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Playing 4 games, then playing until a loss - expected number of losses
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Consider a game where you win with probability $p$. We play 5 such games, and if we win game number 5, we keep playing until we lose.
a) Find the expected number of games played
b) Find the expected number of games lost
Considering the a) part of problem, I defined X to be a random variable that measures number of games, starting at 5 (since we are sure the game is played at least 5 times) and came up with $$E(X) = frac5-4p1-p$$
which appears to be correct.
I have no idea how to approach the second part of the problem.
The official solution states that it is the expected number of games played times the probability of loss, i.e. $E(X)cdot(1-p) = 5-4p$ with no explanation whatsoever. So any sort of intuition or explanation (or an alternative solution) would be greatly appreciated.
$bfedit$ I have arrived at E(X) by defining a discreet random variable with the following distribution
$$Xsim left( beginarray c c c
5 & 6 & 7 & ... & k \
1 - p & p(1-p) & p^2(1-p) & ... & p^k-5(1-p)
endarray right)
$$
and applying the definition of expectation.
$bfedit 2$ I tried defining a random variable $Y$ which measures the number of losses in first four games. So for 0 losses we must win every game, so the probability is p^4, etc. I arrived at the following distribution:
$$Xsim left( beginarray c c c
0 & 1 & 2 & 3 & 4 \
p^4 & p^3(1-p) & p^2(1-p)^2 & p(1-p)^3 & (1-p)^4
endarray right)
$$
However the expectation of this variable doesn't match the solution $5-4p$.
probability expectation
asked Jul 29 at 15:26
barnaby-b
476
add a comment |Â
up vote
4
down vote
favorite
Consider a game where you win with probability $p$. We play 5 such games, and if we win game number 5, we keep playing until we lose.
a) Find the expected number of games played
b) Find the expected number of games lost
Considering the a) part of problem, I defined X to be a random variable that measures number of games, starting at 5 (since we are sure the game is played at least 5 times) and came up with $$E(X) = frac5-4p1-p$$
which appears to be correct.
I have no idea how to approach the second part of the problem.
The official solution states that it is the expected number of games played times the probability of loss, i.e. $E(X)cdot(1-p) = 5-4p$ with no explanation whatsoever. So any sort of intuition or explanation (or an alternative solution) would be greatly appreciated.
$bfedit$ I have arrived at E(X) by defining a discreet random variable with the following distribution
$$Xsim left( beginarray c c c
5 & 6 & 7 & ... & k \
1 - p & p(1-p) & p^2(1-p) & ... & p^k-5(1-p)
endarray right)
$$
and applying the definition of expectation.
$bfedit 2$ I tried defining a random variable $Y$ which measures the number of losses in first four games. So for 0 losses we must win every game, so the probability is p^4, etc. I arrived at the following distribution:
$$Xsim left( beginarray c c c
0 & 1 & 2 & 3 & 4 \
p^4 & p^3(1-p) & p^2(1-p)^2 & p(1-p)^3 & (1-p)^4
endarray right)
$$
However the expectation of this variable doesn't match the solution $5-4p$.
probability expectation
asked Jul 29 at 15:26
barnaby-b
476
2
How do you get $E(X)$? How can we say if you approach is correct if you don´t tell us your thoughts and/or show your calculation.
– callculus
Jul 29 at 15:40
2
After your edit I changed my downvote into an upvote.
– callculus
Jul 29 at 15:53
In your second edit, the probability distribution you've defined has probabilities that, in general, do not add up to one. You may wish to consult the Wikipedia article I link to in my answer below.
– Theoretical Economist
Jul 29 at 16:24
add a comment |Â
up vote
4
down vote
favorite
up vote
4
down vote
favorite
Consider a game where you win with probability $p$. We play 5 such games, and if we win game number 5, we keep playing until we lose.
a) Find the expected number of games played
b) Find the expected number of games lost
Considering the a) part of problem, I defined X to be a random variable that measures number of games, starting at 5 (since we are sure the game is played at least 5 times) and came up with $$E(X) = frac5-4p1-p$$
which appears to be correct.
