Maximum probability of sum occurring in a dice(s) roll in such a way
$begingroup$
One fair die is rolled: let 'x' denote the number that comes up. We then roll 'x' dice(s), and let the sum of the resulting numbers be 'y'. Finally, roll 'y' dice(s) and let 'z' be the sum of the resulting 'y' numbers. Let the expected value of 'z' be 'a', we have to find 'a'.
===================================================================
My approach:-
for 'x' all numbers have same probability of occurring after that if we consider cases for 'y':
1 dice roll
2 dice(s) roll..
.
.
6 dice(s) roll
and sum up the possibilities we get something like this
sum expected probability(out of $6^7$)
1 7776
2 9072
3 10584
4 12348
5 14406
6 16807
7 11832
8 12507
9 13076
10 13482
11 13650
12 13482
13 12852
14 12897
15 12772
16 12453
17 11928
18 11207
19 10332
20 9387
21 8292
22 7101
23 5880
24 4697
25 3612
26 2667
27 1876
28 1251
29 786
30 462
31 252
32 126
33 56
34 21
35 6
36 1
6 is having maximum probability i.e. 16807
and then 'y'=6.
and we know that if we roll 6 dice(s) then the probability of sum that is maximum is 21
so 'z' should be 21 but this is not the answer can someone suggest any different approach
probability permutations binomial-coefficients
asked Dec 26 '18 at 14:28
dank uploaderdank uploader
16
$endgroup$
add a comment |
$begingroup$
One fair die is rolled: let 'x' denote the number that comes up. We then roll 'x' dice(s), and let the sum of the resulting numbers be 'y'. Finally, roll 'y' dice(s) and let 'z' be the sum of the resulting 'y' numbers. Let the expected value of 'z' be 'a', we have to find 'a'.
===================================================================
My approach:-
for 'x' all numbers have same probability of occurring after that if we consider cases for 'y':
1 dice roll
2 dice(s) roll..
.
.
6 dice(s) roll
and sum up the possibilities we get something like this
sum expected probability(out of $6^7$)
1 7776
2 9072
3 10584
4 12348
5 14406
6 16807
7 11832
8 12507
9 13076
10 13482
11 13650
12 13482
13 12852
14 12897
15 12772
16 12453
17 11928
18 11207
19 10332
20 9387
21 8292
22 7101
23 5880
24 4697
25 3612
26 2667
27 1876
28 1251
29 786
30 462
31 252
32 126
33 56
34 21
35 6
36 1
6 is having maximum probability i.e. 16807
and then 'y'=6.
and we know that if we roll 6 dice(s) then the probability of sum that is maximum is 21
so 'z' should be 21 but this is not the answer can someone suggest any different approach
probability permutations binomial-coefficients
asked Dec 26 '18 at 14:28
dank uploaderdank uploader
16
$endgroup$
add a comment |
$begingroup$
One fair die is rolled: let 'x' denote the number that comes up. We then roll 'x' dice(s), and let the sum of the resulting numbers be 'y'. Finally, roll 'y' dice(s) and let 'z' be the sum of the resulting 'y' numbers. Let the expected value of 'z' be 'a', we have to find 'a'.
===================================================================
My approach:-
for 'x' all numbers have same probability of occurring after that if we consider cases for 'y':
1 dice roll
2 dice(s) roll..
.
.
6 dice(s) roll
and sum up the possibilities we get something like this
sum expected probability(out of $6^7$)
1 7776
2 9072
3 10584
4 12348
5 14406
6 16807
7 11832
8 12507
9 13076
10 13482
11 13650
12 13482
13 12852
14 12897
15 12772
16 12453
17 11928
18 11207
19 10332
20 9387
21 8292
22 7101
23 5880
24 4697
25 3612
26 2667
27 1876
28 1251
29 786
30 462
31 252
32 126
33 56
34 21
35 6
36 1
6 is having maximum probability i.e. 16807
and then 'y'=6.
and we know that if we roll 6 dice(s) then the probability of sum that is maximum is 21
so 'z' should be 21 but this is not the answer can someone suggest any different approach
probability permutations binomial-coefficients
asked Dec 26 '18 at 14:28
dank uploaderdank uploader
16
$endgroup$
One fair die is rolled: let 'x' denote the number that comes up. We then roll 'x' dice(s), and let the sum of the resulting numbers be 'y'. Finally, roll 'y' dice(s) and let 'z' be the sum of the resulting 'y' numbers. Let the expected value of 'z' be 'a', we have to find 'a'.
