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Dale M
Dale M
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Combinatorics

Various methods are available depending on how accurate you want to be and the sample space size. For fistfuls of dice, you can brute-force count them: manually, with a spreadsheet, or by writing custom bit of code (like anydice!).

So, for "the highest 2 of 3d10", we have a sample space of \$10^3\$ and an output range of 2-20. So we can get a:

  • 2 in 1 way ([1,1,1]) = \$1 \over1000\$
  • 3 in 3 ways ([2,1,1],[1,2,1],[1,1,2]) = \$3 \over1000\$
  • 4 in 6 ways ([3,1,1],[1,3,1],[1,1,3],[2,2,1],[2,1,2],[1,2,2]) = \$6 \over1000\$
  • ...
  • 20 in 28 ways ([10,10,X],[10,X,10],[X,10,10] remembering to only count [10,10,10] once) = \$28 \over1000\$

Or you can use the formulas generated in response to this question on Math Stack Exchange - be warned, serious mathematics ahead:

Generating Function for sum of N dice [or other multinomial distribution] where lowest N values are "dropped" or removed

Disadvantage is simply the mirror-image of advantage, so once you solved for one, the probabilities apply in the reverse order to the other.

Combinatorics

Various methods are available depending on how accurate you want to be and the sample space size. For fistfuls of dice, you can brute-force count them manually, with a spreadsheet, or by writing custom bit of code (like anydice!).

So, for "the highest 2 of 3d10", we have a sample space of \$10^3\$ and an output range of 2-20. So we can get a:

  • 2 in 1 way ([1,1,1]) = \$1 \over1000\$
  • 3 in 3 ways ([2,1,1],[1,2,1],[1,1,2]) = \$3 \over1000\$
  • 4 in 6 ways ([3,1,1],[1,3,1],[1,1,3],[2,2,1],[2,1,2],[1,2,2]) = \$6 \over1000\$
  • ...
  • 20 in 28 ways ([10,10,X],[10,X,10],[X,10,10] remembering to only count [10,10,10] once) = \$28 \over1000\$

Or you can use the formulas generated in response to this question on Math Stack Exchange - be warned, serious mathematics ahead:

Generating Function for sum of N dice [or other multinomial distribution] where lowest N values are "dropped" or removed

Disadvantage is simply the mirror-image of advantage, so once you solved for one, the probabilities apply in the reverse order to the other.

Combinatorics

Various methods are available depending on how accurate you want to be and the sample space size. For fistfuls of dice, you can brute-force count them: manually, with a spreadsheet, or by writing custom bit of code (like anydice!).

So, for "the highest 2 of 3d10", we have a sample space of \$10^3\$ and an output range of 2-20. So we can get a:

  • 2 in 1 way ([1,1,1]) = \$1 \over1000\$
  • 3 in 3 ways ([2,1,1],[1,2,1],[1,1,2]) = \$3 \over1000\$
  • 4 in 6 ways ([3,1,1],[1,3,1],[1,1,3],[2,2,1],[2,1,2],[1,2,2]) = \$6 \over1000\$
  • ...
  • 20 in 28 ways ([10,10,X],[10,X,10],[X,10,10] remembering to only count [10,10,10] once) = \$28 \over1000\$

Or you can use the formulas generated in response to this question on Math Stack Exchange - be warned, serious mathematics ahead:

Generating Function for sum of N dice [or other multinomial distribution] where lowest N values are "dropped" or removed

Disadvantage is simply the mirror-image of advantage, so once you solved for one, the probabilities apply in the reverse order to the other.

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Thomas Markov
Thomas Markov
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CombinitronicsCombinatorics

Various methods are available depending on how accurate you want to be and the sample space size. For fistfuls of dice, you can brute-force count them manually, with a spreadsheet, or by writing custom bit of code (like anydice!).

