Identical Dice and “Identical” Sum (or Boxes??)
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Say you have two 6 sided dice numbered from 1 to 6. How would you calculate the probability of rolling 2, 3, 4, ..., 12?
My professor told me the generating function $(x+x^2+x^3+x^4+x^5+x^6)^2$ is incorrect for starting the problem because the dice are treated as distinct. That I should be looking for the answer using identical dice and "sum". How would this be solved?
combinatorics dice
asked Dec 21 '18 at 22:51
Math NewbieMath Newbie
428
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add a comment |
$begingroup$
Say you have two 6 sided dice numbered from 1 to 6. How would you calculate the probability of rolling 2, 3, 4, ..., 12?
My professor told me the generating function $(x+x^2+x^3+x^4+x^5+x^6)^2$ is incorrect for starting the problem because the dice are treated as distinct. That I should be looking for the answer using identical dice and "sum". How would this be solved?
combinatorics dice
asked Dec 21 '18 at 22:51
Math NewbieMath Newbie
428
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If the dice are treated as distinct, why does your title describe them as identical?
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– N. F. Taussig
Dec 21 '18 at 23:23
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@N.F.Taussig My professor asked for identical dice while I provided distinct dice.
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– Math Newbie
Dec 22 '18 at 1:58
2
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The coefficients of that generating function (when divided by $6^2$ for normalization), are the probabilities in question.
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– Math1000
Dec 22 '18 at 2:41
add a comment |
$begingroup$
Say you have two 6 sided dice numbered from 1 to 6. How would you calculate the probability of rolling 2, 3, 4, ..., 12?
My professor told me the generating function $(x+x^2+x^3+x^4+x^5+x^6)^2$ is incorrect for starting the problem because the dice are treated as distinct. That I should be looking for the answer using identical dice and "sum". How would this be solved?
combinatorics dice
asked Dec 21 '18 at 22:51
Math NewbieMath Newbie
428
$endgroup$
Say you have two 6 sided dice numbered from 1 to 6. How would you calculate the probability of rolling 2, 3, 4, ..., 12?
My professor told me the generating function $(x+x^2+x^3+x^4+x^5+x^6)^2$ is incorrect for starting the problem because the dice are treated as distinct. That I should be looking for the answer using identical dice and "sum". How would this be solved?
combinatorics dice
combinatorics dice
asked Dec 21 '18 at 22:51
Math NewbieMath Newbie
428
asked Dec 21 '18 at 22:51
Math NewbieMath Newbie
428
asked Dec 21 '18 at 22:51
Math NewbieMath Newbie
428
asked Dec 21 '18 at 22:51
Math NewbieMath Newbie
428
asked Dec 21 '18 at 22:51
Math NewbieMath Newbie
428
428
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If the dice are treated as distinct, why does your title describe them as identical?
$endgroup$
– N. F. Taussig
Dec 21 '18 at 23:23
$begingroup$
@N.F.Taussig My professor asked for identical dice while I provided distinct dice.
$endgroup$
– Math Newbie
Dec 22 '18 at 1:58
2
$begingroup$
The coefficients of that generating function (when divided by $6^2$ for normalization), are the probabilities in question.
$endgroup$
– Math1000
Dec 22 '18 at 2:41
add a comment |
$begingroup$
If the dice are treated as distinct, why does your title describe them as identical?
$endgroup$
– N. F. Taussig
Dec 21 '18 at 23:23
$begingroup$
@N.F.Taussig My professor asked for identical dice while I provided distinct dice.
$endgroup$
– Math Newbie
Dec 22 '18 at 1:58
2
$begingroup$
The coefficients of that generating function (when divided by $6^2$ for normalization), are the probabilities in question.
$endgroup$
– Math1000
Dec 22 '18 at 2:41
$begingroup$
If the dice are treated as distinct, why does your title describe them as identical?
$endgroup$
– N. F. Taussig
Dec 21 '18 at 23:23
$begingroup$
If the dice are treated as distinct, why does your title describe them as identical?
$endgroup$
– N. F. Taussig
Dec 21 '18 at 23:23
$begingroup$
@N.F.Taussig My professor asked for identical dice while I provided distinct dice.