===================================================================
My approach:-
for 'x' all numbers have same probability of occurring after that if we consider cases for 'y':
1 dice roll
2 dice(s) roll..
.
.
6 dice(s) roll
and sum up the possibilities we get something like this
sum expected probability(out of $6^7$)
1 7776
2 9072
3 10584
4 12348
5 14406
6 16807
7 11832
8 12507
9 13076
10 13482
11 13650
12 13482
13 12852
14 12897
15 12772
16 12453
17 11928
18 11207
19 10332
20 9387
21 8292
22 7101
23 5880
24 4697
25 3612
26 2667
27 1876
28 1251
29 786
30 462
31 252
32 126
33 56
34 21
35 6
36 1
6 is having maximum probability i.e. 16807
and then 'y'=6.
and we know that if we roll 6 dice(s) then the probability of sum that is maximum is 21
so 'z' should be 21 but this is not the answer can someone suggest any different approach
probability permutations binomial-coefficients
probability permutations binomial-coefficients
asked Dec 26 '18 at 14:28
dank uploaderdank uploader
16
asked Dec 26 '18 at 14:28
dank uploaderdank uploader
16
asked Dec 26 '18 at 14:28
dank uploaderdank uploader
16
asked Dec 26 '18 at 14:28
dank uploaderdank uploader
16
asked Dec 26 '18 at 14:28
dank uploaderdank uploader
16
16
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Call $x^*$ the outcome of the first dice. Then we throw $x^*$ independent dies whose outcome is $x_i$ and we call $y^*=sum_{i=1}^{x^*}x_i$. Finally, we throw $y^*$ independent dices whose outcome is $y_i$ and we call $z=sum_{i=1}^{y^*}y_i$. We are looking for $mathbb{E}(z)$. So, we can do this using the Law of Iterated Expectations (LIE), as follows:
$$mathbb{E}(z)=mathbb{E}bigg(sum_{i=1}^{y^*}y_ibigg)=mathbb{E}(mathbb{E}(y_1+dots+y_{y^*}mid y^*))=mathbb{E}(y^*mathbb{E}(y_i))=frac{21}{6}mathbb{E}(y^*)$$
Where the second to last equality follows from the fact that each throw is independent and identically distributed, so $y^*$ determines the number of throws but not the outcome of each of them. To finish the problem then we can proceed again using the LIE as follows:
$$mathbb{E}(z)=frac{21}{6}mathbb{E}(y^*)=frac{21}{6}mathbb{E}bigg(sum_{i=1}^{x^*}x_ibigg)=frac{21}{6}mathbb{E}(mathbb{E}(x_1+dots+x_{x^*}mid x^*))=frac{21}{6}mathbb{E}(x^*mathbb{E}(x_i))=bigg(frac{21}{6}bigg)^3$$
answered Dec 26 '18 at 15:25
Zeky MurraZeky Murra
1387
$endgroup$
$begingroup$
I need answer for the expected value of 'a' i.e. the sum which has the maximum probability of occurring and it should be an integer i guess
$endgroup$
– dank uploader
Dec 26 '18 at 15:49
$begingroup$
I´m sorry, but that´s not what the description of the problem say. I think you should rephrase it then.
$endgroup$
– Zeky Murra
Dec 26 '18 at 16:16
$begingroup$
The expected value 'z' is sum of resulting 'y' so 'z' has to be an integer. Can you explain what $(21/6)^3$ is?
