Mixing
Probability dice don't match and expected difference
$begingroup$
The four musketeers decide to send a bouquet of roses to Snow White for
her Birthday. They toss a die each and contribute as many roses to the
bouquet as their die shows.
(i) What is the probability that at least two out of the four musketeers will
contribute different numbers of flowers?
For this part I assumed the probability of two number of flowers not matching is
$$frac{6}{6}times frac{5}{6}timesfrac{4}{6}timesfrac{3}{6}$$
For the first roll some number will be rolled so probability
is 1. For the second roll theres a 1/6 chance the roll will match the first so a 5/6 chance is won't match. For the third there's a 4/6 chance it won't match the 1st or 2nd roll and then with the 4th roll there's a 3/6 chance it will match one of the others.
Similarly for the probability that 3 and 4 choose different numbers of flowers I get
$$frac{6}{6}times frac{5}{6}timesfrac{4}{6}timesfrac{3}{6}$$
and
$$frac{6}{6}times frac{5}{6}timesfrac{4}{6}timesfrac{3}{6}$$I think this isn't right. Do i instead need to work with 1 - probability of same number.
(ii) Let X be a random variable representing the maximal out of the four
contributions, and let Y stands for the minimal one. Calculate the value
of joint probability, PXY (3, 2), that X = 3 AND Y = 2.
For the number of possibilities that satisify this we are selecting 4 outcomes (4 dice rolls) and there are 2 possibilities for each outcome, either a 2 or a 3, so there would be 24 possibilities and a sample space of 64 so the probability is 24/64?
(iii) Calculate the expected difference between the maximal and the minimal contributions, E(X − Y ).
I'm not too sure about this one, but would it be, the probability of having a minimum of 2 would be 54/64 and the probability of the maximum being 3 is 34/64 so would he expected difference be
$$(2)(5^4/6^4)-(3)(3^4/6^4)$$
probability
asked Jan 3 at 22:40
Richard CameronRichard Cameron
71
$endgroup$
add a comment |
$begingroup$
The four musketeers decide to send a bouquet of roses to Snow White for
her Birthday. They toss a die each and contribute as many roses to the
bouquet as their die shows.
(i) What is the probability that at least two out of the four musketeers will
contribute different numbers of flowers?
For this part I assumed the probability of two number of flowers not matching is
$$frac{6}{6}times frac{5}{6}timesfrac{4}{6}timesfrac{3}{6}$$
For the first roll some number will be rolled so probability
is 1. For the second roll theres a 1/6 chance the roll will match the first so a 5/6 chance is won't match. For the third there's a 4/6 chance it won't match the 1st or 2nd roll and then with the 4th roll there's a 3/6 chance it will match one of the others.
Similarly for the probability that 3 and 4 choose different numbers of flowers I get
$$frac{6}{6}times frac{5}{6}timesfrac{4}{6}timesfrac{3}{6}$$
and
$$frac{6}{6}times frac{5}{6}timesfrac{4}{6}timesfrac{3}{6}$$I think this isn't right. Do i instead need to work with 1 - probability of same number.
(ii) Let X be a random variable representing the maximal out of the four
contributions, and let Y stands for the minimal one. Calculate the value
of joint probability, PXY (3, 2), that X = 3 AND Y = 2.
For the number of possibilities that satisify this we are selecting 4 outcomes (4 dice rolls) and there are 2 possibilities for each outcome, either a 2 or a 3, so there would be 24 possibilities and a sample space of 64 so the probability is 24/64?
(iii) Calculate the expected difference between the maximal and the minimal contributions, E(X − Y ).
I'm not too sure about this one, but would it be, the probability of having a minimum of 2 would be 54/64 and the probability of the maximum being 3 is 34/64 so would he expected difference be
$$(2)(5^4/6^4)-(3)(3^4/6^4)$$
probability
asked Jan 3 at 22:40
Richard CameronRichard Cameron
71
$endgroup$
$begingroup$
I got around to (iii) - completing my answer.
$endgroup$
– herb steinberg
Jan 8 at 20:31
add a comment |
$begingroup$
The four musketeers decide to send a bouquet of roses to Snow White for
her Birthday. They toss a die each and contribute as many roses to the
bouquet as their die shows.
(i) What is the probability that at least two out of the four musketeers will
contribute different numbers of flowers?
