Probability of winning roulette wheel
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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
asked Nov 13 at 15:16
Neymar
35613
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up vote
1
down vote
favorite
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
asked Nov 13 at 15:16
Neymar
35613
Hint for the first part: if you win even once you will have a profit.
– lulu
Nov 13 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 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 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 at 2:45
Have you lost all 33 previous bets?
– Mostafa Ayaz
Nov 21 at 19:12
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
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
asked Nov 13 at 15:16
Neymar
35613
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
probability
asked Nov 13 at 15:16
Neymar
35613
asked Nov 13 at 15:16
Neymar
35613
asked Nov 13 at 15:16
Neymar
35613
asked Nov 13 at 15:16
Neymar
35613
asked Nov 13 at 15:16
Neymar
35613
35613
Hint for the first part: if you win even once you will have a profit.
– lulu
Nov 13 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 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 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 at 2:45
Have you lost all 33 previous bets?
– Mostafa Ayaz
Nov 21 at 19:12
add a comment |
Hint for the first part: if you win even once you will have a profit.
– lulu
Nov 13 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 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 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 at 2:45
Have you lost all 33 previous bets?
– Mostafa Ayaz
Nov 21 at 19:12
Hint for the first part: if you win even once you will have a profit.
– lulu
Nov 13 at 15:22
Hint for the first part: if you win even once you will have a profit.
– lulu
Nov 13 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 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 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 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 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 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 at 2:45
Have you lost all 33 previous bets?
– Mostafa Ayaz
Nov 21 at 19:12
Have you lost all 33 previous bets?
– Mostafa Ayaz
Nov 21 at 19:12
add a comment |
1 Answer
1
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votes
up vote
3
down vote
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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 $Z sim N(0,1)$.
answered Nov 20 at 4:19
Siong Thye Goh
95.2k1462115
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
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 $Z sim N(0,1)$.
answered Nov 20 at 4:19
Siong Thye Goh
95.2k1462115
add a comment |
up vote
3
down vote
accepted
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 $Z sim N(0,1)$.
answered Nov 20 at 4:19
Siong Thye Goh
95.2k1462115
add a comment |
up vote
3
down vote
accepted
up vote
3
down vote
accepted
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 $Z sim N(0,1)$.
answered Nov 20 at 4:19
Siong Thye Goh
95.2k1462115
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 $Z sim N(0,1)$.
answered Nov 20 at 4:19
Siong Thye Goh
95.2k1462115
edited Nov 20 at 4:27
answered Nov 20 at 4:19
Siong Thye Goh
95.2k1462115
answered Nov 20 at 4:19
Siong Thye Goh
95.2k1462115
answered Nov 20 at 4:19
Siong Thye Goh
95.2k1462115
95.2k1462115
add a comment |
add a comment |
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Hint for the first part: if you win even once you will have a profit.
– lulu
Nov 13 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 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 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 at 2:45
Have you lost all 33 previous bets?
– Mostafa Ayaz
Nov 21 at 19:12