$endgroup$
– dank uploader
Dec 26 '18 at 16:18
$begingroup$
21/6 is the expected value of a dice throw, which is $1/6(1+2+3+4+5+6)$, it doesn´t have to be integer. If you flip a fair coin where one face is 1 and the other is 0, the expectation is 0.5, which is not integer either. In this case, I defined the sums you´re looking for, and it turns out that the expected sum is $(21/6)^3=42.875$
$endgroup$
– Zeky Murra
Dec 26 '18 at 16:36
add a comment |
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1 Answer
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1 Answer
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oldest
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$begingroup$
Call $x^*$ the outcome of the first dice. Then we throw $x^*$ independent dies whose outcome is $x_i$ and we call $y^*=sum_{i=1}^{x^*}x_i$. Finally, we throw $y^*$ independent dices whose outcome is $y_i$ and we call $z=sum_{i=1}^{y^*}y_i$. We are looking for $mathbb{E}(z)$. So, we can do this using the Law of Iterated Expectations (LIE), as follows:
$$mathbb{E}(z)=mathbb{E}bigg(sum_{i=1}^{y^*}y_ibigg)=mathbb{E}(mathbb{E}(y_1+dots+y_{y^*}mid y^*))=mathbb{E}(y^*mathbb{E}(y_i))=frac{21}{6}mathbb{E}(y^*)$$
Where the second to last equality follows from the fact that each throw is independent and identically distributed, so $y^*$ determines the number of throws but not the outcome of each of them. To finish the problem then we can proceed again using the LIE as follows:
$$mathbb{E}(z)=frac{21}{6}mathbb{E}(y^*)=frac{21}{6}mathbb{E}bigg(sum_{i=1}^{x^*}x_ibigg)=frac{21}{6}mathbb{E}(mathbb{E}(x_1+dots+x_{x^*}mid x^*))=frac{21}{6}mathbb{E}(x^*mathbb{E}(x_i))=bigg(frac{21}{6}bigg)^3$$
answered Dec 26 '18 at 15:25
Zeky MurraZeky Murra
1387
$endgroup$
$begingroup$
I need answer for the expected value of 'a' i.e. the sum which has the maximum probability of occurring and it should be an integer i guess
$endgroup$
– dank uploader
Dec 26 '18 at 15:49
$begingroup$
I´m sorry, but that´s not what the description of the problem say. I think you should rephrase it then.
$endgroup$
– Zeky Murra
Dec 26 '18 at 16:16
$begingroup$
The expected value 'z' is sum of resulting 'y' so 'z' has to be an integer. Can you explain what $(21/6)^3$ is?
$endgroup$
– dank uploader
Dec 26 '18 at 16:18
$begingroup$
21/6 is the expected value of a dice throw, which is $1/6(1+2+3+4+5+6)$, it doesn´t have to be integer. If you flip a fair coin where one face is 1 and the other is 0, the expectation is 0.5, which is not integer either. In this case, I defined the sums you´re looking for, and it turns out that the expected sum is $(21/6)^3=42.875$
$endgroup$
– Zeky Murra
Dec 26 '18 at 16:36
add a comment |
$begingroup$
Call $x^*$ the outcome of the first dice. Then we throw $x^*$ independent dies whose outcome is $x_i$ and we call $y^*=sum_{i=1}^{x^*}x_i$. Finally, we throw $y^*$ independent dices whose outcome is $y_i$ and we call $z=sum_{i=1}^{y^*}y_i$. We are looking for $mathbb{E}(z)$. So, we can do this using the Law of Iterated Expectations (LIE), as follows:
$$mathbb{E}(z)=mathbb{E}bigg(sum_{i=1}^{y^*}y_ibigg)=mathbb{E}(mathbb{E}(y_1+dots+y_{y^*}mid y^*))=mathbb{E}(y^*mathbb{E}(y_i))=frac{21}{6}mathbb{E}(y^*)$$
Where the second to last equality follows from the fact that each throw is independent and identically distributed, so $y^*$ determines the number of throws but not the outcome of each of them. To finish the problem then we can proceed again using the LIE as follows:
$$mathbb{E}(z)=frac{21}{6}mathbb{E}(y^*)=frac{21}{6}mathbb{E}bigg(sum_{i=1}^{x^*}x_ibigg)=frac{21}{6}mathbb{E}(mathbb{E}(x_1+dots+x_{x^*}mid x^*))=frac{21}{6}mathbb{E}(x^*mathbb{E}(x_i))=bigg(frac{21}{6}bigg)^3$$
answered Dec 26 '18 at 15:25
Zeky MurraZeky Murra
1387
$endgroup$
$begingroup$
I need answer for the expected value of 'a' i.e. the sum which has the maximum probability of occurring and it should be an integer i guess
$endgroup$
– dank uploader
Dec 26 '18 at 15:49
$begingroup$
I´m sorry, but that´s not what the description of the problem say. I think you should rephrase it then.