For this part I assumed the probability of two number of flowers not matching is
$$frac{6}{6}times frac{5}{6}timesfrac{4}{6}timesfrac{3}{6}$$
For the first roll some number will be rolled so probability
is 1. For the second roll theres a 1/6 chance the roll will match the first so a 5/6 chance is won't match. For the third there's a 4/6 chance it won't match the 1st or 2nd roll and then with the 4th roll there's a 3/6 chance it will match one of the others.
Similarly for the probability that 3 and 4 choose different numbers of flowers I get
$$frac{6}{6}times frac{5}{6}timesfrac{4}{6}timesfrac{3}{6}$$
and
$$frac{6}{6}times frac{5}{6}timesfrac{4}{6}timesfrac{3}{6}$$I think this isn't right. Do i instead need to work with 1 - probability of same number.
(ii) Let X be a random variable representing the maximal out of the four
contributions, and let Y stands for the minimal one. Calculate the value
of joint probability, PXY (3, 2), that X = 3 AND Y = 2.
For the number of possibilities that satisify this we are selecting 4 outcomes (4 dice rolls) and there are 2 possibilities for each outcome, either a 2 or a 3, so there would be 24 possibilities and a sample space of 64 so the probability is 24/64?
(iii) Calculate the expected difference between the maximal and the minimal contributions, E(X − Y ).
I'm not too sure about this one, but would it be, the probability of having a minimum of 2 would be 54/64 and the probability of the maximum being 3 is 34/64 so would he expected difference be
$$(2)(5^4/6^4)-(3)(3^4/6^4)$$
probability
asked Jan 3 at 22:40
Richard CameronRichard Cameron
71
$endgroup$
The four musketeers decide to send a bouquet of roses to Snow White for
her Birthday. They toss a die each and contribute as many roses to the
bouquet as their die shows.
(i) What is the probability that at least two out of the four musketeers will
contribute different numbers of flowers?
For this part I assumed the probability of two number of flowers not matching is
$$frac{6}{6}times frac{5}{6}timesfrac{4}{6}timesfrac{3}{6}$$
For the first roll some number will be rolled so probability
is 1. For the second roll theres a 1/6 chance the roll will match the first so a 5/6 chance is won't match. For the third there's a 4/6 chance it won't match the 1st or 2nd roll and then with the 4th roll there's a 3/6 chance it will match one of the others.
Similarly for the probability that 3 and 4 choose different numbers of flowers I get
$$frac{6}{6}times frac{5}{6}timesfrac{4}{6}timesfrac{3}{6}$$
and
$$frac{6}{6}times frac{5}{6}timesfrac{4}{6}timesfrac{3}{6}$$I think this isn't right. Do i instead need to work with 1 - probability of same number.
(ii) Let X be a random variable representing the maximal out of the four
contributions, and let Y stands for the minimal one. Calculate the value
of joint probability, PXY (3, 2), that X = 3 AND Y = 2.
For the number of possibilities that satisify this we are selecting 4 outcomes (4 dice rolls) and there are 2 possibilities for each outcome, either a 2 or a 3, so there would be 24 possibilities and a sample space of 64 so the probability is 24/64?
(iii) Calculate the expected difference between the maximal and the minimal contributions, E(X − Y ).
I'm not too sure about this one, but would it be, the probability of having a minimum of 2 would be 54/64 and the probability of the maximum being 3 is 34/64 so would he expected difference be
$$(2)(5^4/6^4)-(3)(3^4/6^4)$$
probability
probability
asked Jan 3 at 22:40
Richard CameronRichard Cameron
71
asked Jan 3 at 22:40
Richard CameronRichard Cameron
71
asked Jan 3 at 22:40
Richard CameronRichard Cameron
71
asked Jan 3 at 22:40
Richard CameronRichard Cameron
71
asked Jan 3 at 22:40
Richard CameronRichard Cameron
71
71
$begingroup$
I got around to (iii) - completing my answer.
$endgroup$
– herb steinberg
Jan 8 at 20:31
add a comment |
$begingroup$
I got around to (iii) - completing my answer.
$endgroup$
– herb steinberg
Jan 8 at 20:31
$begingroup$
I got around to (iii) - completing my answer.
$endgroup$
– herb steinberg
Jan 8 at 20:31
$begingroup$
I got around to (iii) - completing my answer.
$endgroup$
– herb steinberg
Jan 8 at 20:31
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
In part (ii), you also need to be sure that not all the rolls are all $2$ or $3$. As such, your answer should be
$$
(2/6)^4 - (1/6)^4 - (1/6)^4
$$
Since there is a $(1/6)^4$ chance of getting all 2, which is equal to chance of getting all 3.
