Probability of winning roulette wheel












1














enter image description here



Try



Let $X$ be the number of first $n$ bets that we win. $X$ is binomial with $p= frac{1}{38}$ and $n=?$



I knwo that that I need to find $P(X > 34 )$ and approximate with Normal, but how can we find our $n$ ?










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  • Hint for the first part: if you win even once you will have a profit.
    – lulu
    Nov 13 '18 at 15:22










  • I don't understand where $X>34$ comes from, nor why you think you have to determine $n.$ Isn't $n=34$ for part a, and $n=1000$ for part b?
    – saulspatz
    Nov 13 '18 at 16:01










  • The first task reads "you are winning after 34 bets" (not wins). So if $X$ signifies your wins, $P(X > 34)$ can't be the right question.
    – Nearoo
    Nov 13 '18 at 19:19












  • a) you need to find $P(Xge 1| n= 34)$ b) yes, when $n$ is large a binomial distribution begins to look like a normal distribution. What is the mean and standard deviation. With that information how do you find your probabilities?
    – Doug M
    Nov 20 '18 at 2:45










  • Have you lost all 33 previous bets?
    – Mostafa Ayaz
    Nov 21 '18 at 19:12
















1














enter image description here



Try



Let $X$ be the number of first $n$ bets that we win. $X$ is binomial with $p= frac{1}{38}$ and $n=?$



I knwo that that I need to find $P(X > 34 )$ and approximate with Normal, but how can we find our $n$ ?










share|cite|improve this question






















  • Hint for the first part: if you win even once you will have a profit.
    – lulu
    Nov 13 '18 at 15:22










  • I don't understand where $X>34$ comes from, nor why you think you have to determine $n.$ Isn't $n=34$ for part a, and $n=1000$ for part b?
    – saulspatz
    Nov 13 '18 at 16:01










  • The first task reads "you are winning after 34 bets" (not wins). So if $X$ signifies your wins, $P(X > 34)$ can't be the right question.
    – Nearoo
    Nov 13 '18 at 19:19












  • a) you need to find $P(Xge 1| n= 34)$ b) yes, when $n$ is large a binomial distribution begins to look like a normal distribution. What is the mean and standard deviation. With that information how do you find your probabilities?
    – Doug M
    Nov 20 '18 at 2:45










  • Have you lost all 33 previous bets?
    – Mostafa Ayaz
    Nov 21 '18 at 19:12














1












1








1


1





enter image description here



Try



Let $X$ be the number of first $n$ bets that we win. $X$ is binomial with $p= frac{1}{38}$ and $n=?$



I knwo that that I need to find $P(X > 34 )$ and approximate with Normal, but how can we find our $n$ ?










share|cite|improve this question













enter image description here



Try



Let $X$ be the number of first $n$ bets that we win. $X$ is binomial with $p= frac{1}{38}$ and $n=?$



I knwo that that I need to find $P(X > 34 )$ and approximate with Normal, but how can we find our $n$ ?







probability






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share|cite|improve this question











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asked Nov 13 '18 at 15:16









Neymar

374113




374113












  • Hint for the first part: if you win even once you will have a profit.
    – lulu
    Nov 13 '18 at 15:22










  • I don't understand where $X>34$ comes from, nor why you think you have to determine $n.$ Isn't $n=34$ for part a, and $n=1000$ for part b?
    – saulspatz
    Nov 13 '18 at 16:01










  • The first task reads "you are winning after 34 bets" (not wins). So if $X$ signifies your wins, $P(X > 34)$ can't be the right question.
    – Nearoo
    Nov 13 '18 at 19:19












  • a) you need to find $P(Xge 1| n= 34)$ b) yes, when $n$ is large a binomial distribution begins to look like a normal distribution. What is the mean and standard deviation. With that information how do you find your probabilities?
    – Doug M
    Nov 20 '18 at 2:45










  • Have you lost all 33 previous bets?
    – Mostafa Ayaz
    Nov 21 '18 at 19:12


















  • Hint for the first part: if you win even once you will have a profit.
    – lulu
    Nov 13 '18 at 15:22