$endgroup$
– Zeky Murra
Dec 26 '18 at 16:16
$begingroup$
The expected value 'z' is sum of resulting 'y' so 'z' has to be an integer. Can you explain what $(21/6)^3$ is?
$endgroup$
– dank uploader
Dec 26 '18 at 16:18
$begingroup$
21/6 is the expected value of a dice throw, which is $1/6(1+2+3+4+5+6)$, it doesn´t have to be integer. If you flip a fair coin where one face is 1 and the other is 0, the expectation is 0.5, which is not integer either. In this case, I defined the sums you´re looking for, and it turns out that the expected sum is $(21/6)^3=42.875$
$endgroup$
– Zeky Murra
Dec 26 '18 at 16:36
add a comment |
$begingroup$
Call $x^*$ the outcome of the first dice. Then we throw $x^*$ independent dies whose outcome is $x_i$ and we call $y^*=sum_{i=1}^{x^*}x_i$. Finally, we throw $y^*$ independent dices whose outcome is $y_i$ and we call $z=sum_{i=1}^{y^*}y_i$. We are looking for $mathbb{E}(z)$. So, we can do this using the Law of Iterated Expectations (LIE), as follows:
$$mathbb{E}(z)=mathbb{E}bigg(sum_{i=1}^{y^*}y_ibigg)=mathbb{E}(mathbb{E}(y_1+dots+y_{y^*}mid y^*))=mathbb{E}(y^*mathbb{E}(y_i))=frac{21}{6}mathbb{E}(y^*)$$
Where the second to last equality follows from the fact that each throw is independent and identically distributed, so $y^*$ determines the number of throws but not the outcome of each of them. To finish the problem then we can proceed again using the LIE as follows:
$$mathbb{E}(z)=frac{21}{6}mathbb{E}(y^*)=frac{21}{6}mathbb{E}bigg(sum_{i=1}^{x^*}x_ibigg)=frac{21}{6}mathbb{E}(mathbb{E}(x_1+dots+x_{x^*}mid x^*))=frac{21}{6}mathbb{E}(x^*mathbb{E}(x_i))=bigg(frac{21}{6}bigg)^3$$
answered Dec 26 '18 at 15:25
Zeky MurraZeky Murra
1387
$endgroup$
Call $x^*$ the outcome of the first dice. Then we throw $x^*$ independent dies whose outcome is $x_i$ and we call $y^*=sum_{i=1}^{x^*}x_i$. Finally, we throw $y^*$ independent dices whose outcome is $y_i$ and we call $z=sum_{i=1}^{y^*}y_i$. We are looking for $mathbb{E}(z)$. So, we can do this using the Law of Iterated Expectations (LIE), as follows:
$$mathbb{E}(z)=mathbb{E}bigg(sum_{i=1}^{y^*}y_ibigg)=mathbb{E}(mathbb{E}(y_1+dots+y_{y^*}mid y^*))=mathbb{E}(y^*mathbb{E}(y_i))=frac{21}{6}mathbb{E}(y^*)$$
Where the second to last equality follows from the fact that each throw is independent and identically distributed, so $y^*$ determines the number of throws but not the outcome of each of them. To finish the problem then we can proceed again using the LIE as follows:
$$mathbb{E}(z)=frac{21}{6}mathbb{E}(y^*)=frac{21}{6}mathbb{E}bigg(sum_{i=1}^{x^*}x_ibigg)=frac{21}{6}mathbb{E}(mathbb{E}(x_1+dots+x_{x^*}mid x^*))=frac{21}{6}mathbb{E}(x^*mathbb{E}(x_i))=bigg(frac{21}{6}bigg)^3$$
answered Dec 26 '18 at 15:25
Zeky MurraZeky Murra
1387
answered Dec 26 '18 at 15:25
Zeky MurraZeky Murra
1387
answered Dec 26 '18 at 15:25
Zeky MurraZeky Murra
1387
answered Dec 26 '18 at 15:25
Zeky MurraZeky Murra
1387
1387
$begingroup$
I need answer for the expected value of 'a' i.e. the sum which has the maximum probability of occurring and it should be an integer i guess
$endgroup$
– dank uploader
Dec 26 '18 at 15:49
$begingroup$
I´m sorry, but that´s not what the description of the problem say. I think you should rephrase it then.