This can also be done by multinational experiment.
$$
sum_{n=1}^3 frac{4!}{n!(4-n)!}(1/6)^4
$$
In part (iii), the dice rolls is uniform with a pmf of $1/6$ for $i = 1,2,...,6$. As such, the cdf is
$$
F(y) = y/6
$$
Notice that $P(Y_{max} < y)$ occurs only if all other rolls are less than $y$ also. Therefore
$$
P(Y_{max} leq y) = F(y)^4 = (y/6)^4
$$
As such, using $G(y)$ and $H(y)$ to simplify notation, we have
$$
G(y) = P(Y_{max} = y) = P(Y_{max} leq y) - P(Y_{max} leq (y-1)) = (y/6)^4 - ((y-1)/6)^4
$$
Similarly, since $P(Y_{min} geq y) = 1-(1-F(y))^4$, we have
$$
H(y) = P(Y_{min} = y) = P(Y_{min} leq y) - P(Y_{min} leq (y-1)) = ((7-y)/6)^4 - ((6-y)/6)^4
$$
Therefore, summing all possible values of y (1 through 6).
$$E(X-Y) = E(X) - E(Y) = sum yG(y) - sum yH(y) \= 5.24459876543 - 1.75540123457 = 3.48919753086
$$
answered Jan 3 at 23:04
Bryden CBryden C
30418
$endgroup$
$begingroup$
$G(y)$ and $H(y)$ are as defined independently. However don't you need to take into account min $le$ max?
$endgroup$
– herb steinberg
Jan 4 at 6:30
$begingroup$
I finally got around to (iii) and got the same answer as you did.
$endgroup$
– herb steinberg
Jan 8 at 20:28
$begingroup$
Did you use a similar method or completely different?
$endgroup$
– Bryden C
Jan 8 at 22:12
$begingroup$
I didn't try to compare methods, but I presume there must be some similarity, since we got the same answer. As you may have noticed from my previous comments I was concerned about the dependency between $X$ and $Y$, so I tried to explicitly include the connection in my analysis.
$endgroup$
– herb steinberg
Jan 8 at 22:29
$begingroup$
I must have accidentally overlooked that comment, whoops. Do you mind sharing your solution?
$endgroup$
– Bryden C
Jan 8 at 22:41
|
show 1 more comment
$begingroup$
Hint on (iii).
By linearity of expectation: $mathbb E(X-Y)=mathbb EX-mathbb EY$.
What follows makes it more easy to find $mathbb EX$ and $mathbb EY$.
If $Z$ takes values in ${1,2,3,4,5,6}$ then: $$mathbb EZ=sum_{k=1}^6kP(Z=k)=sum_{k=1}^6sum_{r=1}^kP(Z=k)=sum_{r=1}^6sum_{k=r}^6P(Z=k)=sum_{r=1}^6P(Zgeq r)$$
Applying this on $X$ we find:$$mathbb EX=sum_{r=1}^6P(Xgeq r)=sum_{r=1}^6left(1-P(X<r)right)=6-sum_{r=1}^6P(X<r)$$where $P(X<r)$ is quite easy to find.
Also you apply this on $Y$ where $P(Ygeq r)$ is quite easy to find.
answered Jan 4 at 9:39
drhabdrhab
99.1k544130
$endgroup$
add a comment |
$begingroup$
For (i) You started out with the right approach, but the wrong question, although your last remark is correct. The opposite possibility is that all four have the same number which is $P=(frac{1}{6})^3$, so the probability you want is $1-(frac{1}{6})^3=0.99537037037037$
For (ii) You need to subtract the probabilities of all 3 or all 2 which $frac{2}{6^4}$.
For (iii) Let $Z=X-Y, E(Z)=sum_{z=0}^5zP(Z=z)$. To get $P(Z=z)$, consider all pairs $(X,Y)$ where $X-Y=z$. For each $z$ there are $6-z$ such pairs and the probabilities are the same for each pair. To get the probability for each pair for a specific $z$, $P(Y=y,X=x)=P(Yle y,Xge x)-P(Ylt y,Xge x)-P(Yle y,Xgt x)+P(Ylt y,Xgt x)$
$=
{frac{z+1}{6}-2frac{z}{6}+frac{z-1}{6}(see note).}$
Therefore $E(Z)=sum_{z=0}^5 z(6-z)(frac{z+1}{6}-2frac{z}{6}+frac{z-1}{6})=5-frac{1958}{1296}=3.489197530864198.$
Note: The terms where the numerator $=0$ or $-1$ are omitted.
answered Jan 3 at 23:02
herb steinbergherb steinberg
2,5232310
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
In part (ii), you also need to be sure that not all the rolls are all $2$ or $3$. As such, your answer should be
$$
(2/6)^4 - (1/6)^4 - (1/6)^4
$$
Since there is a $(1/6)^4$ chance of getting all 2, which is equal to chance of getting all 3.