I have no idea how to approach the second part of the problem.
The official solution states that it is the expected number of games played times the probability of loss, i.e. $E(X)cdot(1-p) = 5-4p$ with no explanation whatsoever. So any sort of intuition or explanation (or an alternative solution) would be greatly appreciated.
$bfedit$ I have arrived at E(X) by defining a discreet random variable with the following distribution
$$Xsim left( beginarray c c c
5 & 6 & 7 & ... & k \
1 - p & p(1-p) & p^2(1-p) & ... & p^k-5(1-p)
endarray right)
$$
and applying the definition of expectation.
$bfedit 2$ I tried defining a random variable $Y$ which measures the number of losses in first four games. So for 0 losses we must win every game, so the probability is p^4, etc. I arrived at the following distribution:
$$Xsim left( beginarray c c c
0 & 1 & 2 & 3 & 4 \
p^4 & p^3(1-p) & p^2(1-p)^2 & p(1-p)^3 & (1-p)^4
endarray right)
$$
However the expectation of this variable doesn't match the solution $5-4p$.
probability expectation
asked Jul 29 at 15:26
barnaby-b
476
Consider a game where you win with probability $p$. We play 5 such games, and if we win game number 5, we keep playing until we lose.
a) Find the expected number of games played
b) Find the expected number of games lost
Considering the a) part of problem, I defined X to be a random variable that measures number of games, starting at 5 (since we are sure the game is played at least 5 times) and came up with $$E(X) = frac5-4p1-p$$
which appears to be correct.
I have no idea how to approach the second part of the problem.
The official solution states that it is the expected number of games played times the probability of loss, i.e. $E(X)cdot(1-p) = 5-4p$ with no explanation whatsoever. So any sort of intuition or explanation (or an alternative solution) would be greatly appreciated.
$bfedit$ I have arrived at E(X) by defining a discreet random variable with the following distribution
$$Xsim left( beginarray c c c
5 & 6 & 7 & ... & k \
1 - p & p(1-p) & p^2(1-p) & ... & p^k-5(1-p)
endarray right)
$$
and applying the definition of expectation.
$bfedit 2$ I tried defining a random variable $Y$ which measures the number of losses in first four games. So for 0 losses we must win every game, so the probability is p^4, etc. I arrived at the following distribution:
$$Xsim left( beginarray c c c
0 & 1 & 2 & 3 & 4 \
p^4 & p^3(1-p) & p^2(1-p)^2 & p(1-p)^3 & (1-p)^4
endarray right)
$$
However the expectation of this variable doesn't match the solution $5-4p$.
probability expectation
asked Jul 29 at 15:26
barnaby-b
476
edited Jul 29 at 16:09
asked Jul 29 at 15:26
barnaby-b
476
asked Jul 29 at 15:26
barnaby-b
476
asked Jul 29 at 15:26
barnaby-b
476
476
2
How do you get $E(X)$? How can we say if you approach is correct if you don´t tell us your thoughts and/or show your calculation.
– callculus
Jul 29 at 15:40
2
After your edit I changed my downvote into an upvote.
– callculus
Jul 29 at 15:53
In your second edit, the probability distribution you've defined has probabilities that, in general, do not add up to one. You may wish to consult the Wikipedia article I link to in my answer below.
– Theoretical Economist
Jul 29 at 16:24
add a comment |Â
2
How do you get $E(X)$? How can we say if you approach is correct if you don´t tell us your thoughts and/or show your calculation.
– callculus
Jul 29 at 15:40
2
After your edit I changed my downvote into an upvote.
– callculus
Jul 29 at 15:53
In your second edit, the probability distribution you've defined has probabilities that, in general, do not add up to one. You may wish to consult the Wikipedia article I link to in my answer below.