$endgroup$
– Zeky Murra
Dec 26 '18 at 16:16
$begingroup$
The expected value 'z' is sum of resulting 'y' so 'z' has to be an integer. Can you explain what $(21/6)^3$ is?
$endgroup$
– dank uploader
Dec 26 '18 at 16:18
$begingroup$
21/6 is the expected value of a dice throw, which is $1/6(1+2+3+4+5+6)$, it doesn´t have to be integer. If you flip a fair coin where one face is 1 and the other is 0, the expectation is 0.5, which is not integer either. In this case, I defined the sums you´re looking for, and it turns out that the expected sum is $(21/6)^3=42.875$
$endgroup$
– Zeky Murra
Dec 26 '18 at 16:36
add a comment |
$begingroup$
I need answer for the expected value of 'a' i.e. the sum which has the maximum probability of occurring and it should be an integer i guess
$endgroup$
– dank uploader
Dec 26 '18 at 15:49
$begingroup$
I´m sorry, but that´s not what the description of the problem say. I think you should rephrase it then.
$endgroup$
– Zeky Murra
Dec 26 '18 at 16:16
$begingroup$
The expected value 'z' is sum of resulting 'y' so 'z' has to be an integer. Can you explain what $(21/6)^3$ is?
$endgroup$
– dank uploader
Dec 26 '18 at 16:18
$begingroup$
21/6 is the expected value of a dice throw, which is $1/6(1+2+3+4+5+6)$, it doesn´t have to be integer. If you flip a fair coin where one face is 1 and the other is 0, the expectation is 0.5, which is not integer either. In this case, I defined the sums you´re looking for, and it turns out that the expected sum is $(21/6)^3=42.875$
$endgroup$
– Zeky Murra
Dec 26 '18 at 16:36
$begingroup$
I need answer for the expected value of 'a' i.e. the sum which has the maximum probability of occurring and it should be an integer i guess
$endgroup$
– dank uploader
Dec 26 '18 at 15:49
$begingroup$
I need answer for the expected value of 'a' i.e. the sum which has the maximum probability of occurring and it should be an integer i guess
$endgroup$
– dank uploader
Dec 26 '18 at 15:49
$begingroup$
I´m sorry, but that´s not what the description of the problem say. I think you should rephrase it then.
$endgroup$
– Zeky Murra
Dec 26 '18 at 16:16
$begingroup$
I´m sorry, but that´s not what the description of the problem say. I think you should rephrase it then.
$endgroup$
– Zeky Murra
Dec 26 '18 at 16:16
$begingroup$
The expected value 'z' is sum of resulting 'y' so 'z' has to be an integer. Can you explain what $(21/6)^3$ is?
$endgroup$
– dank uploader
Dec 26 '18 at 16:18
$begingroup$
The expected value 'z' is sum of resulting 'y' so 'z' has to be an integer. Can you explain what $(21/6)^3$ is?
$endgroup$
– dank uploader
Dec 26 '18 at 16:18
$begingroup$
21/6 is the expected value of a dice throw, which is $1/6(1+2+3+4+5+6)$, it doesn´t have to be integer. If you flip a fair coin where one face is 1 and the other is 0, the expectation is 0.5, which is not integer either. In this case, I defined the sums you´re looking for, and it turns out that the expected sum is $(21/6)^3=42.875$
$endgroup$
– Zeky Murra
Dec 26 '18 at 16:36
$begingroup$
21/6 is the expected value of a dice throw, which is $1/6(1+2+3+4+5+6)$, it doesn´t have to be integer. If you flip a fair coin where one face is 1 and the other is 0, the expectation is 0.5, which is not integer either. In this case, I defined the sums you´re looking for, and it turns out that the expected sum is $(21/6)^3=42.875$
$endgroup$
– Zeky Murra
Dec 26 '18 at 16:36
add a comment |
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