This can also be done by multinational experiment.
$$
sum_{n=1}^3 frac{4!}{n!(4-n)!}(1/6)^4
$$
In part (iii), the dice rolls is uniform with a pmf of $1/6$ for $i = 1,2,...,6$. As such, the cdf is
$$
F(y) = y/6
$$
Notice that $P(Y_{max} < y)$ occurs only if all other rolls are less than $y$ also. Therefore
$$
P(Y_{max} leq y) = F(y)^4 = (y/6)^4
$$
As such, using $G(y)$ and $H(y)$ to simplify notation, we have
$$
G(y) = P(Y_{max} = y) = P(Y_{max} leq y) - P(Y_{max} leq (y-1)) = (y/6)^4 - ((y-1)/6)^4
$$
Similarly, since $P(Y_{min} geq y) = 1-(1-F(y))^4$, we have
$$
H(y) = P(Y_{min} = y) = P(Y_{min} leq y) - P(Y_{min} leq (y-1)) = ((7-y)/6)^4 - ((6-y)/6)^4
$$
Therefore, summing all possible values of y (1 through 6).
$$E(X-Y) = E(X) - E(Y) = sum yG(y) - sum yH(y) \= 5.24459876543 - 1.75540123457 = 3.48919753086
$$
answered Jan 3 at 23:04
Bryden CBryden C
30418
$endgroup$
$begingroup$
$G(y)$ and $H(y)$ are as defined independently. However don't you need to take into account min $le$ max?
$endgroup$
– herb steinberg
Jan 4 at 6:30
$begingroup$
I finally got around to (iii) and got the same answer as you did.
$endgroup$
– herb steinberg
Jan 8 at 20:28
$begingroup$
Did you use a similar method or completely different?
$endgroup$
– Bryden C
Jan 8 at 22:12
$begingroup$
I didn't try to compare methods, but I presume there must be some similarity, since we got the same answer. As you may have noticed from my previous comments I was concerned about the dependency between $X$ and $Y$, so I tried to explicitly include the connection in my analysis.
$endgroup$
– herb steinberg
Jan 8 at 22:29
$begingroup$
I must have accidentally overlooked that comment, whoops. Do you mind sharing your solution?
$endgroup$
– Bryden C
Jan 8 at 22:41
|
show 1 more comment
$begingroup$
In part (ii), you also need to be sure that not all the rolls are all $2$ or $3$. As such, your answer should be
$$
(2/6)^4 - (1/6)^4 - (1/6)^4
$$
Since there is a $(1/6)^4$ chance of getting all 2, which is equal to chance of getting all 3.
This can also be done by multinational experiment.
$$
sum_{n=1}^3 frac{4!}{n!(4-n)!}(1/6)^4
$$
In part (iii), the dice rolls is uniform with a pmf of $1/6$ for $i = 1,2,...,6$. As such, the cdf is
$$
F(y) = y/6
$$
Notice that $P(Y_{max} < y)$ occurs only if all other rolls are less than $y$ also. Therefore
$$
P(Y_{max} leq y) = F(y)^4 = (y/6)^4
$$
As such, using $G(y)$ and $H(y)$ to simplify notation, we have
$$
G(y) = P(Y_{max} = y) = P(Y_{max} leq y) - P(Y_{max} leq (y-1)) = (y/6)^4 - ((y-1)/6)^4
$$
Similarly, since $P(Y_{min} geq y) = 1-(1-F(y))^4$, we have
$$
H(y) = P(Y_{min} = y) = P(Y_{min} leq y) - P(Y_{min} leq (y-1)) = ((7-y)/6)^4 - ((6-y)/6)^4
$$
Therefore, summing all possible values of y (1 through 6).
$$E(X-Y) = E(X) - E(Y) = sum yG(y) - sum yH(y) \= 5.24459876543 - 1.75540123457 = 3.48919753086
$$
answered Jan 3 at 23:04
Bryden CBryden C
30418
$endgroup$
$begingroup$
$G(y)$ and $H(y)$ are as defined independently. However don't you need to take into account min $le$ max?
