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Message: <blockquote data-quote="Morgenstern" data-source="post: 1000047" data-attributes="member: 5485">Actually, you can get around the massiv summations somewhat... I solved some of this sort of math for L5R (d10) because you use roll & drop for everything in that game (and it has open ended dice, but we don't need to fret over it here).In my head, I deal with this as a series of 6x6x6 'outcome' cubes. Those are the dice you are going to keep. for a 5k3 roll, I have 36 of them. you can then start solving for number of possible outcomes of each result, and then use that to take an average (although the # of outcomes is much more useful since it shows you the -shape- of the probibility curve).With a flat 3k3 you get3: 14: 35: 66: 107: 158: 219: 2510: 2711: 2712: 2513: 2114: 1515: 1016: 617: 318: 1Which is convieniently symetrical, allowing you to casually determine the average is 10.5. That one is easy - it uses only one outcome cube and you can count through it with little trouble.With the first extra dropped die (4k3), the range remains the same, but the shape of the curve distorts rapidly. You now have six cubes, each one reprsenting the above plus the 'extra die than could be used to substitute for any of the dice in above cube. If you build all six cubes and and then sum the possibiltes for each individual result (from 3 to 18), you can again see the curve and take an average. The first cube you have an extra '1' to play with and there is no change- no combination is improved by substituting the 1. In the second cube you have an extra '2', which changes 91 of the 216 outcomes of that cube by +1. Having an extra '3' to play with changes 152 of them (91 by +2, 61 by +1). An extra '4' alters 189 (91 by +3, 61 by +2, 37 by +1). A '5' affects 208 outcomes in it's cube (91 by +4, 61 by + 3, 37 by +2, 19+1), and finally a '6' affects 215 outcomes or all but one the 6+6+6 result (91 by +5, 61 by +4, 37 by +3, 19 by +2, 7 by +1). I may build full results tables on this thread if I get bored <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    :)"  data-smilie="1"data-shortname=":)" />.Even so, computing the average is pretty easy with that information (I'll post how next time <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    :)"  data-smilie="1"data-shortname=":)" />). In any event, the important thing to notice is that a larger extra roll impacts the average more, but each incremental increase affects far fewer new outcomes... This is significant because when you bring a second dropped die into consideration, it actually will have FAR less impact than the first one. The jump in the average between 3k3 and 4k3 and 5k3 isn't linear. It CAN'T be! As a simple example, if the increase in the average between 3k3 and 4k3 were +3 AND it was linear for each dropped die added, then the average would by higher than your maximum possible result by 6k3 (average 19.5 by that logic). Let me fiddle around with the tables a bit and see if I can't get you a specific answer for 5k3 using d6s. It can't be nearly as hard as doing it for d10s was (you are only discussing 7776 possible combinations instead of 100,000 <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    :)"  data-smilie="1"data-shortname=":)" />)...</blockquote>

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