Mixing
Expected value for a sum of two special dice
$begingroup$
I am trying to find the expected value for the sum of two dices, given that if both dices shows an identical number, then the sum is doubled.
Basically if we get $1$ for a dice and $1$ for the other one, this will count as: $4$, similarly: $2+2=8$, $3+3=12$, $4+4=16$, $5+5=20$, $6+6=24$. If any other combination appears then it is counted as for a normal dice.
I know how to find the expected value for two normal dices, since I just find it for one, then double it.
$$E(x)={frac16} (1+2+3+4+5+6)=frac16frac{6cdot 7}{2}=frac72$$
And doubling it gives the expected value for two dices to be $7$. I would expected that in the question case the expected value to be a little higher, but I don't know how to start calculating it.
I don't need a full solution, only some good hints that will help me to calculate it later.
Edit. I think I have gotten an idea to draw a matrix.
$$begin{array}[ht]{|p{2cm}|||p{0.5cm}|p{0.5cm}|p{0.5cm}|p{0.5cm}|p{0.5cm}|p{0.5cm}|p{0.5cm}|p{0.5cm}|p{0.5cm}|} hline text{ x } & 1 &2 &3 &4 &5 &6 \ hline hline hline 1 &4 &3 &4 &5 &6 &7 \ hline 2& 3 & 8 &5 &6 &7&8 \ hline 3& 4 &5 &12 &7 &8&9\ hline 4 &5 &6 &7&16&9&10 \ hline 5 &6 &7&8&9&20&11 \ hline 6&7&8&9&10&11&24 \ hline end{array}$$
And now the expected value in our case is:
$$E=frac1{36}left(l1+l2+l3+l4+l5+l6right)$$
Where $l_k$ is the sum of the numbers in $l_k$.
This gives:
$$E=frac1{36}left(29+37+45+53+61+69right)=frac{294}{36}=8.1(6)$$
Is this fine?
probability dice expected-value
edited Feb 10 at 10:43
jvdhooft
5,65961641
$endgroup$
add a comment |
$begingroup$
I am trying to find the expected value for the sum of two dices, given that if both dices shows an identical number, then the sum is doubled.
Basically if we get $1$ for a dice and $1$ for the other one, this will count as: $4$, similarly: $2+2=8$, $3+3=12$, $4+4=16$, $5+5=20$, $6+6=24$. If any other combination appears then it is counted as for a normal dice.
I know how to find the expected value for two normal dices, since I just find it for one, then double it.
$$E(x)={frac16} (1+2+3+4+5+6)=frac16frac{6cdot 7}{2}=frac72$$
And doubling it gives the expected value for two dices to be $7$. I would expected that in the question case the expected value to be a little higher, but I don't know how to start calculating it.
I don't need a full solution, only some good hints that will help me to calculate it later.
Edit. I think I have gotten an idea to draw a matrix.
$$begin{array}[ht]{|p{2cm}|||p{0.5cm}|p{0.5cm}|p{0.5cm}|p{0.5cm}|p{0.5cm}|p{0.5cm}|p{0.5cm}|p{0.5cm}|p{0.5cm}|} hline text{ x } & 1 &2 &3 &4 &5 &6 \ hline hline hline 1 &4 &3 &4 &5 &6 &7 \ hline 2& 3 & 8 &5 &6 &7&8 \ hline 3& 4 &5 &12 &7 &8&9\ hline 4 &5 &6 &7&16&9&10 \ hline 5 &6 &7&8&9&20&11 \ hline 6&7&8&9&10&11&24 \ hline end{array}$$
And now the expected value in our case is:
$$E=frac1{36}left(l1+l2+l3+l4+l5+l6right)$$
Where $l_k$ is the sum of the numbers in $l_k$.
This gives:
$$E=frac1{36}left(29+37+45+53+61+69right)=frac{294}{36}=8.1(6)$$
Is this fine?
probability dice expected-value
edited Feb 10 at 10:43
jvdhooft
5,65961641
$endgroup$
add a comment |
$begingroup$
I am trying to find the expected value for the sum of two dices, given that if both dices shows an identical number, then the sum is doubled.