$endgroup$
– herb steinberg
Jan 4 at 6:30
$begingroup$
I finally got around to (iii) and got the same answer as you did.
$endgroup$
– herb steinberg
Jan 8 at 20:28
$begingroup$
Did you use a similar method or completely different?
$endgroup$
– Bryden C
Jan 8 at 22:12
$begingroup$
I didn't try to compare methods, but I presume there must be some similarity, since we got the same answer. As you may have noticed from my previous comments I was concerned about the dependency between $X$ and $Y$, so I tried to explicitly include the connection in my analysis.
$endgroup$
– herb steinberg
Jan 8 at 22:29
$begingroup$
I must have accidentally overlooked that comment, whoops. Do you mind sharing your solution?
$endgroup$
– Bryden C
Jan 8 at 22:41
|
show 1 more comment
$begingroup$
In part (ii), you also need to be sure that not all the rolls are all $2$ or $3$. As such, your answer should be
$$
(2/6)^4 - (1/6)^4 - (1/6)^4
$$
Since there is a $(1/6)^4$ chance of getting all 2, which is equal to chance of getting all 3.
This can also be done by multinational experiment.
$$
sum_{n=1}^3 frac{4!}{n!(4-n)!}(1/6)^4
$$
In part (iii), the dice rolls is uniform with a pmf of $1/6$ for $i = 1,2,...,6$. As such, the cdf is
$$
F(y) = y/6
$$
Notice that $P(Y_{max} < y)$ occurs only if all other rolls are less than $y$ also. Therefore
$$
P(Y_{max} leq y) = F(y)^4 = (y/6)^4
$$
As such, using $G(y)$ and $H(y)$ to simplify notation, we have
$$
G(y) = P(Y_{max} = y) = P(Y_{max} leq y) - P(Y_{max} leq (y-1)) = (y/6)^4 - ((y-1)/6)^4
$$
Similarly, since $P(Y_{min} geq y) = 1-(1-F(y))^4$, we have
$$
H(y) = P(Y_{min} = y) = P(Y_{min} leq y) - P(Y_{min} leq (y-1)) = ((7-y)/6)^4 - ((6-y)/6)^4
$$
Therefore, summing all possible values of y (1 through 6).
$$E(X-Y) = E(X) - E(Y) = sum yG(y) - sum yH(y) \= 5.24459876543 - 1.75540123457 = 3.48919753086
$$
answered Jan 3 at 23:04
Bryden CBryden C
30418
$endgroup$
In part (ii), you also need to be sure that not all the rolls are all $2$ or $3$. As such, your answer should be
$$
(2/6)^4 - (1/6)^4 - (1/6)^4
$$
Since there is a $(1/6)^4$ chance of getting all 2, which is equal to chance of getting all 3.
This can also be done by multinational experiment.
$$
sum_{n=1}^3 frac{4!}{n!(4-n)!}(1/6)^4
$$
In part (iii), the dice rolls is uniform with a pmf of $1/6$ for $i = 1,2,...,6$. As such, the cdf is
$$
F(y) = y/6
$$
Notice that $P(Y_{max} < y)$ occurs only if all other rolls are less than $y$ also. Therefore
$$
P(Y_{max} leq y) = F(y)^4 = (y/6)^4
$$
As such, using $G(y)$ and $H(y)$ to simplify notation, we have
$$
G(y) = P(Y_{max} = y) = P(Y_{max} leq y) - P(Y_{max} leq (y-1)) = (y/6)^4 - ((y-1)/6)^4
$$
Similarly, since $P(Y_{min} geq y) = 1-(1-F(y))^4$, we have
$$
H(y) = P(Y_{min} = y) = P(Y_{min} leq y) - P(Y_{min} leq (y-1)) = ((7-y)/6)^4 - ((6-y)/6)^4
$$
Therefore, summing all possible values of y (1 through 6).
$$E(X-Y) = E(X) - E(Y) = sum yG(y) - sum yH(y) \= 5.24459876543 - 1.75540123457 = 3.48919753086
$$
answered Jan 3 at 23:04
Bryden CBryden C
30418
edited Jan 3 at 23:33
answered Jan 3 at 23:04
Bryden CBryden C
30418
answered Jan 3 at 23:04
Bryden CBryden C
30418
answered Jan 3 at 23:04
Bryden CBryden C
30418
30418
$begingroup$
$G(y)$ and $H(y)$ are as defined independently. However don't you need to take into account min $le$ max?
