears there is $0$ chance of landing tails, so you will wait forever.
$endgroup$
– Ross Millikan
Apr 16 at 22:32
$begingroup$
Note: your sums are hard to follow. What's $n$? The upper limit of the sums should be $infty$, the exponent of the probability ought to be a simple function of $i$. Done correctly, your method ought to work (though it's easier to do it the other way).
$endgroup$
– lulu
Apr 16 at 22:58