Basically if we get $1$ for a dice and $1$ for the other one, this will count as: $4$, similarly: $2+2=8$, $3+3=12$, $4+4=16$, $5+5=20$, $6+6=24$. If any other combination appears then it is counted as for a normal dice.
I know how to find the expected value for two normal dices, since I just find it for one, then double it.
$$E(x)={frac16} (1+2+3+4+5+6)=frac16frac{6cdot 7}{2}=frac72$$
And doubling it gives the expected value for two dices to be $7$. I would expected that in the question case the expected value to be a little higher, but I don't know how to start calculating it.
I don't need a full solution, only some good hints that will help me to calculate it later.
Edit. I think I have gotten an idea to draw a matrix.
$$begin{array}[ht]{|p{2cm}|||p{0.5cm}|p{0.5cm}|p{0.5cm}|p{0.5cm}|p{0.5cm}|p{0.5cm}|p{0.5cm}|p{0.5cm}|p{0.5cm}|} hline text{ x } & 1 &2 &3 &4 &5 &6 \ hline hline hline 1 &4 &3 &4 &5 &6 &7 \ hline 2& 3 & 8 &5 &6 &7&8 \ hline 3& 4 &5 &12 &7 &8&9\ hline 4 &5 &6 &7&16&9&10 \ hline 5 &6 &7&8&9&20&11 \ hline 6&7&8&9&10&11&24 \ hline end{array}$$
And now the expected value in our case is:
$$E=frac1{36}left(l1+l2+l3+l4+l5+l6right)$$
Where $l_k$ is the sum of the numbers in $l_k$.
This gives:
$$E=frac1{36}left(29+37+45+53+61+69right)=frac{294}{36}=8.1(6)$$
Is this fine?
probability dice expected-value
edited Feb 10 at 10:43
jvdhooft
5,65961641
$endgroup$
I am trying to find the expected value for the sum of two dices, given that if both dices shows an identical number, then the sum is doubled.
Basically if we get $1$ for a dice and $1$ for the other one, this will count as: $4$, similarly: $2+2=8$, $3+3=12$, $4+4=16$, $5+5=20$, $6+6=24$. If any other combination appears then it is counted as for a normal dice.
I know how to find the expected value for two normal dices, since I just find it for one, then double it.
$$E(x)={frac16} (1+2+3+4+5+6)=frac16frac{6cdot 7}{2}=frac72$$
And doubling it gives the expected value for two dices to be $7$. I would expected that in the question case the expected value to be a little higher, but I don't know how to start calculating it.
I don't need a full solution, only some good hints that will help me to calculate it later.
Edit. I think I have gotten an idea to draw a matrix.
$$begin{array}[ht]{|p{2cm}|||p{0.5cm}|p{0.5cm}|p{0.5cm}|p{0.5cm}|p{0.5cm}|p{0.5cm}|p{0.5cm}|p{0.5cm}|p{0.5cm}|} hline text{ x } & 1 &2 &3 &4 &5 &6 \ hline hline hline 1 &4 &3 &4 &5 &6 &7 \ hline 2& 3 & 8 &5 &6 &7&8 \ hline 3& 4 &5 &12 &7 &8&9\ hline 4 &5 &6 &7&16&9&10 \ hline 5 &6 &7&8&9&20&11 \ hline 6&7&8&9&10&11&24 \ hline end{array}$$
And now the expected value in our case is:
$$E=frac1{36}left(l1+l2+l3+l4+l5+l6right)$$
Where $l_k$ is the sum of the numbers in $l_k$.
This gives:
$$E=frac1{36}left(29+37+45+53+61+69right)=frac{294}{36}=8.1(6)$$
Is this fine?
probability dice expected-value
probability dice expected-value
edited Feb 10 at 10:43
jvdhooft
5,65961641
edited Feb 10 at 10:43
jvdhooft
5,65961641
edited Feb 10 at 10:43
jvdhooft
5,65961641
edited Feb 10 at 10:43
jvdhooft
5,65961641
edited Feb 10 at 10:43
jvdhooft
5,65961641
5,65961641
asked Jan 30 at 17:29
user625055
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
The probability of throwing two $1$s, two $2$s, two $3$s, two $4$s, two $5$s or two $6$s all equal $frac{1}{36}$ each. In these special cases, we must add $2$, $4$, $6$, $8$, $10$ and $12$ to the sum respectively. We thus find:
$$E(X) = 7 + frac{2 + 4 + 6 + 8 + 10 + 12}{36} approx 8.17$$
answered Jan 30 at 17:36
jvdhooftjvdhooft
5,65961641
$endgroup$
$begingroup$
Something doesn't seem right in my head. If we add this to the previous sum, doesn't this mean that we actually count the case of $(1,1)$, $(2,2)$...$(6,6)$ twice?