  • I don't understand where $X>34$ comes from, nor why you think you have to determine $n.$ Isn't $n=34$ for part a, and $n=1000$ for part b?
    – saulspatz
    Nov 13 '18 at 16:01










  • The first task reads "you are winning after 34 bets" (not wins). So if $X$ signifies your wins, $P(X > 34)$ can't be the right question.
    – Nearoo
    Nov 13 '18 at 19:19












  • a) you need to find $P(Xge 1| n= 34)$ b) yes, when $n$ is large a binomial distribution begins to look like a normal distribution. What is the mean and standard deviation. With that information how do you find your probabilities?
    – Doug M
    Nov 20 '18 at 2:45










  • Have you lost all 33 previous bets?
    – Mostafa Ayaz
    Nov 21 '18 at 19:12
















Hint for the first part: if you win even once you will have a profit.
– lulu
Nov 13 '18 at 15:22




Hint for the first part: if you win even once you will have a profit.
– lulu
Nov 13 '18 at 15:22












I don't understand where $X>34$ comes from, nor why you think you have to determine $n.$ Isn't $n=34$ for part a, and $n=1000$ for part b?
– saulspatz
Nov 13 '18 at 16:01




I don't understand where $X>34$ comes from, nor why you think you have to determine $n.$ Isn't $n=34$ for part a, and $n=1000$ for part b?
– saulspatz
Nov 13 '18 at 16:01












The first task reads "you are winning after 34 bets" (not wins). So if $X$ signifies your wins, $P(X > 34)$ can't be the right question.
– Nearoo
Nov 13 '18 at 19:19






The first task reads "you are winning after 34 bets" (not wins). So if $X$ signifies your wins, $P(X > 34)$ can't be the right question.
– Nearoo
Nov 13 '18 at 19:19














a) you need to find $P(Xge 1| n= 34)$ b) yes, when $n$ is large a binomial distribution begins to look like a normal distribution. What is the mean and standard deviation. With that information how do you find your probabilities?
– Doug M
Nov 20 '18 at 2:45




a) you need to find $P(Xge 1| n= 34)$ b) yes, when $n$ is large a binomial distribution begins to look like a normal distribution. What is the mean and standard deviation. With that information how do you find your probabilities?
– Doug M
Nov 20 '18 at 2:45












Have you lost all 33 previous bets?
– Mostafa Ayaz
Nov 21 '18 at 19:12




Have you lost all 33 previous bets?
– Mostafa Ayaz
Nov 21 '18 at 19:12










1 Answer
1






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oldest

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3





+50









Your rewards is $$35W-(N-W)=36W-N$$ where $W$ is the number of times you win out of $N$ game.



We have $W sim Bin(N, frac1{38}).$



Hence, for the first part, we are interested in



We are interested in $P(36W-N > 0)=P(W > frac{N}{36}).$



For the first part $N=34$.



Hence $P(W> frac{34}{36})=1-P(W=0)=1-(1-frac1{38})^{34}$



For the second part $N=1000$,



$P(W > frac{1000}{36})=P(W > 27.777)$



We can use normal approximation to estimate $W$ as $N(frac{N}{38}, frac{37N}{38^2})$ and approximate is as



begin{align}
Pleft( Z>frac{27.5-frac{N}{38}}{sqrt{frac{37N}{38^2}}}right) &= Pleft( Z>frac{38(27.5)-N}{sqrt{37N}}right) \
&= P(Z > 0.23394386021)
end{align}



where $Z sim N(0,1)$.






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    1 Answer
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    1 Answer
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    active

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    3





    +50









    Your rewards is $$35W-(N-W)=36W-N$$ where $W$ is the number of times you win out of $N$ game.



    We have $W sim Bin(N, frac1{38}).$



    Hence, for the first part, we are interested in



    We are interested in $P(36W-N > 0)=P(W > frac{N}{36}).$



    For the first part $N=34$.



