| Klebert L Hall | 18 Sep 2008 9:25 a.m. PST |
I expect many of us have encountered the D&D convention of "roll 4d6 and drop the lowest die". Any math whizzes out there that can tell me the average number this method generates? I reach my "not caring anymore" threshold before I can figure it out, when I try. Thanks, -Kle. |
| Richard Humm | 18 Sep 2008 9:54 a.m. PST |
13 is the modal number if that's what you mean by average, IIRC. I did have a chart showing the modified bell curve it produces somewhere. |
| Jovian1 | 18 Sep 2008 10:00 a.m. PST |
Richard is correct – it modifies the bell curve up from 11 to 13. I had it somewhere too – oh well, not that we use that method much anymore! |
| Bob in Edmonton | 18 Sep 2008 10:12 a.m. PST |
That can be a handy mechanic in generating random force appearances during a scenario--you get a nice spread of possible appearance times (i.e., turn 3 to turn 18) but skewed slightly later in the game. |
| Jovian1 | 18 Sep 2008 11:54 a.m. PST |
Who gets to turn 18 in a game?!? Oh, wait, I've been playing Warhammer 40K for too long – where games only last 5-7 turns – at most, and you usually run out of troops by turn 3 or 4. |
| Bob in Edmonton | 18 Sep 2008 12:54 p.m. PST |
A large DBA game, Blitzkrieg Commander, etc. can all go 18 turns, especially on a large game board. Also, sometimes the reinforcements never show up if the game ends early, which creates tension all of its own for players. |
| Klebert L Hall | 18 Sep 2008 1:23 p.m. PST |
Yes, the modal number. Thanks. So, it's quite similar to 2d6+6. then. Is the latter a flatter curve or peakier? -Kle. |
Parzival  | 18 Sep 2008 1:45 p.m. PST |
So, it's quite similar to 2d6+6. then. Well, no, because that would produce a range of 7-18, vs. 3-18 for 4d6 drop one. Is the latter a flatter curve or peakier? 2d6+6 identical to a 2d6 curve, only the numbers are shifted higher by 6. |
| DS6151 | 18 Sep 2008 3:22 p.m. PST |
This is not an average question, and is in fact, quite hard. |
| Marshal Mark | 19 Sep 2008 5:57 a.m. PST |
It's quite easy to work out on a spreadsheet : Number Percentage 3 0.08% 4 0.31% 5 0.77% 6 1.62% 7 2.93% 8 4.78% 9 7.02% 10 9.41% 11 11.42% 12 12.89% 13 13.27% 14 12.35% 15 10.11% 16 7.25% 17 4.17% 18 1.62% |
| Martin Rapier | 19 Sep 2008 7:58 a.m. PST |
"Who gets to turn 18 in a game?!? " LOL. I have managed to get through 120 turns in one game in an evening, but it was a variable length bound jobbie and most of the participating formations spent most of their time reorganisating and preparing for the next attack before indulging in a few hours of frantic activity prior to the next R&R. |
| Klebert L Hall | 19 Sep 2008 8:16 a.m. PST |
Thanks again, folks. Especially for putting up with my (imbecilic in retrospect) 2d6+6 question. -Kle. |
| Ditto Tango 2 1 | 19 Sep 2008 7:09 p.m. PST |
Who gets to turn 18 in a game?!? Try Crossfire. -- Tim |
| Ditto Tango 2 1 | 19 Sep 2008 7:18 p.m. PST |
I'll have a crack at it, though this is not offered with any kind of authority -I've forgotten most of my 2nd year Stats course, including the probabilities part of it. The average result for a single d6 is 3.5 (1+2+3+4+5+6)/6 For any multiple, x, of d6 rolled, the average result is x times 3.5 I would guess (again, emphasizing this is a guess) that in a 4 d6 take away lowest, that the average of the dice removed would also be a single d6, ie, 3.5 (yes, we're getting "dicey" here as what is taken away is going to be the lowest number, but I'll assume my assumption is correct). Thus, for 3 d6, the average of which is 10.5, you'd add the extra 3.5 being taken away which gives you 14. But 14 is off from the values proffered by Richard and Jovian. Oh well, just a try. -- Tim |
| Klebert L Hall | 20 Sep 2008 9:27 a.m. PST |
But 14 is off from the values proffered by Richard and Jovian. Close enough for rule-of-thumb, though. -Kle. |
| Marshal Mark | 22 Sep 2008 3:51 a.m. PST |
Well, the mode (most common) is clearly 13. Going back to my spreadsheet and calculating the mean, it comes out at 12.24. |