May 25, 2012  Posted by Martin Ralya 
For my Bleakstone hex crawl, I decided I wanted a random encounter system that met the following criteria:
 Simple
 Conforms roughly to the baseline 1/6 chance of an encounter/day in Basic D&D
 Can be weighted by terrain, region, PC activity, etc.
 Not entirely predictable by my players
I couldn’t find a system like that in any of my books, or online, so I built one.
The spectrum
My baseline was Labyrinth Lord’s system (which is the same as Moldvay Basic D&D, which LL clones): Roll d6 once a day, and there’s an encounter on a 1. That looked too simple and too predictable for my tastes, because it ignores terrain (etc.) and players can count on 0 or 1 encounter/day when traveling.
On the other end of the spectrum was AD&D 1e’s system, which involves up to 6 rolls/day at different times of day, is weighted by terrain type (forests are more dangerous than plains, etc.), and produces encounters on a 1 on a d10, d12, or d20 roll (depending on region type). That sounded a bit too fussy for me, and the time of day element has the potential to be predictable.
I like aspects of both of those systems, but I wanted to land somewhere between them.
My system
The roll is d10+d10, and a modifier of +1 to +5 can be applied depending on terrain type, region, or other factors (wagon trains, 30 mercenaries in tow, etc.). d10+d10 gives a pyramidshaped distribution, and out of the various rolls I tried it was the easiest to map out.
Below are three tables showing how this works in practice: d10+d10 at the top, d10+d10+3 in the middle, and d10+d10+5 at the bottom. (You could go higher than +5, but +5 gives a 50/50 chance of 1+ encounters, and that’s where I wanted to stop.)
The middle — a forest, maybe — corresponds roughly to a 2/6 chance of having an encounter, but there’s a 10% chance of two encounters. And the bottom is the most dangerous region, the Haunted Moor of Doom, wherein there’s a 3/6 chance of having at 1+ encounters and a possibility (6% chance) of having as many as three.
I like that this system has a small footprint (in my rules document, it’s just one table, the first one; the other two are for illustrative purposes), scales easily based on whatever factors I want it to, isn’t immediately predictable but also isn’t wildly unpredictable (players should be able to catch the increased frequency of encounters in some areas, but not know for sure what the cap is), and aggregates what could be multiple rolls into a single roll.
Time of day
If I need to randomize when an encounter will occur, that’s as simple as rolling d6 for each encounter: 1 Morning, 2 Noon, 3 Evening, 4 Night, 5 Midnight, 6 PreDawn. (Those are the same times AD&D 1e uses.)
The math
If you want to see the math behind this system, check out AnyDice, my favorite online probability calculator. Here are the results for d10+d10, +3, and +5. My thanks to our own Matt Neagley — who’s a probability wizard — for helping me crunch and verify the numbers, presenting an alternate system, and talking me through a lot of the math.
If this system is useful to you, or looks horribly flawed in some way, sound off in the comments!
16 Responses to A Lightweight System for Random Encounters: d10+d10

Samardan Press  Isles of the Saharan Sea: Exploring Hexes…
[…] encounter checks, I am currently considering the use the system outlined in Martin Rayla’s A Lightweight System for Random Encounters: d10+d10 article on Gnome Stew. In this article, he proposes a simple 2d10 table that, with modifiers, can […]

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[…] My new system for hex crawl encounters is going to be something I can roll up ahead of time, setting up whole days at a time. I will roll 6d6, and for each pair of numbers, the players will face an encounter during that sixth of the day. Stealing from Drow in the comments here: […]
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May 25, 2012 at 8:59 am  Link
So I haven’t checked the numbers on this or anything, but to borrow an idea from Legends of the Wulin: What if you roll Nd6 and count the number of dice showing the same face? With 2d6, the chances of getting two identical faces is exactly 1/6; with higher amount of dice the probability of getting two identical results increases dramatically – at 6d6 it’s almost a certainty. Then, simply count the highest number of identical results, and have a number of encounters be equal to the identical results minus one.
That system is intuitive, unpredictable, and fairly easy to adjust (although it would be a bit fiddly to predict the probabilities), and it doesn’t require a table. If you roll two fours or two sixes or two threes, that means one encounter occurs. If you roll four fours or four sixes or four threes, that means three encounters occur – and so on.
The thought just occurred to me as I looked at the probabilities outlined in this table, because I think they would look something similar for the “Yahtzee method”.
May 25, 2012 at 9:56 am  Link
That’s a neat idea, and I’d love to see the math worked out for the probabilities. It’s similar to the alternate system Matt proposed while I was working on d10+d10, which used multiple d20s with each die indicating a potential encounter.
“Yahtzee method” is a really clever moniker for your Xd6 idea, too.
One of the things I like about d10+d10 is that a cardcarrying math buffoon like me can wrap my head around the probabilities involved. That’s important to me, and may be much less so to others.
