Finding the expected number of flips needed for a coin having probability $p$ of landing on heads ...
3 doors, three guards, one stone
How does debian/ubuntu knows a package has a updated version
What is Wonderstone and are there any references to it pre-1982?
Understanding Ceva's Theorem
Ring Automorphisms that fix 1.
Why are both D and D# fitting into my E minor key?
How to find out what spells would be useless to a blind NPC spellcaster?
2001: A Space Odyssey's use of the song "Daisy Bell" (Bicycle Built for Two); life imitates art or vice-versa?
Why do people hide their license plates in the EU?
What causes the vertical darker bands in my photo?
Withdrew £2800, but only £2000 shows as withdrawn on online banking; what are my obligations?
Why didn't this character "real die" when they blew their stack out in Altered Carbon?
What is Arya's weapon design?
Why aren't air breathing engines used as small first stages
Why is "Consequences inflicted." not a sentence?
How to tell that you are a giant?
Should I discuss the type of campaign with my players?
Should I use a zero-interest credit card for a large one-time purchase?
How much time will it take to get my passport back if I am applying for multiple Schengen visa countries?
In predicate logic, does existential quantification (∃) include universal quantification (∀), i.e. can 'some' imply 'all'?
Can an alien society believe that their star system is the universe?
String `!23` is replaced with `docker` in command line
Is it ethical to give a final exam after the professor has quit before teaching the remaining chapters of the course?
How to find all the available tools in mac terminal?
Finding the expected number of flips needed for a coin having probability $p$ 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 farHow does one calculate the expected number of coin flips for this game to last?Expected deviation of a coin that obeys the gambler's fallacyCoin-flipping experiment: the expected number of flips that land on headsFormula for consecutive heads or tails in x coin flips.Finding expected number coin flips to get 2 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 heads6 fair coin flips: probability of exactly 3 headsIs there a higher chance of a coin landing heads if it just landed on tails?
$begingroup$
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:
$$E(N)=sum_{i=0}^{infty}ip^{n-1}q+sum_{i=0}^{n}iq^{n-1}p=frac{1}{q}+frac{1}{p}.$$
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
edited 4 hours ago
Saad
20.8k92452
asked 7 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.
$endgroup$
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:
$$E(N)=sum_{i=0}^{infty}ip^{n-1}q+sum_{i=0}^{n}iq^{n-1}p=frac{1}{q}+frac{1}{p}.$$
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
edited 4 hours ago
Saad
20.8k92452
asked 7 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
7 hours ago
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
7 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
7 hours ago
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:
$$E(N)=sum_{i=0}^{infty}ip^{n-1}q+sum_{i=0}^{n}iq^{n-1}p=frac{1}{q}+frac{1}{p}.$$
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
edited 4 hours ago
Saad
20.8k92452
asked 7 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 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:
$$E(N)=sum_{i=0}^{infty}ip^{n-1}q+sum_{i=0}^{n}iq^{n-1}p=frac{1}{q}+frac{1}{p}.$$
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
edited 4 hours ago
Saad
20.8k92452
asked 7 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 4 hours ago
Saad
20.8k92452
asked 7 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 4 hours ago
Saad
20.8k92452
edited 4 hours ago
Saad
20.8k92452
edited 4 hours ago
Saad
20.8k92452
20.8k92452
asked 7 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 7 hours ago
Mistah WhiteMistah White
62
asked 7 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
7 hours ago
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
7 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
7 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
7 hours ago
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
7 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
7 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
7 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
7 hours ago
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
7 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
7 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
7 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
7 hours ago
add a comment |
2 Answers
2
active
oldest
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 $1+E(Y)$.
answered 7 hours ago
Peter ForemanPeter Foreman
7,9221320
$endgroup$
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 7 hours ago
leonbloyleonbloy
42.5k647108
$endgroup$
add a comment |
Your Answer
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Mistah White is a new contributor. Be nice, and check out our Code of Conduct.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3190435%2ffinding-the-expected-number-of-flips-needed-for-a-coin-having-probability-p-of%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
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 $1+E(Y)$.
answered 7 hours ago
Peter ForemanPeter Foreman
7,9221320
$endgroup$
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 7 hours ago
Peter ForemanPeter Foreman
7,9221320
$endgroup$
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 7 hours ago
Peter ForemanPeter Foreman
7,9221320
$endgroup$
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 7 hours ago
Peter ForemanPeter Foreman
7,9221320
edited 7 hours ago
answered 7 hours ago
Peter ForemanPeter Foreman
7,9221320
answered 7 hours ago
Peter ForemanPeter Foreman
7,9221320
answered 7 hours ago
Peter ForemanPeter Foreman
7,9221320
7,9221320
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 7 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 7 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 7 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 7 hours ago
leonbloyleonbloy
42.5k647108
edited 7 hours ago
answered 7 hours ago
leonbloyleonbloy
42.5k647108
answered 7 hours ago
leonbloyleonbloy
42.5k647108
answered 7 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.
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3190435%2ffinding-the-expected-number-of-flips-needed-for-a-coin-having-probability-p-of%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
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
7 hours ago
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
7 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
7 hours ago