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D&D 5th ed. gives the following instructions for determining your “ability scores.”

Roll four 6-sided dice and record the total of the highest three dice

If I repeat the “roll-toss-and-total” 6 times (generating a set of 6 totals), what is the probability that my resulting set will contain

  • 4 or more totals ≥ 10 and
  • 3 or more totals ≥ 12 and
  • 2 or more totals ≥ 15?

Note that this problem contains and statements.

I already figured out the probabilities the totals of these rolls. (3: 1/1296; 4: 1/324; 5: 5/648; 6: 7/432; 7: 19/648; 8: 31/648; 9: 91/1296; 10: 61/648; 11: 37/324; 12: 167/1296; 13: 43/324; 14: 10/81; 15: 131/1296; 16: 47/648; 17: 1/24; 18: 7/432)

I can’t remember how to use these probabilities to ask more than basic questions. :-/

Nik
  • 41

1 Answers1

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For something with this many rules you would likely need to manually add up the probabilities that satisfy your conditions.

Fortunately we have the technology! if you go to http://topps.diku.dk/torbenm/troll.msp

and plug x := (6 # (sum largest 3 4d6)); if (2 <= (count 15 <= x)) & (3 <= (count 12 <= x)) & (4 <= (count 10 <= x)) then 1 else 0 With a probability of 39%.

That is assuming I have the code correct.

Snark
  • 377
  • +1 The probability is ${ 4346417192834375\over 10968475320188928 }\approx 0.3962644822$. –  Apr 04 '17 at 01:02