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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 2 hours 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


















4












$begingroup$

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

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      4












      $begingroup$

      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











      $endgroup$

















        4












        $begingroup$

        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











        $endgroup$















          4












          4








          4





          $begingroup$

          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 2 hours ago









          Peter ForemanPeter Foreman

          7,8931320




          7,8931320





















              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 2 hours ago

























                  answered 2 hours ago









                  leonbloyleonbloy

                  42.5k647108




                  42.5k647108




















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