– Theoretical Economist
Jul 29 at 16:24
2
2
How do you get $E(X)$? How can we say if you approach is correct if you don´t tell us your thoughts and/or show your calculation.
– callculus
Jul 29 at 15:40
How do you get $E(X)$? How can we say if you approach is correct if you don´t tell us your thoughts and/or show your calculation.
– callculus
Jul 29 at 15:40
2
2
After your edit I changed my downvote into an upvote.
– callculus
Jul 29 at 15:53
After your edit I changed my downvote into an upvote.
– callculus
Jul 29 at 15:53
In your second edit, the probability distribution you've defined has probabilities that, in general, do not add up to one. You may wish to consult the Wikipedia article I link to in my answer below.
– Theoretical Economist
Jul 29 at 16:24
In your second edit, the probability distribution you've defined has probabilities that, in general, do not add up to one. You may wish to consult the Wikipedia article I link to in my answer below.
– Theoretical Economist
Jul 29 at 16:24
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
3
down vote
Your expected number of losses is the expected number of losses in the first four games, plus the expected number of losses in all games played after the first four.
The number of losses in the first four games is binomially distributed with "success" probability $1-p$ and four trials. Thus, the expected number of losses in the first four games is $4(1-p)$. The expected number of losses in all games played after the first four is $1$. (Since you stop immediately after a loss.)
Hence, the expected number of losses is
$$ 4(1-p) + 1 = 5 - 4p. $$
answered Jul 29 at 15:50
Theoretical Economist
3,1352626
Ok, so since after the fourth game, the expected number of losses is 1, I get that now. But how is the number of losses in first four games $4(1-p)$?
– barnaby-b
Jul 29 at 16:04
1
@barnaby-b This is a property of the binomial distribution, which I linked to in my edit. Alternatively, the expected number of losses from a single game is just the probability of losing, which is $1-p$. Since expectation is linear, the expected number of losses in four games is $4(1-p)$.
– Theoretical Economist
Jul 29 at 16:16
add a comment |Â
up vote
1
down vote
Another answer has already derived the expected number of losses in an elementary manner. To derive the same result in the manner in which the official solution derives it, apply the general version of Wald's identity. You have an infinite sequence of (i.i.d.) binary random variables $X_n$ that take the value $1$ for a loss and the value $0$ for a win, as well as a random number $N$ of them that you sum, and they satisfy the premises of the identity. In particular, even though $N$ is not independent of the $X_n$ (as required for the basic version of the identity), we have
$$
mathsf Eleft[X_n1_Nge nright]=mathsf Eleft[X_nright]mathsf P(Nge n);.
$$
Thus, the expected number of losses is the expected number of games played times the probability of a loss.
answered Jul 29 at 16:17
joriki
164k10179328
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
Your expected number of losses is the expected number of losses in the first four games, plus the expected number of losses in all games played after the first four.
The number of losses in the first four games is binomially distributed with "success" probability $1-p$ and four trials. Thus, the expected number of losses in the first four games is $4(1-p)$. The expected number of losses in all games played after the first four is $1$. (Since you stop immediately after a loss.)
Hence, the expected number of losses is
$$ 4(1-p) + 1 = 5 - 4p. $$
answered Jul 29 at 15:50
Theoretical Economist
3,1352626
Ok, so since after the fourth game, the expected number of losses is 1, I get that now. But how is the number of losses in first four games $4(1-p)$?
– barnaby-b
Jul 29 at 16:04
1
@barnaby-b This is a property of the binomial distribution, which I linked to in my edit. Alternatively, the expected number of losses from a single game is just the probability of losing, which is $1-p$. Since expectation is linear, the expected number of losses in four games is $4(1-p)$.
– Theoretical Economist
Jul 29 at 16:16
add a comment |Â
up vote
3
down vote
Your expected number of losses is the expected number of losses in the first four games, plus the expected number of losses in all games played after the first four.