So, for "the highest 2 of 3d10", we have a sample space of \$10^3\$ and an output range of 2-20. So we can get a:

  • 2 in 1 way ([1,1,1]) = \$1 \over1000\$
  • 3 in 3 ways ([2,1,1],[1,2,1],[1,1,2]) = \$3 \over1000\$
  • 4 in 6 ways ([3,1,1],[1,3,1],[1,1,3],[2,2,1],[2,1,2],[1,2,2]) = \$6 \over1000\$
  • ...
  • 20 in 28 ways ([10,10,X],[10,X,10],[X,10,10] remembering to only count [10,10,10] once) = \$28 \over1000\$

Or you can use the formulas generated in response to this question on Math Stack Exchange - be warned, serious mathematics ahead:

Generating Function for sum of N dice [or other multinomial distribution] where lowest N values are "dropped" or removed

Disadvantage is simply the mirror-image of advantage, so once you solved for one, the probabilities apply in the reverse order to the other.

Combinitronics

Various methods are available depending on how accurate you want to be and the sample space size. For fistfuls of dice, you can brute-force count them manually, with a spreadsheet, or by writing custom bit of code (like anydice!).

So, for "the highest 2 of 3d10", we have a sample space of \$10^3\$ and an output range of 2-20. So we can get a:

  • 2 in 1 way ([1,1,1]) = \$1 \over1000\$
  • 3 in 3 ways ([2,1,1],[1,2,1],[1,1,2]) = \$3 \over1000\$
  • 4 in 6 ways ([3,1,1],[1,3,1],[1,1,3],[2,2,1],[2,1,2],[1,2,2]) = \$6 \over1000\$
  • ...
  • 20 in 28 ways ([10,10,X],[10,X,10],[X,10,10] remembering to only count [10,10,10] once) = \$28 \over1000\$

Or you can use the formulas generated in response to this question on Math Stack Exchange - be warned, serious mathematics ahead:

Generating Function for sum of N dice [or other multinomial distribution] where lowest N values are "dropped" or removed

Disadvantage is simply the mirror-image of advantage, so once you solved for one, the probabilities apply in the reverse order to the other.

Combinatorics

Various methods are available depending on how accurate you want to be and the sample space size. For fistfuls of dice, you can brute-force count them manually, with a spreadsheet, or by writing custom bit of code (like anydice!).

So, for "the highest 2 of 3d10", we have a sample space of \$10^3\$ and an output range of 2-20. So we can get a:

  • 2 in 1 way ([1,1,1]) = \$1 \over1000\$
  • 3 in 3 ways ([2,1,1],[1,2,1],[1,1,2]) = \$3 \over1000\$
  • 4 in 6 ways ([3,1,1],[1,3,1],[1,1,3],[2,2,1],[2,1,2],[1,2,2]) = \$6 \over1000\$
  • ...
  • 20 in 28 ways ([10,10,X],[10,X,10],[X,10,10] remembering to only count [10,10,10] once) = \$28 \over1000\$

Or you can use the formulas generated in response to this question on Math Stack Exchange - be warned, serious mathematics ahead:

Generating Function for sum of N dice [or other multinomial distribution] where lowest N values are "dropped" or removed

Disadvantage is simply the mirror-image of advantage, so once you solved for one, the probabilities apply in the reverse order to the other.

Source Link
Dale M
Dale M
  • 216k
  • 42
  • 545
  • 912

Combinitronics

Various methods are available depending on how accurate you want to be and the sample space size. For fistfuls of dice, you can brute-force count them manually, with a spreadsheet, or by writing custom bit of code (like anydice!).

So, for "the highest 2 of 3d10", we have a sample space of \$10^3\$ and an output range of 2-20. So we can get a:

  • 2 in 1 way ([1,1,1]) = \$1 \over1000\$
  • 3 in 3 ways ([2,1,1],[1,2,1],[1,1,2]) = \$3 \over1000\$
  • 4 in 6 ways ([3,1,1],[1,3,1],[1,1,3],[2,2,1],[2,1,2],[1,2,2]) = \$6 \over1000\$
  • ...
  • 20 in 28 ways ([10,10,X],[10,X,10],[X,10,10] remembering to only count [10,10,10] once) = \$28 \over1000\$

Or you can use the formulas generated in response to this question on Math Stack Exchange - be warned, serious mathematics ahead:

Generating Function for sum of N dice [or other multinomial distribution] where lowest N values are "dropped" or removed

Disadvantage is simply the mirror-image of advantage, so once you solved for one, the probabilities apply in the reverse order to the other.