$endgroup$
– herb steinberg
Jan 4 at 6:30
$begingroup$
I finally got around to (iii) and got the same answer as you did.
$endgroup$
– herb steinberg
Jan 8 at 20:28
$begingroup$
Did you use a similar method or completely different?
$endgroup$
– Bryden C
Jan 8 at 22:12
$begingroup$
I didn't try to compare methods, but I presume there must be some similarity, since we got the same answer. As you may have noticed from my previous comments I was concerned about the dependency between $X$ and $Y$, so I tried to explicitly include the connection in my analysis.
$endgroup$
– herb steinberg
Jan 8 at 22:29
$begingroup$
I must have accidentally overlooked that comment, whoops. Do you mind sharing your solution?
$endgroup$
– Bryden C
Jan 8 at 22:41
|
show 1 more comment
$begingroup$
$G(y)$ and $H(y)$ are as defined independently. However don't you need to take into account min $le$ max?
$endgroup$
– herb steinberg
Jan 4 at 6:30
$begingroup$
I finally got around to (iii) and got the same answer as you did.
$endgroup$
– herb steinberg
Jan 8 at 20:28
$begingroup$
Did you use a similar method or completely different?
$endgroup$
– Bryden C
Jan 8 at 22:12
$begingroup$
I didn't try to compare methods, but I presume there must be some similarity, since we got the same answer. As you may have noticed from my previous comments I was concerned about the dependency between $X$ and $Y$, so I tried to explicitly include the connection in my analysis.
$endgroup$
– herb steinberg
Jan 8 at 22:29
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I must have accidentally overlooked that comment, whoops. Do you mind sharing your solution?
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– Bryden C
Jan 8 at 22:41
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$G(y)$ and $H(y)$ are as defined independently. However don't you need to take into account min $le$ max?
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– herb steinberg
Jan 4 at 6:30
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$G(y)$ and $H(y)$ are as defined independently. However don't you need to take into account min $le$ max?
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– herb steinberg
Jan 4 at 6:30
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I finally got around to (iii) and got the same answer as you did.
$endgroup$
– herb steinberg
Jan 8 at 20:28
$begingroup$
I finally got around to (iii) and got the same answer as you did.
$endgroup$
– herb steinberg
Jan 8 at 20:28
$begingroup$
Did you use a similar method or completely different?
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– Bryden C
Jan 8 at 22:12
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Did you use a similar method or completely different?
$endgroup$
– Bryden C
Jan 8 at 22:12
$begingroup$
I didn't try to compare methods, but I presume there must be some similarity, since we got the same answer. As you may have noticed from my previous comments I was concerned about the dependency between $X$ and $Y$, so I tried to explicitly include the connection in my analysis.
$endgroup$
– herb steinberg
Jan 8 at 22:29
$begingroup$
I didn't try to compare methods, but I presume there must be some similarity, since we got the same answer. As you may have noticed from my previous comments I was concerned about the dependency between $X$ and $Y$, so I tried to explicitly include the connection in my analysis.
$endgroup$
– herb steinberg
Jan 8 at 22:29
$begingroup$
I must have accidentally overlooked that comment, whoops. Do you mind sharing your solution?
$endgroup$
– Bryden C
Jan 8 at 22:41
$begingroup$
I must have accidentally overlooked that comment, whoops. Do you mind sharing your solution?
$endgroup$
– Bryden C
Jan 8 at 22:41
|
show 1 more comment
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Hint on (iii).
By linearity of expectation: $mathbb E(X-Y)=mathbb EX-mathbb EY$.
What follows makes it more easy to find $mathbb EX$ and $mathbb EY$.
If $Z$ takes values in ${1,2,3,4,5,6}$ then: $$mathbb EZ=sum_{k=1}^6kP(Z=k)=sum_{k=1}^6sum_{r=1}^kP(Z=k)=sum_{r=1}^6sum_{k=r}^6P(Z=k)=sum_{r=1}^6P(Zgeq r)$$
Applying this on $X$ we find:$$mathbb EX=sum_{r=1}^6P(Xgeq r)=sum_{r=1}^6left(1-P(X<r)right)=6-sum_{r=1}^6P(X<r)$$where $P(X<r)$ is quite easy to find.
Also you apply this on $Y$ where $P(Ygeq r)$ is quite easy to find.
answered Jan 4 at 9:39
drhabdrhab
99.1k544130
$endgroup$
add a comment |
$begingroup$
Hint on (iii).
