A coin, having probability p of landing heads and probability of q=(1-p) of landing on heads. Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Flipping a special coin: probability of getting heads equals the proportion of heads in the flips so farBiased coin flipped until $r$ heads appearBiased coin probabilityCoin-flipping experiment: the expected number of flips that land on headsWhy are odds of a coin landing heads $50%$ after $'n'$ consecutive headsWhat is the probability of a biased coin flipping heads (probability of heads is $frac 35$) exactly $65$ times in $100$ trials?Flipping rigged coin, calculating most common number of flips between headsChernoff bound probability: value of $n$ so that with probability $.999$ at least half of the coin flips come out headsFlip a coin 6 times. Probability with past results and probability without past results are different?Probability density function of flipping until heads and tails
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A coin, having probability p of landing heads and probability of q=(1-p) of landing on heads.
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Flipping a special coin: probability of getting heads equals the proportion of heads in the flips so farBiased coin flipped until $r$ heads appearBiased coin probabilityCoin-flipping experiment: the expected number of flips that land on headsWhy are odds of a coin landing heads $50%$ after $'n'$ consecutive headsWhat is the probability of a biased coin flipping heads (probability of heads is $frac 35$) exactly $65$ times in $100$ trials?Flipping rigged coin, calculating most common number of flips between headsChernoff bound probability: value of $n$ so that with probability $.999$ at least half of the coin flips come out headsFlip a coin 6 times. Probability with past results and probability without past results are different?Probability density function of flipping until heads and tails
$begingroup$
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)=$Sigma_i=0^inftyip^n-1q+Sigma_i=0^niq^n-1p=frac1q+frac1p$.
However, I am completely wrong.
The answer is
E(N)=$p(1+frac1q)+q(1+frac1p)$
For example, consider we flip for heads first. Then we have:
E(N|H)=$p+pSigma_i=0^inftynp^n-1q$... 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
New contributor
$endgroup$
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)=$Sigma_i=0^inftyip^n-1q+Sigma_i=0^niq^n-1p=frac1q+frac1p$.
However, I am completely wrong.
The answer is
E(N)=$p(1+frac1q)+q(1+frac1p)$
For example, consider we flip for heads first. Then we have:
E(N|H)=$p+pSigma_i=0^inftynp^n-1q$... 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
New contributor
$endgroup$
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
3 hours ago
2
$begingroup$
In both the title and first paragraph it appears there is $0$ chance of landing tails, so you will wait forever.
$endgroup$
– Ross Millikan
3 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
2 hours 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)=$Sigma_i=0^inftyip^n-1q+Sigma_i=0^niq^n-1p=frac1q+frac1p$.
However, I am completely wrong.
The answer is
E(N)=$p(1+frac1q)+q(1+frac1p)$
For example, consider we flip for heads first. Then we have:
E(N|H)=$p+pSigma_i=0^inftynp^n-1q$... 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
New contributor
$endgroup$
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)=$Sigma_i=0^inftyip^n-1q+Sigma_i=0^niq^n-1p=frac1q+frac1p$.
However, I am completely wrong.
The answer is
E(N)=$p(1+frac1q)+q(1+frac1p)$
For example, consider we flip for heads first. Then we have:
E(N|H)=$p+pSigma_i=0^inftynp^n-1q$... 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
New contributor
New contributor
edited 3 hours ago
Mistah White
New contributor
asked 3 hours ago
Mistah WhiteMistah White
62
62
New contributor
New contributor
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
3 hours ago
2
$begingroup$
In both the title and first paragraph it appears there is $0$ chance of landing tails, so you will wait forever.
$endgroup$
– Ross Millikan
3 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
2 hours 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
3 hours ago
2
$begingroup$
In both the title and first paragraph it appears there is $0$ chance of landing tails, so you will wait forever.
$endgroup$
– Ross Millikan
3 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
2 hours 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
3 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
3 hours ago
2
2
$begingroup$
In both the title and first paragraph it appears there is $0$ chance of landing tails, so you will wait forever.
$endgroup$
– Ross Millikan
3 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
3 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
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
2 hours 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 $1+E(Y)$.
$endgroup$
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 $E[X]=frac1p$ and $E[Y]=frac1q$, 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 $$E[W]=E[E[W | A ]] = P(A=1) E[W|A=1]+P(A=0) E[W|A=0]=\=p(E[Y]+1)+q(E[X]+1)$$
$endgroup$
add a comment |
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2 Answers
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2 Answers
2
active
oldest
votes
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oldest
votes
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votes
$begingroup$
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 $1+E(Y)$.
$endgroup$
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 $1+E(Y)$.
$endgroup$
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 $1+E(Y)$.
$endgroup$
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 $1+E(Y)$.
edited 2 hours ago
answered 3 hours ago
Peter ForemanPeter Foreman
7,8921320
7,8921320
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 $E[X]=frac1p$ and $E[Y]=frac1q$, 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 $$E[W]=E[E[W | A ]] = P(A=1) E[W|A=1]+P(A=0) E[W|A=0]=\=p(E[Y]+1)+q(E[X]+1)$$
$endgroup$
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 $E[X]=frac1p$ and $E[Y]=frac1q$, 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 $$E[W]=E[E[W | A ]] = P(A=1) E[W|A=1]+P(A=0) E[W|A=0]=\=p(E[Y]+1)+q(E[X]+1)$$
$endgroup$
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 $E[X]=frac1p$ and $E[Y]=frac1q$, 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 $$E[W]=E[E[W | A ]] = P(A=1) E[W|A=1]+P(A=0) E[W|A=0]=\=p(E[Y]+1)+q(E[X]+1)$$
$endgroup$
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 $E[X]=frac1p$ and $E[Y]=frac1q$, 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 $$E[W]=E[E[W | A ]] = P(A=1) E[W|A=1]+P(A=0) E[W|A=0]=\=p(E[Y]+1)+q(E[X]+1)$$
edited 2 hours ago
answered 3 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
3 hours ago
2
$begingroup$
In both the title and first paragraph it appears there is $0$ chance of landing tails, so you will wait forever.
$endgroup$
– Ross Millikan
3 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
2 hours ago