
Post by keith418 on Jan 14, 2012 16:10:39 GMT 6
"If the party encounters orcs, is the DM supposed to roll 30d10? 1d10 x30? 5d10 x6? Or something else of his own concoction?" I saw this question over: www.wizards.com/default.asp?x=dnd/alumni/20070302aIs there any way to actually generate a 30300 range with the dice that isn't super cumbersome? I am going through the OD&D rules and have no idea how to get the number of "brigands" I might need for a random encounter.



Post by Harbinger on Jan 14, 2012 16:28:05 GMT 6
I'd probably do 3d10 x 10.



Post by murquhart72 on Jan 14, 2012 16:38:54 GMT 6
You're supposed to just choose, using the numbers as a loose guideline. They're given in certain ranges so that if you can't decide and/or don't care, you can determine the number randomly. The actual method doesn't matter as it'd be all random in the end anyway.



Post by talysman on Jan 14, 2012 18:32:32 GMT 6
I'd probably do 3d10 x 10. themattjon's response is a good one, but this is a good default for random rolls, if desired. I've noticed that in the treasure tables, there aren't very many rolls of large numbers of dice, even when the maximum end of the range is pretty large; anything above 5dX to 7dX is usually expressed as a multipliers, with lots of entries of the form "10 x 5d6". So I would probably factor out 10 or 20 for any "number appearing" range that looks pretty large before figuring out how many dice to roll.



Post by Harbinger on Jan 26, 2012 12:34:03 GMT 6
3d10 x 10 gives an nice bell curve centered on 160/170 orcs.
If you want something with a linear probability (all results are equally likely) then I would roll d100 and a d6. The d6 indicates what '100s' the number is: 12:1100, 34: 101200, 56: 201300. Reroll any result < 10. This sounds complicated but is really just as fast and you only have a 3% chance of having to reroll.



Post by Stormcrow on Jan 26, 2012 13:04:58 GMT 6
This is why Gary added all that stuff about dice and probabilities to the beginning of the DMG: understanding probability curves lets you understand how to generate numbers like this.
Yes, there are 30–300 orcs. There are some number of orcs between 30 and 300. How you choose that number is entirely up to you. You can select it yourself, or roll 30d10, or 3d10×10, or write a program, or throw darts at a dartboard, or look at tea leaves, or consult the bones...
(It just so happens that, if you don't want to choose, rolling 3d10×10 is particularly easy.)



Post by Mr. Darke on Jan 27, 2012 9:15:35 GMT 6
I use this as a general guideline. Obviously a party of 1st level adventurers are not going to be able to take 30 orcs let alone 300. I tend to use this number for tribes rather than encounters. Thus, the Black Skull tribe would have that number but not groups designed for encounters.



Post by Harbinger on Jan 27, 2012 9:36:34 GMT 6
I use this as a general guideline. Obviously a party of 1st level adventurers are not going to be able to take 30 orcs let alone 300. I tend to use this number for tribes rather than encounters. Thus, the Black Skull tribe would have that number but not groups designed for encounters. A little epiphany I had recently since I've been rediscovering the 'Old Ways' is that's where the 'evasion and pursuit' rules come into play. I had always ignored them in the past. If a band of 1stlevel adventurers are silly enough to wander out of the 'civilized' radius around their home base, they can still hope to escape detection. If they're not surprise, it's pretty easy to evade a large party of monsters.



Post by aldarron on Jan 27, 2012 9:59:36 GMT 6



Post by makofan on Jan 27, 2012 12:39:44 GMT 6
1d10 * 30



Post by jmccann on Jan 28, 2012 12:18:25 GMT 6
I would avoid this as multiple dice give a bell curve and weight toward the center of the range.


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Post by busman on Jan 28, 2012 13:37:50 GMT 6
I would avoid this as multiple dice give a bell curve and weight toward the center of the range. That's assuming you want middle of the range for the result. =) mako's gives a nice flat result across the whole range, which provides more randomness to the result. I've used both methods in campaigns depending upon what I was looking to do.



Post by makofan on Jan 28, 2012 15:52:15 GMT 6
d10*30 also matches the text in the monster entry very closely. I had never even considered that people did it any other way. Weird!



