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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










1












$begingroup$


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?










share|cite|improve this question









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















1












$begingroup$


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?










share|cite|improve this question









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













1












1








1


1



$begingroup$


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?










share|cite|improve this question









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)=$Sigma_i=0^inftyip^n-1q+Sigma_i=0^niq^n-1p=frac1q+frac1p$.



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






share|cite|improve this question









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.











share|cite|improve this question









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.









share|cite|improve this question




share|cite|improve this question








edited 1 hour ago







Mistah White













New contributor




Mistah White is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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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.
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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












  • 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










2 Answers
2






active

oldest

votes


















3












$begingroup$

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 $1+E(Y)$.






share|cite|improve this answer











$endgroup$




















    2












    $begingroup$

    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 $$E[W]=E[E[W | A ]] = P(A=1) E[W|A=1]+P(A=0) E[W|A=0]=\=p(E[Y]+1)+q(E[X]+1)$$






    share|cite|improve this answer











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      2 Answers
      2






      active

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      2 Answers
      2






      active

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      active

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      active

      oldest

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      3












      $begingroup$

      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 $1+E(Y)$.






      share|cite|improve this answer











      $endgroup$

















        3












        $begingroup$

        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 $1+E(Y)$.






        share|cite|improve this answer











        $endgroup$















          3












          3








          3





          $begingroup$

          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 $1+E(Y)$.






          share|cite|improve this answer











          $endgroup$



          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 $1+E(Y)$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited 1 hour ago

























          answered 1 hour ago









          Peter ForemanPeter Foreman

          7,8731320




          7,8731320





















              2












              $begingroup$

              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 $$E[W]=E[E[W | A ]] = P(A=1) E[W|A=1]+P(A=0) E[W|A=0]=\=p(E[Y]+1)+q(E[X]+1)$$






              share|cite|improve this answer











              $endgroup$

















                2












                $begingroup$

                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 $$E[W]=E[E[W | A ]] = P(A=1) E[W|A=1]+P(A=0) E[W|A=0]=\=p(E[Y]+1)+q(E[X]+1)$$






                share|cite|improve this answer











                $endgroup$















                  2












                  2








                  2





                  $begingroup$

                  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 $$E[W]=E[E[W | A ]] = P(A=1) E[W|A=1]+P(A=0) E[W|A=0]=\=p(E[Y]+1)+q(E[X]+1)$$






                  share|cite|improve this answer











                  $endgroup$



                  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 $$E[W]=E[E[W | A ]] = P(A=1) E[W|A=1]+P(A=0) E[W|A=0]=\=p(E[Y]+1)+q(E[X]+1)$$







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited 1 hour ago

























                  answered 1 hour ago









                  leonbloyleonbloy

                  42.5k647108




                  42.5k647108




















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                      Are there any AGPL-style licences that require source code modifications to be public? Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern) Announcing the arrival of Valued Associate #679: Cesar Manara Unicorn Meta Zoo #1: Why another podcast?Force derivative works to be publicAre there any GPL like licenses for Apple App Store?Do you violate the GPL if you provide source code that cannot be compiled?GPL - is it distribution to use libraries in an appliance loaned to customers?Distributing App for free which uses GPL'ed codeModifications of server software under GPL, with web/CLI interfaceDoes using an AGPLv3-licensed library prevent me from dual-licensing my own source code?Can I publish only select code under GPLv3 from a private project?Is there published precedent regarding the scope of covered work that uses AGPL software?If MIT licensed code links to GPL licensed code what should be the license of the resulting binary program?If I use a public API endpoint that has its source code licensed under AGPL in my app, do I need to disclose my source?

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                      Mortes em março de 2019 Referências Menu de navegação«Zhores Alferov, Nobel de Física bielorrusso, morre aos 88 anos - Ciência»«Fallece Rafael Torija, o bispo emérito de Ciudad Real»«Peter Hurford dies at 88»«Keith Flint, vocalista do The Prodigy, morre aos 49 anos»«Luke Perry, ator de 'Barrados no baile' e 'Riverdale', morre aos 52 anos»«Former Rangers and Scotland captain Eric Caldow dies, aged 84»«Morreu, aos 61 anos, a antiga lenda do wrestling King Kong Bundy»«Fallece el actor y director teatral Abraham Stavans»«In Memoriam Guillaume Faye»«Sidney Sheinberg, a Force Behind Universal and Spielberg, Is Dead at 84»«Carmine Persico, Colombo Crime Family Boss, Is Dead at 85»«Dirigent Michael Gielen gestorben»«Ciclista tricampeã mundial e prata na Rio 2016 é encontrada morta em casa aos 23 anos»«Pagan Community Notes: Raven Grimassi dies, Indianapolis pop-up event cancelled, Circle Sanctuary announces new podcast, and more!»«Hal Blaine, Wrecking Crew Drummer, Dies at 90»«Morre Coutinho, que editou dupla lendária com Pelé no Santos»«Cantor Demétrius, ídolo da Jovem Guarda, morre em SP»«Ex-presidente do Vasco, Eurico Miranda morre no Rio de Janeiro»«Bronze no Mundial de basquete de 1971, Laís Elena morre aos 76 anos»«Diretor de Corridas da F1, Charlie Whiting morre aos 66 anos às vésperas do GP da Austrália»«Morreu o cardeal Danneels, da Bélgica»«Morreu o cartoonista Augusto Cid»«Morreu a atriz Maria Isabel de Lizandra, de "Vale Tudo" e novelas da Tupi»«WS Merwin, prize-winning poet of nature, dies at 91»«Atriz Márcia Real morre em São Paulo aos 88 anos»«Mauritanie: décès de l'ancien président Mohamed Mahmoud ould Louly»«Morreu Dick Dale, o rei da surf guitar e de "Pulp Fiction"»«Falleció Víctor Genes»«João Carlos Marinho, autor de 'O Gênio do Crime', morre em SP»«Legendary Horror Director and SFX Artist John Carl Buechler Dies at 66»«Morre em Salvador a religiosa Makota Valdina»«مرگ بازیکن‌ سابق نساجی بر اثر سقوط سنگ در مازندران»«Domingos Oliveira morre no Rio»«Morre Airton Ravagniani, ex-São Paulo, Fla, Vasco, Grêmio e Sport - Notícias»«Morre o escritor Flavio Moreira da Costa»«Larry Cohen, Writer-Director of 'It's Alive' and 'Hell Up in Harlem,' Dies at 77»«Scott Walker, experimental singer-songwriter, dead at 76»«Joseph Pilato, Day of the Dead Star and Horror Favorite, Dies at 70»«Sheffield United set to pay tribute to legendary goalkeeper Ted Burgin who has died at 91»«Morre Rafael Henzel, sobrevivente de acidente aéreo da Chapecoense»«Morre Valery Bykovsky, um dos primeiros cosmonautas da União Soviética»«Agnès Varda, cineasta da Nouvelle Vague, morre aos 90 anos»«Agnès Varda, cineasta francesa, morre aos 90 anos»«Tania Mallet, James Bond Actress and Helen Mirren's Cousin, Dies at 77»e