$endgroup$
– user625055
Jan 30 at 17:40
$begingroup$
A better way to think of it is that when we added all the possibilities once and got $7$ we were short by this much because we didn't count the doubles enough.
$endgroup$
– Ross Millikan
Jan 30 at 17:46
1
$begingroup$
@Porogami Luckily, we did not. The probability of each possible combination of dice stays the same (that is, $frac{1}{36}$) and we count each element once; we only double the value of the corresponding sum (e.g., $2$ becomes $4$). We can thus calculate the expected value of the two dice separately, and add the missing numbers afterwards (e.g., $2$ for the case in which we throw two $1$s).
$endgroup$
– jvdhooft
Jan 30 at 17:47
$begingroup$
I think we got the same results, right?
$endgroup$
– user625055
Jan 30 at 17:57
$begingroup$
You've made a calculation error. The answer is $frac{294}{36} = frac{49}{6} approx 8.17$.
$endgroup$
– jvdhooft
Jan 30 at 18:13
add a comment |
$begingroup$
Linearity of expectation works. Each of the die is worth $1$ with probability $frac 16 times frac 56$, worth $2$ with probability $frac 16 times frac56+left(frac 16right)^2$ and so on Keep in mind that it might have value $8,10,12$ as well as the usual.
answered Jan 30 at 17:43
lulululu
43.3k25080
$endgroup$
$begingroup$
I googled linearity of expectation and it gave me a matrix. I have updated my question now, can you take a look? // Sorry I can't seem to understand what you did there. Why $frac16cdot frac56$ and $frac16cdot frac56 +frac1{6^2}$?
$endgroup$
– user625055
Jan 30 at 17:56
1
$begingroup$
The only way the first die can be worth $1$ is if it comes up $1$ ($frac 16$) and the second die comes up something else ($frac 56$). The first die can be worth $2$ in two ways, either it comes up $2$ and the second comes up something else (probability $frac 16times frac 56$) or both dice can come up $1$ (probability $frac 1{6^2}$).
$endgroup$
– lulu
Jan 30 at 18:06
$begingroup$
I can't really follow your method The correct answer is $8.1666...$ so your answer is close but not correct.
$endgroup$
– lulu
Jan 30 at 18:08
$begingroup$
Ah, I did the calculation in my head, and got for the fourth line $54$ instead of $53$. Now it is $8.1(6)$.
$endgroup$
– user625055
Jan 30 at 18:13
$begingroup$
Fair enough. Nevertheless, it's worth looking at Linearity methods. Suppose we had $100$ dice...your matrix is quite unwieldy but the linearity method is more or less the same.
$endgroup$
– lulu
Jan 30 at 18:20
add a comment |
$begingroup$
Given that there are two dice with 6 sidesand we are considering all possible pairs of outcomes, there will be a total of 36 possible sum outcomes ranging from 2 to 12. You can find the average of these outcomes by $$(2 + 3 + 3 +4 + 4 + 4 +...+12)/36$$ To get the proper average with the new weighted outcome you could simply multiply by 36 (to just get the sum of the outcomes) add the sum of the like pairs in (doubling their original weight), then divide by 36 (since the number of actual outcomes is still 36) again. However, we can do the same thing in less steps by using the linearity of expectation.
answered Jan 30 at 17:43
marlowmarlow
254
$endgroup$
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
$begingroup$
The probability of throwing two $1$s, two $2$s, two $3$s, two $4$s, two $5$s or two $6$s all equal $frac{1}{36}$ each. In these special cases, we must add $2$, $4$, $6$, $8$, $10$ and $12$ to the sum respectively. We thus find:
$$E(X) = 7 + frac{2 + 4 + 6 + 8 + 10 + 12}{36} approx 8.17$$
answered Jan 30 at 17:36
jvdhooftjvdhooft
5,65961641
$endgroup$
$begingroup$
Something doesn't seem right in my head. If we add this to the previous sum, doesn't this mean that we actually count the case of $(1,1)$, $(2,2)$...$(6,6)$ twice?