The number of losses in the first four games is binomially distributed with "success" probability $1-p$ and four trials. Thus, the expected number of losses in the first four games is $4(1-p)$. The expected number of losses in all games played after the first four is $1$. (Since you stop immediately after a loss.)
Hence, the expected number of losses is
$$ 4(1-p) + 1 = 5 - 4p. $$
answered Jul 29 at 15:50
Theoretical Economist
3,1352626
Ok, so since after the fourth game, the expected number of losses is 1, I get that now. But how is the number of losses in first four games $4(1-p)$?
– barnaby-b
Jul 29 at 16:04
1
@barnaby-b This is a property of the binomial distribution, which I linked to in my edit. Alternatively, the expected number of losses from a single game is just the probability of losing, which is $1-p$. Since expectation is linear, the expected number of losses in four games is $4(1-p)$.
– Theoretical Economist
Jul 29 at 16:16
add a comment |Â
up vote
3
down vote
up vote
3
down vote
Your expected number of losses is the expected number of losses in the first four games, plus the expected number of losses in all games played after the first four.
The number of losses in the first four games is binomially distributed with "success" probability $1-p$ and four trials. Thus, the expected number of losses in the first four games is $4(1-p)$. The expected number of losses in all games played after the first four is $1$. (Since you stop immediately after a loss.)
Hence, the expected number of losses is
$$ 4(1-p) + 1 = 5 - 4p. $$
answered Jul 29 at 15:50
Theoretical Economist
3,1352626
Your expected number of losses is the expected number of losses in the first four games, plus the expected number of losses in all games played after the first four.
The number of losses in the first four games is binomially distributed with "success" probability $1-p$ and four trials. Thus, the expected number of losses in the first four games is $4(1-p)$. The expected number of losses in all games played after the first four is $1$. (Since you stop immediately after a loss.)
Hence, the expected number of losses is
$$ 4(1-p) + 1 = 5 - 4p. $$
answered Jul 29 at 15:50
Theoretical Economist
3,1352626
edited Jul 29 at 16:14
answered Jul 29 at 15:50
Theoretical Economist
3,1352626
answered Jul 29 at 15:50
Theoretical Economist
3,1352626
answered Jul 29 at 15:50
Theoretical Economist
3,1352626
3,1352626
Ok, so since after the fourth game, the expected number of losses is 1, I get that now. But how is the number of losses in first four games $4(1-p)$?
– barnaby-b
Jul 29 at 16:04
1
@barnaby-b This is a property of the binomial distribution, which I linked to in my edit. Alternatively, the expected number of losses from a single game is just the probability of losing, which is $1-p$. Since expectation is linear, the expected number of losses in four games is $4(1-p)$.
– Theoretical Economist
Jul 29 at 16:16
add a comment |Â
Ok, so since after the fourth game, the expected number of losses is 1, I get that now. But how is the number of losses in first four games $4(1-p)$?
– barnaby-b
Jul 29 at 16:04
1
@barnaby-b This is a property of the binomial distribution, which I linked to in my edit. Alternatively, the expected number of losses from a single game is just the probability of losing, which is $1-p$. Since expectation is linear, the expected number of losses in four games is $4(1-p)$.
– Theoretical Economist
Jul 29 at 16:16
Ok, so since after the fourth game, the expected number of losses is 1, I get that now. But how is the number of losses in first four games $4(1-p)$?
– barnaby-b
Jul 29 at 16:04
Ok, so since after the fourth game, the expected number of losses is 1, I get that now. But how is the number of losses in first four games $4(1-p)$?
– barnaby-b
Jul 29 at 16:04
1
1
@barnaby-b This is a property of the binomial distribution, which I linked to in my edit. Alternatively, the expected number of losses from a single game is just the probability of losing, which is $1-p$. Since expectation is linear, the expected number of losses in four games is $4(1-p)$.
