Mixing
Die probability and net gain, expected result
Clash Royale CLAN TAG#URR8PPP
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The question is:
Suppose you roll a fair 6-sided die. If you roll a 2 you win $$4$, and
if you roll a 3 you win $$6$. If you roll any other number, you win
nothing. You must pay $1 to roll the die once. What is your expected
net gain (in dollars) for a single roll?
and the solution is as follows:
Let X be a random variable representing your gain from a single roll.
If you roll a 2, $X = 3$.
If you roll a 3, $X = 5$.
If you roll a $1, 4, 5,$ or $6$, $X = −1$.
It’s a fair die, so $P(X = 1) = 1/6, P(X = 2) = 1/6 , P(X
= −1) = 4/6$.
$E(X) = 3·P(X =1) + 5·P(X =2) − 1·P(X =−1) = 2/3$
What I am not understanding is where $P(X = 1)$ and $P(X = 2)$ came from or what they are.
probability
edited Jul 20 at 15:24
callculus
16.4k31427
asked Jul 20 at 15:12
ib-
153
add a comment |Â
up vote
1
down vote
favorite
The question is:
Suppose you roll a fair 6-sided die. If you roll a 2 you win $$4$, and
if you roll a 3 you win $$6$. If you roll any other number, you win
nothing. You must pay $1 to roll the die once. What is your expected
net gain (in dollars) for a single roll?
and the solution is as follows:
Let X be a random variable representing your gain from a single roll.
If you roll a 2, $X = 3$.
If you roll a 3, $X = 5$.
If you roll a $1, 4, 5,$ or $6$, $X = −1$.
It’s a fair die, so $P(X = 1) = 1/6, P(X = 2) = 1/6 , P(X
= −1) = 4/6$.
$E(X) = 3·P(X =1) + 5·P(X =2) − 1·P(X =−1) = 2/3$
What I am not understanding is where $P(X = 1)$ and $P(X = 2)$ came from or what they are.
probability
edited Jul 20 at 15:24
callculus
16.4k31427
asked Jul 20 at 15:12
ib-
153
"fair" here means that the probability of throwing any value among $1,2,3,4,5,6$ is equal (hence $frac 16$).
– lulu
Jul 20 at 15:21
2
Looks like a mere typo, thankfully. (so it should be $P(X=3) = 1/6$ in the line about fair dice and $3cdot P(X=3)$ et al in the final expectation calculation.
– Dan Uznanski
Jul 20 at 15:22
@DanUznanski Oh okay it makes sense now, thank you so much!
– ib-
Jul 20 at 15:27
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
The question is:
Suppose you roll a fair 6-sided die. If you roll a 2 you win $$4$, and
if you roll a 3 you win $$6$. If you roll any other number, you win
nothing. You must pay $1 to roll the die once. What is your expected
net gain (in dollars) for a single roll?
and the solution is as follows:
Let X be a random variable representing your gain from a single roll.
If you roll a 2, $X = 3$.
If you roll a 3, $X = 5$.
If you roll a $1, 4, 5,$ or $6$, $X = −1$.
It’s a fair die, so $P(X = 1) = 1/6, P(X = 2) = 1/6 , P(X
= −1) = 4/6$.
$E(X) = 3·P(X =1) + 5·P(X =2) − 1·P(X =−1) = 2/3$
What I am not understanding is where $P(X = 1)$ and $P(X = 2)$ came from or what they are.
probability
edited Jul 20 at 15:24
callculus
16.4k31427
asked Jul 20 at 15:12
ib-
153
The question is:
Suppose you roll a fair 6-sided die. If you roll a 2 you win $$4$, and
if you roll a 3 you win $$6$. If you roll any other number, you win
nothing. You must pay $1 to roll the die once. What is your expected
net gain (in dollars) for a single roll?
and the solution is as follows:
Let X be a random variable representing your gain from a single roll.
If you roll a 2, $X = 3$.
If you roll a 3, $X = 5$.
If you roll a $1, 4, 5,$ or $6$, $X = −1$.
It’s a fair die, so $P(X = 1) = 1/6, P(X = 2) = 1/6 , P(X
= −1) = 4/6$.
$E(X) = 3·P(X =1) + 5·P(X =2) − 1·P(X =−1) = 2/3$
What I am not understanding is where $P(X = 1)$ and $P(X = 2)$ came from or what they are.
probability
edited Jul 20 at 15:24
callculus
16.4k31427
asked Jul 20 at 15:12
ib-
153
edited Jul 20 at 15:24
callculus
16.4k31427
edited Jul 20 at 15:24
callculus
16.4k31427
edited Jul 20 at 15:24
callculus
16.4k31427
16.4k31427
asked Jul 20 at 15:12
ib-
153
asked Jul 20 at 15:12
ib-
153
asked Jul 20 at 15:12
ib-
153
153
"fair" here means that the probability of throwing any value among $1,2,3,4,5,6$ is equal (hence $frac 16$).
– lulu
Jul 20 at 15:21
2
Looks like a mere typo, thankfully. (so it should be $P(X=3) = 1/6$ in the line about fair dice and $3cdot P(X=3)$ et al in the final expectation calculation.