    Hence $P(W> frac{34}{36})=1-P(W=0)=1-(1-frac1{38})^{34}$



    For the second part $N=1000$,



    $P(W > frac{1000}{36})=P(W > 27.777)$



    We can use normal approximation to estimate $W$ as $N(frac{N}{38}, frac{37N}{38^2})$ and approximate is as



    begin{align}
    Pleft( Z>frac{27.5-frac{N}{38}}{sqrt{frac{37N}{38^2}}}right) &= Pleft( Z>frac{38(27.5)-N}{sqrt{37N}}right) \
    &= P(Z > 0.23394386021)
    end{align}



    where $Z sim N(0,1)$.






    share|cite|improve this answer




























      3





      +50









      Your rewards is $$35W-(N-W)=36W-N$$ where $W$ is the number of times you win out of $N$ game.



      We have $W sim Bin(N, frac1{38}).$



      Hence, for the first part, we are interested in



      We are interested in $P(36W-N > 0)=P(W > frac{N}{36}).$



      For the first part $N=34$.



      Hence $P(W> frac{34}{36})=1-P(W=0)=1-(1-frac1{38})^{34}$



      For the second part $N=1000$,



      $P(W > frac{1000}{36})=P(W > 27.777)$



      We can use normal approximation to estimate $W$ as $N(frac{N}{38}, frac{37N}{38^2})$ and approximate is as



      begin{align}
      Pleft( Z>frac{27.5-frac{N}{38}}{sqrt{frac{37N}{38^2}}}right) &= Pleft( Z>frac{38(27.5)-N}{sqrt{37N}}right) \
      &= P(Z > 0.23394386021)
      end{align}



      where $Z sim N(0,1)$.






      share|cite|improve this answer


























        3





        +50







        3





        +50



        3




        +50




        Your rewards is $$35W-(N-W)=36W-N$$ where $W$ is the number of times you win out of $N$ game.



        We have $W sim Bin(N, frac1{38}).$



        Hence, for the first part, we are interested in



        We are interested in $P(36W-N > 0)=P(W > frac{N}{36}).$



        For the first part $N=34$.



        Hence $P(W> frac{34}{36})=1-P(W=0)=1-(1-frac1{38})^{34}$



        For the second part $N=1000$,



        $P(W > frac{1000}{36})=P(W > 27.777)$



        We can use normal approximation to estimate $W$ as $N(frac{N}{38}, frac{37N}{38^2})$ and approximate is as



        begin{align}
        Pleft( Z>frac{27.5-frac{N}{38}}{sqrt{frac{37N}{38^2}}}right) &= Pleft( Z>frac{38(27.5)-N}{sqrt{37N}}right) \
        &= P(Z > 0.23394386021)
        end{align}



        where $Z sim N(0,1)$.






        share|cite|improve this answer














        Your rewards is $$35W-(N-W)=36W-N$$ where $W$ is the number of times you win out of $N$ game.



        We have $W sim Bin(N, frac1{38}).$



        Hence, for the first part, we are interested in



        We are interested in $P(36W-N > 0)=P(W > frac{N}{36}).$



        For the first part $N=34$.



        Hence $P(W> frac{34}{36})=1-P(W=0)=1-(1-frac1{38})^{34}$



        For the second part $N=1000$,



        $P(W > frac{1000}{36})=P(W > 27.777)$



        We can use normal approximation to estimate $W$ as $N(frac{N}{38}, frac{37N}{38^2})$ and approximate is as



        begin{align}
        Pleft( Z>frac{27.5-frac{N}{38}}{sqrt{frac{37N}{38^2}}}right) &= Pleft( Z>frac{38(27.5)-N}{sqrt{37N}}right) \
        &= P(Z > 0.23394386021)
        end{align}



        where $Z sim N(0,1)$.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Nov 20 '18 at 4:27

























        answered Nov 20 '18 at 4:19









        Siong Thye Goh

        99.3k1464117




        99.3k1464117






























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