Post by jmccann on Jan 28, 2012 17:15:39 GMT 6
I would avoid this as multiple dice give a bell curve and weight toward the center of the range. That's assuming you want middle of the range for the result. =) mako's gives a nice flat result across the whole range, which provides more randomness to the result. I've used both methods in campaigns depending upon what I was looking to do. Yes it depends on what you are trying to do. If you are interested in simulating real world population size phenomena often a normal probability curve will be called for, and using multiple dice is a good way to simulate this. It is still random although it is less uniformly random. If you don't care it doesn't matter. I like verisimilitude, if not realism. YMMV.



Post by strangebrew on Jan 30, 2012 13:27:37 GMT 6
From the WotC article:



Post by aher on Jan 30, 2012 22:23:34 GMT 6
I have an inexpensive TI30XIIS calculator. It has a function called RANDI, used to generate a random integer between two integers you specify: RANDI(a,b). Presumably, these random integers are drawn from the discrete uniform distribution: RANDI(a,b) ~ DU(ba+1,a,1), where the notation DU(k,c,h) means a discrete uniform distribution with the probability range consisting of the k distinct values x_j = c + h*j for j = 0,1,...,k1 with equal probabilities 1/k.
The keystrokes for RANDI(30,300) are PRB, left arrow, Enter, 30, 2nd, right parens, 300, right parens, Enter. To repeat the calculation (draw another random number from the same range), you can type up arrow, Enter. I suspect most calculators have similar random number generating capabilities.
The other dice methods suggested here all have the same endpoints (30 and 300) and the same mean (165). But they all have different probability ranges (i.e., possible outcomes) and different standard deviations. The mean value is not always part of their probability range. Let's look at some of the possibilities in detail:
RANDI(30,300) This is the discrete uniform distribution: DU(271,30,1). The probability range is 30 to 300 in steps of 1. There are 271 possible outcomes. Each outcome has the same probability 0.369%. The mean is 165 and this is a possible outcome. The standard deviation is 78.2304288624.
d10x30 This is the discrete uniform distribution DU(10,30,30). The probability range is 30 to 300 in steps of 30. There are 10 possible outcomes. The probability of each outcome is 10%. The mean is 165, but this outcome is not in the range. The standard deviation is 86.1684396981.
d30x10 reroll on a 1 or a 2 This is the discrete uniform distribution DU(28,30,10). The probability range is 30 to 300 in steps of 10. The probability of each possible outcome is 3.571%. The mean is 165, but this outcome is impossible. The standard deviation is 80.777472107. The chance of one or more rerolls is approximately 7.14%.
3d10x10 This is a belllike distribution. The probability range is 30 to 300 in steps of 10. There are 28 possible outcomes. The mean value of 165 is not a possible outcome. Probabilities are symmetric about the mean. The mode is both 160 and 170, each with probability 7.5%. The probability of 30 (or 300) is 0.1%. The standard deviation is 49.7493718553.
5d10x6 Another belllike distribution. Range is 30 to 300 in steps of 6. 46 possible outcomes. Mean value is not a possible outcome. Mode is 162 and 168 w/prob 6% ea. Probability of 30 (or 300) is 0.001% ea. Standard deviation is 38.535697736.
30d10 Belllike. Range is all integers from 30 to 300, so the mean value of 165 is possible. P(X=165) = 2.523%. P(X=300) approx 0. Spread is 15.7321327226. So this distribution has the smallest standard deviation of all those considered.


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Post by Aplus on Mar 15, 2012 0:05:00 GMT 6
Good lord!



Post by nizrad on Apr 11, 2012 16:52:19 GMT 6
I also simply used 1d10*30 to generate a number of orcs, additionally I would tend to put that many orcs in the 1st level of the dungeon and or around the surface or 2nd level.



Post by Finarvyn on Apr 14, 2012 4:46:28 GMT 6
1. Wow. That's cool. I don't do much with calculator programming anymore, but I remember old calculators that would generate a random threedigit decimal from 0.000 to 1.000 and then I could just multiply by whatever dice type I wanted. Seems similar, and often a pain. Nice post, though. 2. I always found it interesting that there was no "official" way to do this. It could be 1d10*30 (as per nizrad's suggestion) or 30d10 if a person was more into accuracy. Heck, with the funky dice out there I suppose it could be 1d30*10. 3. It is also interesting to me that there is a single number range given, apparently most approapriate for wilderness adventures rather than the dungeons. (I suppose there could be 30300 orcs in a dungeon, but that really messes up lowlevel characters.) There really should be three number ranges: wilderness wandering, wilderness lair, dungeon encounter.