By linearity of expectation: $mathbb E(X-Y)=mathbb EX-mathbb EY$.
What follows makes it more easy to find $mathbb EX$ and $mathbb EY$.
If $Z$ takes values in ${1,2,3,4,5,6}$ then: $$mathbb EZ=sum_{k=1}^6kP(Z=k)=sum_{k=1}^6sum_{r=1}^kP(Z=k)=sum_{r=1}^6sum_{k=r}^6P(Z=k)=sum_{r=1}^6P(Zgeq r)$$
Applying this on $X$ we find:$$mathbb EX=sum_{r=1}^6P(Xgeq r)=sum_{r=1}^6left(1-P(X<r)right)=6-sum_{r=1}^6P(X<r)$$where $P(X<r)$ is quite easy to find.
Also you apply this on $Y$ where $P(Ygeq r)$ is quite easy to find.
answered Jan 4 at 9:39
drhabdrhab
99.1k544130
$endgroup$
add a comment |
$begingroup$
Hint on (iii).
By linearity of expectation: $mathbb E(X-Y)=mathbb EX-mathbb EY$.
What follows makes it more easy to find $mathbb EX$ and $mathbb EY$.
If $Z$ takes values in ${1,2,3,4,5,6}$ then: $$mathbb EZ=sum_{k=1}^6kP(Z=k)=sum_{k=1}^6sum_{r=1}^kP(Z=k)=sum_{r=1}^6sum_{k=r}^6P(Z=k)=sum_{r=1}^6P(Zgeq r)$$
Applying this on $X$ we find:$$mathbb EX=sum_{r=1}^6P(Xgeq r)=sum_{r=1}^6left(1-P(X<r)right)=6-sum_{r=1}^6P(X<r)$$where $P(X<r)$ is quite easy to find.
Also you apply this on $Y$ where $P(Ygeq r)$ is quite easy to find.
answered Jan 4 at 9:39
drhabdrhab
99.1k544130
$endgroup$
Hint on (iii).
By linearity of expectation: $mathbb E(X-Y)=mathbb EX-mathbb EY$.
What follows makes it more easy to find $mathbb EX$ and $mathbb EY$.
If $Z$ takes values in ${1,2,3,4,5,6}$ then: $$mathbb EZ=sum_{k=1}^6kP(Z=k)=sum_{k=1}^6sum_{r=1}^kP(Z=k)=sum_{r=1}^6sum_{k=r}^6P(Z=k)=sum_{r=1}^6P(Zgeq r)$$
Applying this on $X$ we find:$$mathbb EX=sum_{r=1}^6P(Xgeq r)=sum_{r=1}^6left(1-P(X<r)right)=6-sum_{r=1}^6P(X<r)$$where $P(X<r)$ is quite easy to find.
Also you apply this on $Y$ where $P(Ygeq r)$ is quite easy to find.
answered Jan 4 at 9:39
drhabdrhab
99.1k544130
answered Jan 4 at 9:39
drhabdrhab
99.1k544130
answered Jan 4 at 9:39
drhabdrhab
99.1k544130
answered Jan 4 at 9:39
drhabdrhab
99.1k544130
99.1k544130
add a comment |
add a comment |
$begingroup$
For (i) You started out with the right approach, but the wrong question, although your last remark is correct. The opposite possibility is that all four have the same number which is $P=(frac{1}{6})^3$, so the probability you want is $1-(frac{1}{6})^3=0.99537037037037$
For (ii) You need to subtract the probabilities of all 3 or all 2 which $frac{2}{6^4}$.
For (iii) Let $Z=X-Y, E(Z)=sum_{z=0}^5zP(Z=z)$. To get $P(Z=z)$, consider all pairs $(X,Y)$ where $X-Y=z$. For each $z$ there are $6-z$ such pairs and the probabilities are the same for each pair. To get the probability for each pair for a specific $z$, $P(Y=y,X=x)=P(Yle y,Xge x)-P(Ylt y,Xge x)-P(Yle y,Xgt x)+P(Ylt y,Xgt x)$
$=
{frac{z+1}{6}-2frac{z}{6}+frac{z-1}{6}(see note).}$
Therefore $E(Z)=sum_{z=0}^5 z(6-z)(frac{z+1}{6}-2frac{z}{6}+frac{z-1}{6})=5-frac{1958}{1296}=3.489197530864198.$
Note: The terms where the numerator $=0$ or $-1$ are omitted.
answered Jan 3 at 23:02
herb steinbergherb steinberg
2,5232310
$endgroup$
add a comment |
$begingroup$
For (i) You started out with the right approach, but the wrong question, although your last remark is correct. The opposite possibility is that all four have the same number which is $P=(frac{1}{6})^3$, so the probability you want is $1-(frac{1}{6})^3=0.99537037037037$
For (ii) You need to subtract the probabilities of all 3 or all 2 which $frac{2}{6^4}$.
