Thu Oct 31 11:09:03 2013 by Anon |

Hello, I can't find solution to calculate this kind of die roll: ---- This is a roll of two six-sided dice (2d6) but with an additional bonus: if any individual die roll is “6,” one is subtracted, and then that die is re-rolled and added to the result. This is referred to as an “open-ended” 2d6 roll. Example: Two dice are rolled and they come up 2,6. Because one of the dice was a “6,” one is subtracted from the total (making 7), and the die is rolled again. But this die is also a 6. So one is subtracted from the total (now up to 12) and a die is rolled again. It is a 4. The final result for this DRN is 16. Note that if both original dice came up as 6, both would be re-rolled and added as above. If a die keeps coming up 6, it keeps getting re-rolled and added, which can very occasionally lead to large numbers. ----- Can someone help there? Thank you |

Fri Nov 1 10:03:18 2013 by Torben |

The easiest is to treat each die separately and then add them at the end. And it is also easier to roll the dice as z5's (dice with values from 0 to 5) and then add two to the result at the end. So this should work:`2 + sum 2#(accumulate x := z5 while x=5)` |