A coin, having probability p of landing heads and probability of q=(1-p) of landing on heads.
$begingroup$
A coin, having probability p of landing heads and probability of q=(1-p) of landing on heads. It is continuously flipped until at least one head and one tail have been flipped.
This is not part of a homework assignment. I am studying for a final and don't understand the professors solutions.
a.) Find the expected number of flips needed.
Since this is clearly geometric, I would think the solution would be:
E(N)=$Sigma_{i=0}^{infty}ip^{n-1}q+Sigma_{i=0}^{n}iq^{n-1}p=frac{1}{q}+frac{1}{p}$.
However, I am completely wrong.
The answer is
E(N)=$p(1+frac{1}{q})+q(1+frac{1}{p})$
For example, consider we flip for heads first. Then we have:
E(N|H)=$p+pSigma_{i=0}^{infty}np^{n-1}q$... 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 probability-theory probability-distributions expected-value
asked 2 hours ago
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.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
A coin, having probability p of landing heads and probability of q=(1-p) of landing on heads. It is continuously flipped until at least one head and one tail have been flipped.
This is not part of a homework assignment. I am studying for a final and don't understand the professors solutions.
a.) Find the expected number of flips needed.
Since this is clearly geometric, I would think the solution would be:
E(N)=$Sigma_{i=0}^{infty}ip^{n-1}q+Sigma_{i=0}^{n}iq^{n-1}p=frac{1}{q}+frac{1}{p}$.
However, I am completely wrong.
The answer is
E(N)=$p(1+frac{1}{q})+q(1+frac{1}{p})$
For example, consider we flip for heads first. Then we have:
E(N|H)=$p+pSigma_{i=0}^{infty}np^{n-1}q$... 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 probability-theory probability-distributions expected-value
asked 2 hours ago
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
2 hours ago
$begingroup$
In both the title and first paragraph it appears there is $0$ chance of landing tails, so you will wait forever.
$endgroup$
– Ross Millikan
2 hours ago
$begingroup$
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
1 hour ago
add a comment |
$begingroup$
A coin, having probability p of landing heads and probability of q=(1-p) of landing on heads. It is continuously flipped until at least one head and one tail have been flipped.
This is not part of a homework assignment. I am studying for a final and don't understand the professors solutions.
a.) Find the expected number of flips needed.
Since this is clearly geometric, I would think the solution would be:
E(N)=$Sigma_{i=0}^{infty}ip^{n-1}q+Sigma_{i=0}^{n}iq^{n-1}p=frac{1}{q}+frac{1}{p}$.
However, I am completely wrong.
The answer is
E(N)=$p(1+frac{1}{q})+q(1+frac{1}{p})$
For example, consider we flip for heads first. Then we have:
E(N|H)=$p+pSigma_{i=0}^{infty}np^{n-1}q$... 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 probability-theory probability-distributions expected-value
asked 2 hours ago
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 heads and probability of q=(1-p) of landing on heads. It is continuously flipped until at least one head and one tail have been flipped.
This is not part of a homework assignment. I am studying for a final and don't understand the professors solutions.
a.) Find the expected number of flips needed.
Since this is clearly geometric, I would think the solution would be:
E(N)=$Sigma_{i=0}^{infty}ip^{n-1}q+Sigma_{i=0}^{n}iq^{n-1}p=frac{1}{q}+frac{1}{p}$.
However, I am completely wrong.
The answer is
E(N)=$p(1+frac{1}{q})+q(1+frac{1}{p})$
For example, consider we flip for heads first. Then we have:
E(N|H)=$p+pSigma_{i=0}^{infty}np^{n-1}q$... 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 probability-theory probability-distributions expected-value
probability probability-theory probability-distributions expected-value
asked 2 hours ago
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 2 hours ago
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.
edited 2 hours ago
Mistah White
asked 2 hours ago
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 2 hours ago
Mistah WhiteMistah White
62
asked 2 hours ago
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.
Mistah White is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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
2 hours ago
$begingroup$
In both the title and first paragraph it appears there is $0$ chance of landing tails, so you will wait forever.
$endgroup$
– Ross Millikan
2 hours ago
$begingroup$
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
1 hour ago
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
2 hours ago
$begingroup$
In both the title and first paragraph it appears there is $0$ chance of landing tails, so you will wait forever.
$endgroup$
– Ross Millikan
2 hours ago
$begingroup$
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
1 hour ago
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
2 hours ago
$begingroup$
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
2 hours ago
$begingroup$
In both the title and first paragraph it appears there is $0$ chance of landing tails, so you will wait forever.
$endgroup$
– Ross Millikan
2 hours ago
$begingroup$
In both the title and first paragraph it appears there is $0$ chance of landing tails, so you will wait forever.
$endgroup$
– Ross Millikan
2 hours ago
$begingroup$
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
1 hour ago
$begingroup$
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
1 hour ago
add a comment |
2 Answers
2
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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 2 hours ago
Peter ForemanPeter Foreman
7,8931320
$endgroup$
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 2 hours ago
leonbloyleonbloy
42.5k647108
$endgroup$
add a comment |
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2 Answers
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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 2 hours ago
Peter ForemanPeter Foreman
7,8931320
$endgroup$
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 2 hours ago
Peter ForemanPeter Foreman
7,8931320
$endgroup$
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 2 hours ago
Peter ForemanPeter Foreman
7,8931320
$endgroup$
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 2 hours ago
Peter ForemanPeter Foreman
7,8931320
edited 2 hours ago
answered 2 hours ago
Peter ForemanPeter Foreman
7,8931320
answered 2 hours ago
Peter ForemanPeter Foreman
7,8931320
answered 2 hours ago
Peter ForemanPeter Foreman
7,8931320
7,8931320
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 2 hours ago
leonbloyleonbloy
42.5k647108
$endgroup$
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 2 hours ago
leonbloyleonbloy
42.5k647108
$endgroup$
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 2 hours ago
leonbloyleonbloy
42.5k647108
$endgroup$
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 2 hours ago
leonbloyleonbloy
42.5k647108
edited 2 hours ago
answered 2 hours ago
leonbloyleonbloy
42.5k647108
answered 2 hours ago
leonbloyleonbloy
42.5k647108
answered 2 hours ago
leonbloyleonbloy
42.5k647108
42.5k647108
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.
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
2 hours ago
$begingroup$
In both the title and first paragraph it appears there is $0$ chance of landing tails, so you will wait forever.
$endgroup$
– Ross Millikan
2 hours ago
$begingroup$
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
1 hour ago