May 25, 2012 at 4:54 pm  Link
@Martin Ralya – I seem to remember AD&D 2nd Edition using a similar approach for single encounters on an encounter table with 1d8+1d12, resulting in a number between two and twenty. I wonder what using two different types of dice does to probability for dice that provide results in the same number range, if anything….
May 25, 2012 at 9:48 pm  Link
My individual encounter tables are d8+d12 and based on AD&D.
I love the trapezoidal distribution of d8+d12 for those because 913 are all equally likely and 2 and 20 are vanishingly rare (1% each). That lets me stock a table with encounters that are really iconic to the terrain/region (“Whenever we go to the haunted moor, we seem to meet undead…”) in the 913 slots and still have a couple of apex predators like dragons show up once in a blue moon (2 and 20).
d10+d10 is a pyramidal distribution, with 11 being the most likely and no flat spot in the middle. That makes the math easy for me, and the percentages line up pretty neatly with my baseline 1/6 chance, but in practical terms I could have used d8+d12 as the baseline here. Since I don’t care about the probability of any individual number as much as I do about the aggregate probability of a range of numbers, I’m pretty sure the math would work.
All of that said, I’m not mathinclined and could easily be wrong about that whole second paragraph. 😉
May 26, 2012 at 7:20 pm  Link
I suggest a simpler system that achieves much the same results:
Each day roll 1d6. on a 1 there is a random encounter (or a 12 if it is very likely, or a 13 if it’s extremely likely). If the result indicates a random encounter, roll again, using the same probability, to see if there’s another one. Stop when you hit 3.
The probabilities aren’t exactly the same (especially at higher likelyhoods — the odds of 23 encounters goes up compared to your chart) but it has the advantage that there’s no table to memorize or refer to.
— 77IM
May 26, 2012 at 8:10 pm  Link
@77IM – Sure, it’s simpler, but if you want to have an increased chance based on terrain (for example) you have to look up a table anyway. 😉
May 28, 2012 at 9:32 am  Link
roll 2d6. if both dice show the same number, the party encounters something in that time window. for example, if you roll two 3’s, the party encounters something that evening. the overall odds of an encounter on any particular day are 1/6.
to increase the probability of an encounter, and possibly the number of encounters in a day, roll more dice. if at least two show the same number, etc.
May 28, 2012 at 6:32 pm  Link
@drow – Is the chance of doubles on 2d6 the same as the 1/6 chance of rolling a given number on 1d6? That doesn’t sound quite right, but I don’t know how to test it.
I like the use of the double rolled to determine the time slot, though — that’s a neat trick!
May 28, 2012 at 8:56 pm  Link
@martin yeah, it’s 1/6 exactly. there are 36 possible combinations for 2d6, of equal probability, and 6 doubles (1,1, 2,2, 3,3, 4,4, 5,5, 6,6). 6/36 = 1/6. you can also think of the dice separately. no matter what the first number is, the odds of the second die being that number is 1/6.
May 28, 2012 at 10:35 pm  Link
@drow – Neat! I can wrap my head around that. How does it scale for multiple encounters, then? Adding a third die ups the chance two will match, so would a second encounter occur if all three matched?
May 30, 2012 at 3:28 pm  Link
i put together some code to crunch numbers… provided a set of dice (d6), rolled once per day, an encounter occurs at a particular time of day (numbered 1 through 6) if at least two dice show that number.
for 2d6…
0 encounters … 30 in 36 … 83.3 %
1 encounter … 6 in 36 … 16.7 %
for 3d6…
0 encounters … 120 in 216 … 55.6 %
1 encounter … 96 in 216 … 44.4 %
for 4d6…
0 encounters … 360 in 1296 … 27.8 %
1 encounter … 846 in 1296 … 65.3 %
2 encounters … 90 in 1296 … 6.9 %
for 5d6…
0 encounters … 720 in 7776 … 9.3 %
1 encounter … 4956 in 7776 … 63.7 %
2 encounters … 2100 in 7776 … 27.0 %
for 6d6…
0 encounters … 720 in 46656 … 1.5 %
1 encounter … 19986 in 46656 … 42.8 %
2 encounters … 24150 in 46656 … 51.8 %
3 encounters … 1800 in 46656 … 3.9 %
for 7d6…
1 encounter … 55656 in 279936 … 19.9 %
2 encounters … 173880 in 279936 … 62.1 %
3 encounters … 50400 in 279936 … 18.0 %
May 31, 2012 at 9:55 pm  Link
Okay, that’s pretty awesome. I love the idea of rolling 4d6 and looking for pairs, and of having a randomlydetermined time of occurrence as part of the mix. Rolling dice is inherently fun, so rolling more dice within reason is more fun.
This sounds like more fun than my method!
June 12, 2012 at 1:37 pm  Link
I really like this method, Martin. I’m recommending it to my newsletter subscribers (as well as combining it with an AD&Dstyle d8+d12 encounter table). Kudos!
June 13, 2012 at 9:30 pm  Link
@Mark Chance – Right on!