For (iii) Let $Z=X-Y, E(Z)=sum_{z=0}^5zP(Z=z)$. To get $P(Z=z)$, consider all pairs $(X,Y)$ where $X-Y=z$. For each $z$ there are $6-z$ such pairs and the probabilities are the same for each pair. To get the probability for each pair for a specific $z$, $P(Y=y,X=x)=P(Yle y,Xge x)-P(Ylt y,Xge x)-P(Yle y,Xgt x)+P(Ylt y,Xgt x)$
$=
{frac{z+1}{6}-2frac{z}{6}+frac{z-1}{6}(see note).}$
Therefore $E(Z)=sum_{z=0}^5 z(6-z)(frac{z+1}{6}-2frac{z}{6}+frac{z-1}{6})=5-frac{1958}{1296}=3.489197530864198.$
Note: The terms where the numerator $=0$ or $-1$ are omitted.
answered Jan 3 at 23:02
herb steinbergherb steinberg
2,5232310
$endgroup$
add a comment |
$begingroup$
For (i) You started out with the right approach, but the wrong question, although your last remark is correct. The opposite possibility is that all four have the same number which is $P=(frac{1}{6})^3$, so the probability you want is $1-(frac{1}{6})^3=0.99537037037037$
For (ii) You need to subtract the probabilities of all 3 or all 2 which $frac{2}{6^4}$.
For (iii) Let $Z=X-Y, E(Z)=sum_{z=0}^5zP(Z=z)$. To get $P(Z=z)$, consider all pairs $(X,Y)$ where $X-Y=z$. For each $z$ there are $6-z$ such pairs and the probabilities are the same for each pair. To get the probability for each pair for a specific $z$, $P(Y=y,X=x)=P(Yle y,Xge x)-P(Ylt y,Xge x)-P(Yle y,Xgt x)+P(Ylt y,Xgt x)$
$=
{frac{z+1}{6}-2frac{z}{6}+frac{z-1}{6}(see note).}$
Therefore $E(Z)=sum_{z=0}^5 z(6-z)(frac{z+1}{6}-2frac{z}{6}+frac{z-1}{6})=5-frac{1958}{1296}=3.489197530864198.$
Note: The terms where the numerator $=0$ or $-1$ are omitted.
answered Jan 3 at 23:02
herb steinbergherb steinberg
2,5232310
$endgroup$
For (i) You started out with the right approach, but the wrong question, although your last remark is correct. The opposite possibility is that all four have the same number which is $P=(frac{1}{6})^3$, so the probability you want is $1-(frac{1}{6})^3=0.99537037037037$
For (ii) You need to subtract the probabilities of all 3 or all 2 which $frac{2}{6^4}$.
For (iii) Let $Z=X-Y, E(Z)=sum_{z=0}^5zP(Z=z)$. To get $P(Z=z)$, consider all pairs $(X,Y)$ where $X-Y=z$. For each $z$ there are $6-z$ such pairs and the probabilities are the same for each pair. To get the probability for each pair for a specific $z$, $P(Y=y,X=x)=P(Yle y,Xge x)-P(Ylt y,Xge x)-P(Yle y,Xgt x)+P(Ylt y,Xgt x)$
$=
{frac{z+1}{6}-2frac{z}{6}+frac{z-1}{6}(see note).}$
Therefore $E(Z)=sum_{z=0}^5 z(6-z)(frac{z+1}{6}-2frac{z}{6}+frac{z-1}{6})=5-frac{1958}{1296}=3.489197530864198.$
Note: The terms where the numerator $=0$ or $-1$ are omitted.
answered Jan 3 at 23:02
herb steinbergherb steinberg
2,5232310
edited Jan 8 at 20:25
answered Jan 3 at 23:02
herb steinbergherb steinberg
2,5232310
answered Jan 3 at 23:02
herb steinbergherb steinberg
2,5232310
answered Jan 3 at 23:02
herb steinbergherb steinberg
2,5232310
2,5232310
add a comment |
add a comment |
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$begingroup$
I got around to (iii) - completing my answer.
$endgroup$
– herb steinberg
Jan 8 at 20:31