Post by Stormcrow on Apr 14, 2012 8:24:13 GMT 6
2. I always found it interesting that there was no "official" way to do this. The point is to give you a convenient range, not to specify how you generate a number within that range. Rolling dice is really only secondary; you're best off choosing the number you want. The Number Appearing is explicitly only for wilderness encounters, whether lair or not. Starting with the '81 revision of the rules, the Number Appearing statistic changes. Goblins, for instance, have "2–8 (6–60)." The first range is the number appearing in a dungeon and not in a lair. The range in parentheses is both the number appearing in a dungeon lair and the number appearing in the wilderness but not in a lair. Multiply the second range by 5 to get the number appearing in a wilderness lair. Dungeon NonLair: 2–8 Dungeon Lair: 6–60 Wilderness NonLair: 6–60 Wilderness Lair: 30–300 Number Appearing in a dungeon is modified by distance from the dungeon level corresponding to the monster's level. Monster level is determined by hit dice, not counting pluses or minuses, but adding one level for each special ability bonus (the latter not mentioned in the rules but generally applied in the wandering monster tables). Goblins have 11 HD, so they're level 1 monsters. You could therefore find them on levels 1–3 of the dungeon, roughly doubling or trebling (exact numbers are not mentioned) them on levels 2 or 3, respectively. Monsters are usually not encountered more than two levels away from the dungeon level corresponding to their monster level. Goblins have Treasure Type "R (C)" (individuals R; C in lair or wilderness). The amount of lair treasure rolled (if determined randomly) assumes the maximum number of monsters in the lair. If there are fewer monsters, adjust the treasure by the ratio of monsters present to the maximum number. Wilderness lairs therefore have more guardians of the same amount of treasure than dungeon lairs.



Post by aldarron on Apr 14, 2012 11:58:01 GMT 6
2. I always found it interesting that there was no "official" way to do this. The point is to give you a convenient range, not to specify how you generate a number within that range. Rolling dice is really only secondary; you're best off choosing the number you want. The Number Appearing is explicitly only for wilderness encounters, whether lair or not. Starting with the '81 revision of the rules, the Number Appearing statistic changes. Goblins, for instance, have "2–8 (6–60)." The first range is the number appearing in a dungeon and not in a lair. The range in parentheses is both the number appearing in a dungeon lair and the number appearing in the wilderness but not in a lair. Multiply the second range by 5 to get the number appearing in a wilderness lair. Dungeon NonLair: 2–8 Dungeon Lair: 6–60 Wilderness NonLair: 6–60 Wilderness Lair: 30–300 Great post Stromcrow! I hadn't realized that the B/X multiply by 5 rule would get you the OD&D numbers. That's cool. Number Appearing in a dungeon is modified by distance from the dungeon level corresponding to the monster's level. Monster level is determined by hit dice, not counting pluses or minuses, but adding one level for each special ability bonus (the latter not mentioned in the rules but generally applied in the wandering monster tables). Goblins have 11 HD, so they're level 1 monsters. You could therefore find them on levels 1–3 of the dungeon, roughly doubling or trebling (exact numbers are not mentioned) them on levels 2 or 3, respectively. Monsters are usually not encountered more than two levels away from the dungeon level corresponding to their monster level. With the above paragraph I assume we are back to OD&D?: especialy the quote from pg 11 of UW&A "If the level beneath the surface roughly corresponds with the level of the monster then the number of monsters will be based on a single creature, modified by type (that is Orcs and the like will be in groups) and the number of adventurers in the party. A party of from 13 would draw the basic number of monsters, 46 would bring about twice as many, and so on." So, assuming I'm reading you right, an adventuring pary of 3 would encounter 39 goblins on level 1, 618 on level 2 and 1236 on level 3. Is that consistent with your reading? Goblins have Treasure Type "R (C)" (individuals R; C in lair or wilderness). The amount of lair treasure rolled (if determined randomly) assumes the maximum number of monsters in the lair. If there are fewer monsters, adjust the treasure by the ratio of monsters present to the maximum number. Wilderness lairs therefore have more guardians of the same amount of treasure than dungeon lairs. That makes sense, but I'm wondering what you are drawing on here. I've seen a lot of commentary that assumes the the treasure tables amount is independant of the actual number of monsters present.



