0
$\begingroup$

Looking how to calculate (1) the number of permutations for different number ranges on each dice; and (2) the number of permutations for number ranges shared by two or more dice.

All dice are fair and 6 sided.

I can do the easy calculations where how many rolls have exactly one 6, or at least one 6, two 6's, or at least two 6's etc.

'm looking for a more general approach where I have m dice, and find permutations where say where:

at least one die is 5+ and at least another is 4+.

in general I would like to be able to find a formula or counting approach what are the number of permutations where x die are greater than 5, when rolling m dice where a constant c can be distributed across the die.

$\endgroup$
  • $\begingroup$ Any help in setting up the problem would be useful as well. Currently looking at defining it as a set S, with the elements S={1,2,3,4,5,6}. Each dice "roll" corresponds a selection from the set. So thinking the problem could be phrased as how many permutations exist where on three selections (s1,s2,s3) at least one selection is a {4,5,6} and another is {5,6}. $\endgroup$ – MikeX Nov 6 '17 at 15:34
  • $\begingroup$ A more enumeration approach I am now trying makes use of generating functions. Let each 6-sided die have its own generating function. I assign generating functions as follows. For the die I want a 4,5 or 6 on it is 3x+3y. For the other two die I want at least 5+, so I assign for their generating functions 4x+2y. Then I just sum the coefficients of any term containing y. I've tired this for 2 dice and 3 dice with a simpler case (6+ on at least one die) and it seems to work. Since the math is beyond me, going to try brute force enumeration of all 3 dice permutations and check. $\endgroup$ – MikeX Nov 8 '17 at 2:51
0
$\begingroup$

I found a method to answer this question using generating functions. Represent set each die by the equation $3x+2y+z$ where x represents the numbers 1-3, y represents the numbers 5-6, and z the number 4.

For m dice calculate the polynomial $(3x+2y+z)^m$

Use the coefficients of the various terms to determine the number of combinations that yield.

for example, with 4 dice the sum of the coefficients of the terms $y^4$ and $y^3z$ will give you the number of combinations you can have 4 dice with a 5 or higher when you can add +1 to one of the rolls.

The sum of the coefficients of the terms $xy^3+xy^2z+y^2z^2$ will provide the number of combinations with 3 dice 5 or higher where you can add +1 to one die.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.