– Theoretical Economist
Jul 29 at 16:16
@barnaby-b This is a property of the binomial distribution, which I linked to in my edit. Alternatively, the expected number of losses from a single game is just the probability of losing, which is $1-p$. Since expectation is linear, the expected number of losses in four games is $4(1-p)$.
– Theoretical Economist
Jul 29 at 16:16
add a comment |Â
up vote
1
down vote
Another answer has already derived the expected number of losses in an elementary manner. To derive the same result in the manner in which the official solution derives it, apply the general version of Wald's identity. You have an infinite sequence of (i.i.d.) binary random variables $X_n$ that take the value $1$ for a loss and the value $0$ for a win, as well as a random number $N$ of them that you sum, and they satisfy the premises of the identity. In particular, even though $N$ is not independent of the $X_n$ (as required for the basic version of the identity), we have
$$
mathsf Eleft[X_n1_Nge nright]=mathsf Eleft[X_nright]mathsf P(Nge n);.
$$
Thus, the expected number of losses is the expected number of games played times the probability of a loss.
answered Jul 29 at 16:17
joriki
164k10179328
add a comment |Â
up vote
1
down vote
Another answer has already derived the expected number of losses in an elementary manner. To derive the same result in the manner in which the official solution derives it, apply the general version of Wald's identity. You have an infinite sequence of (i.i.d.) binary random variables $X_n$ that take the value $1$ for a loss and the value $0$ for a win, as well as a random number $N$ of them that you sum, and they satisfy the premises of the identity. In particular, even though $N$ is not independent of the $X_n$ (as required for the basic version of the identity), we have
$$
mathsf Eleft[X_n1_Nge nright]=mathsf Eleft[X_nright]mathsf P(Nge n);.
$$
Thus, the expected number of losses is the expected number of games played times the probability of a loss.
answered Jul 29 at 16:17
joriki
164k10179328
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Another answer has already derived the expected number of losses in an elementary manner. To derive the same result in the manner in which the official solution derives it, apply the general version of Wald's identity. You have an infinite sequence of (i.i.d.) binary random variables $X_n$ that take the value $1$ for a loss and the value $0$ for a win, as well as a random number $N$ of them that you sum, and they satisfy the premises of the identity. In particular, even though $N$ is not independent of the $X_n$ (as required for the basic version of the identity), we have
$$
mathsf Eleft[X_n1_Nge nright]=mathsf Eleft[X_nright]mathsf P(Nge n);.
$$
Thus, the expected number of losses is the expected number of games played times the probability of a loss.
answered Jul 29 at 16:17
joriki
164k10179328
Another answer has already derived the expected number of losses in an elementary manner. To derive the same result in the manner in which the official solution derives it, apply the general version of Wald's identity. You have an infinite sequence of (i.i.d.) binary random variables $X_n$ that take the value $1$ for a loss and the value $0$ for a win, as well as a random number $N$ of them that you sum, and they satisfy the premises of the identity. In particular, even though $N$ is not independent of the $X_n$ (as required for the basic version of the identity), we have
$$
mathsf Eleft[X_n1_Nge nright]=mathsf Eleft[X_nright]mathsf P(Nge n);.
$$
Thus, the expected number of losses is the expected number of games played times the probability of a loss.
answered Jul 29 at 16:17
joriki
164k10179328
answered Jul 29 at 16:17
joriki
164k10179328
answered Jul 29 at 16:17
joriki
164k10179328
answered Jul 29 at 16:17
joriki
164k10179328
164k10179328
add a comment |Â
add a comment |Â
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2
How do you get $E(X)$? How can we say if you approach is correct if you don´t tell us your thoughts and/or show your calculation.
– callculus
Jul 29 at 15:40
2
After your edit I changed my downvote into an upvote.
– callculus
Jul 29 at 15:53
In your second edit, the probability distribution you've defined has probabilities that, in general, do not add up to one. You may wish to consult the Wikipedia article I link to in my answer below.
– Theoretical Economist
Jul 29 at 16:24