– Dan Uznanski
Jul 20 at 15:22
@DanUznanski Oh okay it makes sense now, thank you so much!
– ib-
Jul 20 at 15:27
add a comment |Â
"fair" here means that the probability of throwing any value among $1,2,3,4,5,6$ is equal (hence $frac 16$).
– lulu
Jul 20 at 15:21
2
Looks like a mere typo, thankfully. (so it should be $P(X=3) = 1/6$ in the line about fair dice and $3cdot P(X=3)$ et al in the final expectation calculation.
– Dan Uznanski
Jul 20 at 15:22
@DanUznanski Oh okay it makes sense now, thank you so much!
– ib-
Jul 20 at 15:27
"fair" here means that the probability of throwing any value among $1,2,3,4,5,6$ is equal (hence $frac 16$).
– lulu
Jul 20 at 15:21
"fair" here means that the probability of throwing any value among $1,2,3,4,5,6$ is equal (hence $frac 16$).
– lulu
Jul 20 at 15:21
2
2
Looks like a mere typo, thankfully. (so it should be $P(X=3) = 1/6$ in the line about fair dice and $3cdot P(X=3)$ et al in the final expectation calculation.
– Dan Uznanski
Jul 20 at 15:22
Looks like a mere typo, thankfully. (so it should be $P(X=3) = 1/6$ in the line about fair dice and $3cdot P(X=3)$ et al in the final expectation calculation.
– Dan Uznanski
Jul 20 at 15:22
@DanUznanski Oh okay it makes sense now, thank you so much!
– ib-
Jul 20 at 15:27
@DanUznanski Oh okay it makes sense now, thank you so much!
– ib-
Jul 20 at 15:27
add a comment |Â
1 Answer
1
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oldest
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0
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Where did you copy this from? There is a typo, the author should have either been consistently using the variable $X$ as he defines it, and express everything with it, or introduce another variable for the roll of the dice. He or she somehow mixed the two.
What should have been written is $P(X=3)=1/6$, $P(X=5)=1/6$ and $P(X=-1)=2/3$, so $E(X)= 3cdot P(X=3)+5cdot P(X=5)+(-1)cdot P(X=-1) = 2/3$.
answered Jul 20 at 15:29
A. Pongrácz
2,319221
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Where did you copy this from? There is a typo, the author should have either been consistently using the variable $X$ as he defines it, and express everything with it, or introduce another variable for the roll of the dice. He or she somehow mixed the two.
What should have been written is $P(X=3)=1/6$, $P(X=5)=1/6$ and $P(X=-1)=2/3$, so $E(X)= 3cdot P(X=3)+5cdot P(X=5)+(-1)cdot P(X=-1) = 2/3$.
answered Jul 20 at 15:29
A. Pongrácz
2,319221
add a comment |Â
up vote
0
down vote
Where did you copy this from? There is a typo, the author should have either been consistently using the variable $X$ as he defines it, and express everything with it, or introduce another variable for the roll of the dice. He or she somehow mixed the two.
What should have been written is $P(X=3)=1/6$, $P(X=5)=1/6$ and $P(X=-1)=2/3$, so $E(X)= 3cdot P(X=3)+5cdot P(X=5)+(-1)cdot P(X=-1) = 2/3$.
answered Jul 20 at 15:29
A. Pongrácz
2,319221
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Where did you copy this from? There is a typo, the author should have either been consistently using the variable $X$ as he defines it, and express everything with it, or introduce another variable for the roll of the dice. He or she somehow mixed the two.
What should have been written is $P(X=3)=1/6$, $P(X=5)=1/6$ and $P(X=-1)=2/3$, so $E(X)= 3cdot P(X=3)+5cdot P(X=5)+(-1)cdot P(X=-1) = 2/3$.
answered Jul 20 at 15:29
A. Pongrácz
2,319221
Where did you copy this from? There is a typo, the author should have either been consistently using the variable $X$ as he defines it, and express everything with it, or introduce another variable for the roll of the dice. He or she somehow mixed the two.
What should have been written is $P(X=3)=1/6$, $P(X=5)=1/6$ and $P(X=-1)=2/3$, so $E(X)= 3cdot P(X=3)+5cdot P(X=5)+(-1)cdot P(X=-1) = 2/3$.
answered Jul 20 at 15:29
A. Pongrácz
2,319221
answered Jul 20 at 15:29
A. Pongrácz
2,319221
answered Jul 20 at 15:29
A. Pongrácz
2,319221
answered Jul 20 at 15:29
A. Pongrácz
2,319221
2,319221
add a comment |Â
add a comment |Â
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"fair" here means that the probability of throwing any value among $1,2,3,4,5,6$ is equal (hence $frac 16$).
– lulu
Jul 20 at 15:21
2
Looks like a mere typo, thankfully. (so it should be $P(X=3) = 1/6$ in the line about fair dice and $3cdot P(X=3)$ et al in the final expectation calculation.
– Dan Uznanski
Jul 20 at 15:22
@DanUznanski Oh okay it makes sense now, thank you so much!
– ib-
Jul 20 at 15:27