$endgroup$
– user625055
Jan 30 at 17:40
$begingroup$
A better way to think of it is that when we added all the possibilities once and got $7$ we were short by this much because we didn't count the doubles enough.
$endgroup$
– Ross Millikan
Jan 30 at 17:46
1
$begingroup$
@Porogami Luckily, we did not. The probability of each possible combination of dice stays the same (that is, $frac{1}{36}$) and we count each element once; we only double the value of the corresponding sum (e.g., $2$ becomes $4$). We can thus calculate the expected value of the two dice separately, and add the missing numbers afterwards (e.g., $2$ for the case in which we throw two $1$s).
$endgroup$
– jvdhooft
Jan 30 at 17:47
$begingroup$
I think we got the same results, right?
$endgroup$
– user625055
Jan 30 at 17:57
$begingroup$
You've made a calculation error. The answer is $frac{294}{36} = frac{49}{6} approx 8.17$.
$endgroup$
– jvdhooft
Jan 30 at 18:13
add a comment |
$begingroup$
The probability of throwing two $1$s, two $2$s, two $3$s, two $4$s, two $5$s or two $6$s all equal $frac{1}{36}$ each. In these special cases, we must add $2$, $4$, $6$, $8$, $10$ and $12$ to the sum respectively. We thus find:
$$E(X) = 7 + frac{2 + 4 + 6 + 8 + 10 + 12}{36} approx 8.17$$
answered Jan 30 at 17:36
jvdhooftjvdhooft
5,65961641
$endgroup$
$begingroup$
Something doesn't seem right in my head. If we add this to the previous sum, doesn't this mean that we actually count the case of $(1,1)$, $(2,2)$...$(6,6)$ twice?
$endgroup$
– user625055
Jan 30 at 17:40
$begingroup$
A better way to think of it is that when we added all the possibilities once and got $7$ we were short by this much because we didn't count the doubles enough.
$endgroup$
– Ross Millikan
Jan 30 at 17:46
1
$begingroup$
@Porogami Luckily, we did not. The probability of each possible combination of dice stays the same (that is, $frac{1}{36}$) and we count each element once; we only double the value of the corresponding sum (e.g., $2$ becomes $4$). We can thus calculate the expected value of the two dice separately, and add the missing numbers afterwards (e.g., $2$ for the case in which we throw two $1$s).
$endgroup$
– jvdhooft
Jan 30 at 17:47
$begingroup$
I think we got the same results, right?
$endgroup$
– user625055
Jan 30 at 17:57
$begingroup$
You've made a calculation error. The answer is $frac{294}{36} = frac{49}{6} approx 8.17$.
$endgroup$
– jvdhooft
Jan 30 at 18:13
add a comment |
$begingroup$
The probability of throwing two $1$s, two $2$s, two $3$s, two $4$s, two $5$s or two $6$s all equal $frac{1}{36}$ each. In these special cases, we must add $2$, $4$, $6$, $8$, $10$ and $12$ to the sum respectively. We thus find:
$$E(X) = 7 + frac{2 + 4 + 6 + 8 + 10 + 12}{36} approx 8.17$$
answered Jan 30 at 17:36
jvdhooftjvdhooft
5,65961641
$endgroup$
The probability of throwing two $1$s, two $2$s, two $3$s, two $4$s, two $5$s or two $6$s all equal $frac{1}{36}$ each. In these special cases, we must add $2$, $4$, $6$, $8$, $10$ and $12$ to the sum respectively. We thus find:
$$E(X) = 7 + frac{2 + 4 + 6 + 8 + 10 + 12}{36} approx 8.17$$
answered Jan 30 at 17:36
jvdhooftjvdhooft
5,65961641
edited Jan 30 at 17:40
answered Jan 30 at 17:36
jvdhooftjvdhooft
5,65961641
answered Jan 30 at 17:36
jvdhooftjvdhooft
5,65961641
answered Jan 30 at 17:36
jvdhooftjvdhooft
5,65961641
5,65961641
$begingroup$
Something doesn't seem right in my head. If we add this to the previous sum, doesn't this mean that we actually count the case of $(1,1)$, $(2,2)$...$(6,6)$ twice?
