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<blockquote data-quote="LazarusLong42" data-source="post: 1000146" data-attributes="member: 9620">Correct on both counts; it is extremely hard to do.&nbsp; I figured I'd put my formula out there for reference and as a starting point, as someone had already mentioned 3d6 + 1d8 drop lowest.I suppose it depends on your situation and what you're trying to do.&nbsp; If you're trying to find the average, then that might be slightly faster.&nbsp; If, however, you want to find the probability of rolling, say, 65 on 20d6, I'll guarantee you that one summation/nested loop (mathematical formula) is faster than 20 nested loops (your way).For the dropping problem... the verdict is out, I'd say.&nbsp; I still haven't properly understood what Nathan's written above. <img src="https://cdn.jsdelivr.net/joypixels/assets/8.0/png/unicode/64/1f642.png" class="smilie smilie--emoji" loading="lazy" width="64" height="64" alt=":)" title="Smile&nbsp; &nbsp; :)"&nbsp; data-smilie="1"data-shortname=":)" />&nbsp; But even if it takes three extra nested loops for dropping, four nested loops (3 + 1) to run 5d6 drop 2 is faster than your five nested loops (one per die).Edit:&nbsp; I've just realized I'm mixing terminology.&nbsp; What you describe may be easier to write a program for.&nbsp; The difference, I suppose, would be that my program will finish today for calculating the probability of getting 65 on 20d6 (the loop will run through the same few operations 8 times).Your suggestion will run through the same set of operations 3,656,158,440,062,976 (6^20) times.&nbsp; Assuming you only have one operation in your loop, and you are lucky enough to eke out a gigaflop from a Pentium machine, your program will finish running middle of next month.So, yes, what you describe is easier to write, but it would be cumbersome to run. <img src="https://cdn.jsdelivr.net/joypixels/assets/8.0/png/unicode/64/1f642.png" class="smilie smilie--emoji" loading="lazy" width="64" height="64" alt=":)" title="Smile&nbsp; &nbsp; :)"&nbsp; data-smilie="1"data-shortname=":)" /></blockquote>

[QUOTE="LazarusLong42, post: 1000146, member: 9620"] Correct on both counts; it is extremely hard to do. I figured I'd put my formula out there for reference and as a starting point, as someone had already mentioned 3d6 + 1d8 drop lowest. I suppose it depends on your situation and what you're trying to do. If you're trying to find the average, then that might be slightly faster. If, however, you want to find the probability of rolling, say, 65 on 20d6, I'll guarantee you that one summation/nested loop (mathematical formula) is faster than 20 nested loops (your way). For the dropping problem... the verdict is out, I'd say. I still haven't properly understood what Nathan's written above. :) But even if it takes three extra nested loops for dropping, four nested loops (3 + 1) to run 5d6 drop 2 is faster than your five nested loops (one per die). Edit: I've just realized I'm mixing terminology. What you describe may be easier to [i]write[/i] a program for. The difference, I suppose, would be that my program will finish today for calculating the probability of getting 65 on 20d6 (the loop will run through the same few operations 8 times). Your suggestion will run through the same set of operations 3,656,158,440,062,976 (6^20) times. Assuming you only have one operation in your loop, and you are lucky enough to eke out a gigaflop from a Pentium machine, your program will finish running middle of next month. So, yes, what you describe is easier to write, but it would be cumbersome to run. :) [/QUOTE]

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