Post by Stormcrow on Apr 14, 2012 15:11:05 GMT 6
I hadn't realized that the B/X multiply by 5 rule would get you the OD&D numbers. That's cool. Only kindasorta. They obviously weren't trying to reproduce them exactly, just generally, as convenient. No, still in the '81 rules. Not according to the '81 rules. In those rules, monster numbers are not adjusted according to party size. Strictly speaking, a dungeon encounter with goblins (No. Appearing 2–8 (6–60)) on level 1 will have 2–8 goblins, no matter the size of your party. You will encounter more goblins on levels 2 or 3. Exact numbers are not defined for deeper (or less deep) levels, but I believe that in general double the number one level deeper or triple the number two levels deeper are good values. Let's take a more interesting example. Ogres have 4+1 HD and No. Appearing 1–6 (2–12). On dungeon level 4, a party will encounter 1–6 of them outside of lairs, or 2–12 of them in a lair. On dungeon level 5, a party will encounter 2–12 of them, or 4–24 in a lair. On dungeon level 6, a party will encounter 3–18 of them, or 6–36 of them in a lair. On dungeon level 3, a party will encounter 1–4 (1–6); on dungeon level 2, there will be 1–2 (1–4). Ogres are very rare on dungeon level 1, and if they appear there there will only be 1 of them at a time. Ogres on dungeon levels deeper than 6 will rare, but if they are found there will be many of them, and they will probably be in the employ of a greater power (to explain their presence). There will be no lairs outside of levels 2–6. Basic Set p. B45: "An encounter with less than a full lair should yield less treasure." An example of this is illustrated on p. B56, where the fraction of monsters out of a full lair yields an equal fraction of rolled treasure. Note that in AD&D the treasure types assume an average lair, not a full lair. The difference is probably due to the fact that the original rules don't specify clearly how the adjustment should be made.



Post by aldarron on Apr 14, 2012 20:39:52 GMT 6
Ah, okay. Thanks for clarifying that.



Post by blackbarn on Jul 3, 2012 9:45:22 GMT 6
If you are using a computer during prep (or use on during play) go to www.random.org and type in the range of 30300, or whatever else you want. It generates a true random number based on any parameters you choose.



Post by makofan on Aug 5, 2012 20:06:36 GMT 6
An update to this thread: I was reading the MM2 today, and Gygax writes that 1040 is 1d4x10, and 20160 is 2d8x10, so 30300 was obviously 1d10x30



Post by Harbinger on Aug 9, 2012 7:57:08 GMT 6
As an aside  one interesting take on the number appearing I've seen suggested is to use the wilderness number and apply it to the whole HDappropriate dungeon level.
So if you encounter hobgoblins on wandering, you'll encounter a 'party and level appropriate' quantity, but then the DM can also create a lair on the HD equivalent level that has the wilderness amount of hobgoblins scattered about the level in groups of varying sizes to be encountered. Of course they will coordinate to defend their lair.
I use this on levels I haven't fleshed out yet to help me populate them  once the PCs encounter something on the wandering monster table, I build out the lair on the right level.


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Post by Deleted on May 2, 2013 15:36:17 GMT 6
If you have python installed (maybe running some sort of *nix), then you could use the random library to generate numbers.
bsd52$ python2.7 Python 2.7.3 (default, Aug 3 2012, 05:33:56) [GCC 4.2.1 20070719 ] on openbsd5 Type "help", "copyright", "credits" or "license" for more information. >>> import random >>> random.randint(30,300) 162 >>> ^D


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Post by luc on May 11, 2013 6:10:28 GMT 6
An update to this thread: I was reading the MM2 today, and Gygax writes that 1040 is 1d4x10, and 20160 is 2d8x10, so 30300 was obviously 1d10x30 Wouldn't that be 3d10x10?



Post by Stormcrow on May 11, 2013 10:01:45 GMT 6
The monster tables were designed at a time when Chainmail was still large in the players' minds. Since Chainmail primarily uses a 1:10 figure ratio, and since the Number Appearing column refers only to wilderness adventuring, it might make sense if those numbers were designed to make scaling for Chainmail easy.
Thus, for nonfantastic figures, you'd need No. App. in multiples of 10—which is exactly what we have! Hence when goblins are encountered, there aren't just 40–400 of them; there are 4–40 figures. This means a random determination of goblins would have to be 4d10 figures, or 4d10×10 goblins, not 40d10 goblins.
Naturally, monsters that don't come in multiples of 10 are not "fantastic" types, and use a figure ratio of 1:1.


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Post by Deleted on May 11, 2013 11:08:46 GMT 6
Actually, CHAINMAIL is 1:20