$endgroup$
– Math Newbie
Dec 22 '18 at 1:58
$begingroup$
@N.F.Taussig My professor asked for identical dice while I provided distinct dice.
$endgroup$
– Math Newbie
Dec 22 '18 at 1:58
2
2
$begingroup$
The coefficients of that generating function (when divided by $6^2$ for normalization), are the probabilities in question.
$endgroup$
– Math1000
Dec 22 '18 at 2:41
$begingroup$
The coefficients of that generating function (when divided by $6^2$ for normalization), are the probabilities in question.
$endgroup$
– Math1000
Dec 22 '18 at 2:41
add a comment |
1 Answer
1
active
oldest
votes
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Well if your dice are identical then you can simply say that for finding the probability of rolling a certain sum $s$ we need to find the number of ways $x_1 + x_2$ can be equal to $s$ or rather we need to find the number of integral solutions to the equation $$x_1 + x_2 = s$$ where $$x_1,x_2 equiv 1,2,3,4,5,6$$
We can solve this rather easily and then we can use the argument for probability.
answered Dec 22 '18 at 16:42
Prakhar NagpalPrakhar Nagpal
752318
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add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Well if your dice are identical then you can simply say that for finding the probability of rolling a certain sum $s$ we need to find the number of ways $x_1 + x_2$ can be equal to $s$ or rather we need to find the number of integral solutions to the equation $$x_1 + x_2 = s$$ where $$x_1,x_2 equiv 1,2,3,4,5,6$$
We can solve this rather easily and then we can use the argument for probability.
answered Dec 22 '18 at 16:42
Prakhar NagpalPrakhar Nagpal
752318
$endgroup$
add a comment |
$begingroup$
Well if your dice are identical then you can simply say that for finding the probability of rolling a certain sum $s$ we need to find the number of ways $x_1 + x_2$ can be equal to $s$ or rather we need to find the number of integral solutions to the equation $$x_1 + x_2 = s$$ where $$x_1,x_2 equiv 1,2,3,4,5,6$$
We can solve this rather easily and then we can use the argument for probability.
answered Dec 22 '18 at 16:42
Prakhar NagpalPrakhar Nagpal
752318
$endgroup$
add a comment |
$begingroup$
Well if your dice are identical then you can simply say that for finding the probability of rolling a certain sum $s$ we need to find the number of ways $x_1 + x_2$ can be equal to $s$ or rather we need to find the number of integral solutions to the equation $$x_1 + x_2 = s$$ where $$x_1,x_2 equiv 1,2,3,4,5,6$$
We can solve this rather easily and then we can use the argument for probability.
answered Dec 22 '18 at 16:42
Prakhar NagpalPrakhar Nagpal
752318
$endgroup$
Well if your dice are identical then you can simply say that for finding the probability of rolling a certain sum $s$ we need to find the number of ways $x_1 + x_2$ can be equal to $s$ or rather we need to find the number of integral solutions to the equation $$x_1 + x_2 = s$$ where $$x_1,x_2 equiv 1,2,3,4,5,6$$
We can solve this rather easily and then we can use the argument for probability.
answered Dec 22 '18 at 16:42
Prakhar NagpalPrakhar Nagpal
752318
answered Dec 22 '18 at 16:42
Prakhar NagpalPrakhar Nagpal
752318
answered Dec 22 '18 at 16:42
Prakhar NagpalPrakhar Nagpal
752318
answered Dec 22 '18 at 16:42
Prakhar NagpalPrakhar Nagpal
752318
752318
add a comment |
add a comment |
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$begingroup$
If the dice are treated as distinct, why does your title describe them as identical?
$endgroup$
– N. F. Taussig
Dec 21 '18 at 23:23
$begingroup$
@N.F.Taussig My professor asked for identical dice while I provided distinct dice.
$endgroup$
– Math Newbie
Dec 22 '18 at 1:58
2
$begingroup$
The coefficients of that generating function (when divided by $6^2$ for normalization), are the probabilities in question.
$endgroup$
– Math1000
Dec 22 '18 at 2:41