$endgroup$
– user625055
Jan 30 at 17:40
$begingroup$
A better way to think of it is that when we added all the possibilities once and got $7$ we were short by this much because we didn't count the doubles enough.
$endgroup$
– Ross Millikan
Jan 30 at 17:46
1
$begingroup$
@Porogami Luckily, we did not. The probability of each possible combination of dice stays the same (that is, $frac{1}{36}$) and we count each element once; we only double the value of the corresponding sum (e.g., $2$ becomes $4$). We can thus calculate the expected value of the two dice separately, and add the missing numbers afterwards (e.g., $2$ for the case in which we throw two $1$s).
$endgroup$
– jvdhooft
Jan 30 at 17:47
$begingroup$
I think we got the same results, right?
$endgroup$
– user625055
Jan 30 at 17:57
$begingroup$
You've made a calculation error. The answer is $frac{294}{36} = frac{49}{6} approx 8.17$.
$endgroup$
– jvdhooft
Jan 30 at 18:13
add a comment |
$begingroup$
Something doesn't seem right in my head. If we add this to the previous sum, doesn't this mean that we actually count the case of $(1,1)$, $(2,2)$...$(6,6)$ twice?
$endgroup$
– user625055
Jan 30 at 17:40
$begingroup$
A better way to think of it is that when we added all the possibilities once and got $7$ we were short by this much because we didn't count the doubles enough.
$endgroup$
– Ross Millikan
Jan 30 at 17:46
1
$begingroup$
@Porogami Luckily, we did not. The probability of each possible combination of dice stays the same (that is, $frac{1}{36}$) and we count each element once; we only double the value of the corresponding sum (e.g., $2$ becomes $4$). We can thus calculate the expected value of the two dice separately, and add the missing numbers afterwards (e.g., $2$ for the case in which we throw two $1$s).
$endgroup$
– jvdhooft
Jan 30 at 17:47
$begingroup$
I think we got the same results, right?
$endgroup$
– user625055
Jan 30 at 17:57
$begingroup$
You've made a calculation error. The answer is $frac{294}{36} = frac{49}{6} approx 8.17$.
$endgroup$
– jvdhooft
Jan 30 at 18:13
$begingroup$
Something doesn't seem right in my head. If we add this to the previous sum, doesn't this mean that we actually count the case of $(1,1)$, $(2,2)$...$(6,6)$ twice?
$endgroup$
– user625055
Jan 30 at 17:40
$begingroup$
Something doesn't seem right in my head. If we add this to the previous sum, doesn't this mean that we actually count the case of $(1,1)$, $(2,2)$...$(6,6)$ twice?
$endgroup$
– user625055
Jan 30 at 17:40
$begingroup$
A better way to think of it is that when we added all the possibilities once and got $7$ we were short by this much because we didn't count the doubles enough.
$endgroup$
– Ross Millikan
Jan 30 at 17:46
$begingroup$
A better way to think of it is that when we added all the possibilities once and got $7$ we were short by this much because we didn't count the doubles enough.
$endgroup$
– Ross Millikan
Jan 30 at 17:46
1
1
$begingroup$
@Porogami Luckily, we did not. The probability of each possible combination of dice stays the same (that is, $frac{1}{36}$) and we count each element once; we only double the value of the corresponding sum (e.g., $2$ becomes $4$). We can thus calculate the expected value of the two dice separately, and add the missing numbers afterwards (e.g., $2$ for the case in which we throw two $1$s).
$endgroup$
– jvdhooft
Jan 30 at 17:47
$begingroup$
@Porogami Luckily, we did not. The probability of each possible combination of dice stays the same (that is, $frac{1}{36}$) and we count each element once; we only double the value of the corresponding sum (e.g., $2$ becomes $4$). We can thus calculate the expected value of the two dice separately, and add the missing numbers afterwards (e.g., $2$ for the case in which we throw two $1$s).
$endgroup$
– jvdhooft
Jan 30 at 17:47
$begingroup$
I think we got the same results, right?
$endgroup$
– user625055
Jan 30 at 17:57
$begingroup$
I think we got the same results, right?
$endgroup$
– user625055
Jan 30 at 17:57
$begingroup$
You've made a calculation error. The answer is $frac{294}{36} = frac{49}{6} approx 8.17$.
$endgroup$
– jvdhooft
Jan 30 at 18:13
$begingroup$
You've made a calculation error. The answer is $frac{294}{36} = frac{49}{6} approx 8.17$.
$endgroup$
– jvdhooft
Jan 30 at 18:13
add a comment |
$begingroup$
Linearity of expectation works. Each of the die is worth $1$ with probability $frac 16 times frac 56$, worth $2$ with probability $frac 16 times frac56+left(frac 16right)^2$ and so on Keep in mind that it might have value $8,10,12$ as well as the usual.
answered Jan 30 at 17:43
lulululu
43.3k25080
$endgroup$
$begingroup$
I googled linearity of expectation and it gave me a matrix. I have updated my question now, can you take a look? // Sorry I can't seem to understand what you did there. Why $frac16cdot frac56$ and $frac16cdot frac56 +frac1{6^2}$?
$endgroup$
– user625055
Jan 30 at 17:56
1
$begingroup$
The only way the first die can be worth $1$ is if it comes up $1$ ($frac 16$) and the second die comes up something else ($frac 56$). The first die can be worth $2$ in two ways, either it comes up $2$ and the second comes up something else (probability $frac 16times frac 56$) or both dice can come up $1$ (probability $frac 1{6^2}$).
$endgroup$
– lulu
Jan 30 at 18:06
$begingroup$
I can't really follow your method The correct answer is $8.1666...$ so your answer is close but not correct.
$endgroup$
– lulu
Jan 30 at 18:08
$begingroup$
Ah, I did the calculation in my head, and got for the fourth line $54$ instead of $53$. Now it is $8.1(6)$.
$endgroup$
– user625055
Jan 30 at 18:13
$begingroup$
Fair enough. Nevertheless, it's worth looking at Linearity methods. Suppose we had $100$ dice...your matrix is quite unwieldy but the linearity method is more or less the same.
$endgroup$
– lulu
Jan 30 at 18:20
add a comment |
$begingroup$
Linearity of expectation works. Each of the die is worth $1$ with probability $frac 16 times frac 56$, worth $2$ with probability $frac 16 times frac56+left(frac 16right)^2$ and so on Keep in mind that it might have value $8,10,12$ as well as the usual.
answered Jan 30 at 17:43
lulululu
43.3k25080
$endgroup$
$begingroup$
I googled linearity of expectation and it gave me a matrix. I have updated my question now, can you take a look? // Sorry I can't seem to understand what you did there. Why $frac16cdot frac56$ and $frac16cdot frac56 +frac1{6^2}$?
$endgroup$
– user625055
Jan 30 at 17:56
1
$begingroup$
The only way the first die can be worth $1$ is if it comes up $1$ ($frac 16$) and the second die comes up something else ($frac 56$). The first die can be worth $2$ in two ways, either it comes up $2$ and the second comes up something else (probability $frac 16times frac 56$) or both dice can come up $1$ (probability $frac 1{6^2}$).
$endgroup$
– lulu
Jan 30 at 18:06
$begingroup$
I can't really follow your method The correct answer is $8.1666...$ so your answer is close but not correct.
$endgroup$
– lulu
Jan 30 at 18:08
$begingroup$
Ah, I did the calculation in my head, and got for the fourth line $54$ instead of $53$. Now it is $8.1(6)$.
$endgroup$
– user625055
Jan 30 at 18:13
$begingroup$
Fair enough. Nevertheless, it's worth looking at Linearity methods. Suppose we had $100$ dice...your matrix is quite unwieldy but the linearity method is more or less the same.
$endgroup$
– lulu
Jan 30 at 18:20
add a comment |
$begingroup$
Linearity of expectation works. Each of the die is worth $1$ with probability $frac 16 times frac 56$, worth $2$ with probability $frac 16 times frac56+left(frac 16right)^2$ and so on Keep in mind that it might have value $8,10,12$ as well as the usual.
answered Jan 30 at 17:43
lulululu
43.3k25080
$endgroup$
Linearity of expectation works. Each of the die is worth $1$ with probability $frac 16 times frac 56$, worth $2$ with probability $frac 16 times frac56+left(frac 16right)^2$ and so on Keep in mind that it might have value $8,10,12$ as well as the usual.
answered Jan 30 at 17:43
lulululu
43.3k25080
answered Jan 30 at 17:43
lulululu
43.3k25080
answered Jan 30 at 17:43
lulululu
43.3k25080
answered Jan 30 at 17:43
lulululu
43.3k25080
43.3k25080
$begingroup$
I googled linearity of expectation and it gave me a matrix. I have updated my question now, can you take a look? // Sorry I can't seem to understand what you did there. Why $frac16cdot frac56$ and $frac16cdot frac56 +frac1{6^2}$?
$endgroup$
– user625055
Jan 30 at 17:56
1
$begingroup$
The only way the first die can be worth $1$ is if it comes up $1$ ($frac 16$) and the second die comes up something else ($frac 56$). The first die can be worth $2$ in two ways, either it comes up $2$ and the second comes up something else (probability $frac 16times frac 56$) or both dice can come up $1$ (probability $frac 1{6^2}$).
$endgroup$
– lulu
Jan 30 at 18:06
$begingroup$
I can't really follow your method The correct answer is $8.1666...$ so your answer is close but not correct.
$endgroup$
– lulu
Jan 30 at 18:08
$begingroup$
Ah, I did the calculation in my head, and got for the fourth line $54$ instead of $53$. Now it is $8.1(6)$.
$endgroup$
– user625055
Jan 30 at 18:13
$begingroup$
Fair enough. Nevertheless, it's worth looking at Linearity methods. Suppose we had $100$ dice...your matrix is quite unwieldy but the linearity method is more or less the same.
$endgroup$
– lulu
Jan 30 at 18:20
add a comment |
$begingroup$
I googled linearity of expectation and it gave me a matrix. I have updated my question now, can you take a look? // Sorry I can't seem to understand what you did there. Why $frac16cdot frac56$ and $frac16cdot frac56 +frac1{6^2}$?
$endgroup$
– user625055
Jan 30 at 17:56
1
$begingroup$
The only way the first die can be worth $1$ is if it comes up $1$ ($frac 16$) and the second die comes up something else ($frac 56$). The first die can be worth $2$ in two ways, either it comes up $2$ and the second comes up something else (probability $frac 16times frac 56$) or both dice can come up $1$ (probability $frac 1{6^2}$).
$endgroup$
– lulu
Jan 30 at 18:06
$begingroup$
I can't really follow your method The correct answer is $8.1666...$ so your answer is close but not correct.
$endgroup$
– lulu
Jan 30 at 18:08
$begingroup$
Ah, I did the calculation in my head, and got for the fourth line $54$ instead of $53$. Now it is $8.1(6)$.
$endgroup$
– user625055
Jan 30 at 18:13
$begingroup$
Fair enough. Nevertheless, it's worth looking at Linearity methods. Suppose we had $100$ dice...your matrix is quite unwieldy but the linearity method is more or less the same.
$endgroup$
– lulu
Jan 30 at 18:20
$begingroup$
I googled linearity of expectation and it gave me a matrix. I have updated my question now, can you take a look? // Sorry I can't seem to understand what you did there. Why $frac16cdot frac56$ and $frac16cdot frac56 +frac1{6^2}$?
$endgroup$
– user625055
Jan 30 at 17:56
$begingroup$
I googled linearity of expectation and it gave me a matrix. I have updated my question now, can you take a look? // Sorry I can't seem to understand what you did there. Why $frac16cdot frac56$ and $frac16cdot frac56 +frac1{6^2}$?
$endgroup$
– user625055
Jan 30 at 17:56
1
1
$begingroup$
The only way the first die can be worth $1$ is if it comes up $1$ ($frac 16$) and the second die comes up something else ($frac 56$). The first die can be worth $2$ in two ways, either it comes up $2$ and the second comes up something else (probability $frac 16times frac 56$) or both dice can come up $1$ (probability $frac 1{6^2}$).
$endgroup$
– lulu
Jan 30 at 18:06
$begingroup$
The only way the first die can be worth $1$ is if it comes up $1$ ($frac 16$) and the second die comes up something else ($frac 56$). The first die can be worth $2$ in two ways, either it comes up $2$ and the second comes up something else (probability $frac 16times frac 56$) or both dice can come up $1$ (probability $frac 1{6^2}$).
$endgroup$
– lulu
Jan 30 at 18:06
$begingroup$
I can't really follow your method The correct answer is $8.1666...$ so your answer is close but not correct.
$endgroup$
– lulu
Jan 30 at 18:08
$begingroup$
I can't really follow your method The correct answer is $8.1666...$ so your answer is close but not correct.
$endgroup$
– lulu
Jan 30 at 18:08
$begingroup$
Ah, I did the calculation in my head, and got for the fourth line $54$ instead of $53$. Now it is $8.1(6)$.
$endgroup$
– user625055
Jan 30 at 18:13
$begingroup$
Ah, I did the calculation in my head, and got for the fourth line $54$ instead of $53$. Now it is $8.1(6)$.
$endgroup$
– user625055
Jan 30 at 18:13
$begingroup$
Fair enough. Nevertheless, it's worth looking at Linearity methods. Suppose we had $100$ dice...your matrix is quite unwieldy but the linearity method is more or less the same.
$endgroup$
– lulu
Jan 30 at 18:20
$begingroup$
Fair enough. Nevertheless, it's worth looking at Linearity methods. Suppose we had $100$ dice...your matrix is quite unwieldy but the linearity method is more or less the same.
$endgroup$
– lulu
Jan 30 at 18:20
add a comment |
$begingroup$
Given that there are two dice with 6 sidesand we are considering all possible pairs of outcomes, there will be a total of 36 possible sum outcomes ranging from 2 to 12. You can find the average of these outcomes by $$(2 + 3 + 3 +4 + 4 + 4 +...+12)/36$$ To get the proper average with the new weighted outcome you could simply multiply by 36 (to just get the sum of the outcomes) add the sum of the like pairs in (doubling their original weight), then divide by 36 (since the number of actual outcomes is still 36) again. However, we can do the same thing in less steps by using the linearity of expectation.
answered Jan 30 at 17:43
marlowmarlow
254
$endgroup$
add a comment |
$begingroup$
Given that there are two dice with 6 sidesand we are considering all possible pairs of outcomes, there will be a total of 36 possible sum outcomes ranging from 2 to 12. You can find the average of these outcomes by $$(2 + 3 + 3 +4 + 4 + 4 +...+12)/36$$ To get the proper average with the new weighted outcome you could simply multiply by 36 (to just get the sum of the outcomes) add the sum of the like pairs in (doubling their original weight), then divide by 36 (since the number of actual outcomes is still 36) again. However, we can do the same thing in less steps by using the linearity of expectation.
answered Jan 30 at 17:43
marlowmarlow
254
$endgroup$
add a comment |
$begingroup$
Given that there are two dice with 6 sidesand we are considering all possible pairs of outcomes, there will be a total of 36 possible sum outcomes ranging from 2 to 12. You can find the average of these outcomes by $$(2 + 3 + 3 +4 + 4 + 4 +...+12)/36$$ To get the proper average with the new weighted outcome you could simply multiply by 36 (to just get the sum of the outcomes) add the sum of the like pairs in (doubling their original weight), then divide by 36 (since the number of actual outcomes is still 36) again. However, we can do the same thing in less steps by using the linearity of expectation.
answered Jan 30 at 17:43
marlowmarlow
254
$endgroup$
Given that there are two dice with 6 sidesand we are considering all possible pairs of outcomes, there will be a total of 36 possible sum outcomes ranging from 2 to 12. You can find the average of these outcomes by $$(2 + 3 + 3 +4 + 4 + 4 +...+12)/36$$ To get the proper average with the new weighted outcome you could simply multiply by 36 (to just get the sum of the outcomes) add the sum of the like pairs in (doubling their original weight), then divide by 36 (since the number of actual outcomes is still 36) again. However, we can do the same thing in less steps by using the linearity of expectation.
answered Jan 30 at 17:43
marlowmarlow
254
answered Jan 30 at 17:43
marlowmarlow
254
answered Jan 30 at 17:43
marlowmarlow
254
answered Jan 30 at 17:43
marlowmarlow
254
254
add a comment |
add a comment |
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