Random Encounter maths

[Boci]

So my players had to find their way through some relatively unknown mountains. There's no map or chart and they couldn't find the guide, the enemy had managed to kidnap them. To figure out how they do this, I assembled a random encounter table of 19 encounters (1d8+1d12). Most were simple random encounters, combat, obstacles, and other noteworthy thing, with 20 being the pass through the mountains, and 2 and 3 being a local tribe that knew about the pass and could lead them there.

They've been fumbling through the mountains for several sessions now, I don't necessarily think this is a problem, its an uncharted mountain range, getting through it without significant magic should be a chore, but I am wondering, mathematically, what would be the expected number of encounters before they find their way through? Is it just as simple as the chance to roll 2, 3 or 20 on a d8+1d12, which I think would be about 4.1%. So you'd expect about 12 encounters before they found the way through? Also each encounter could only occur once, so repeats were rerolled.

[Mastikator]

There is a 1.04% chance they'll roll 2 or 20. The odds of rolling 20 is greater than 50% if you roll 67 times. (and less than 50% if you roll 66 times)

The math for 2 and 20 is the same 1/(8*12)

You could add clues/navigational information that improve their odds of navigating.

[Boci]

Okay but they don't need to roll a 20, they need to roll a 20, a 2 or a 3. Any of those of those three results gets them through the mountain

Also I don't think you're accounting for the no duplicates in your maths.

[pragma]

https://anydice.com/program/2532

You have about a 4% chance of clearing the mountains on any roll.

If you don't trust any dice, note that there are 8*12=96 possible results on d8+d12, and for ways to roll 2, 3 or 12. 4/96 is a bit more than 4%

Repeated trials are described by something more favorable than a binomial distribution (which doesn't account for your rerolls). Here's a calculator for binomial distributions: https://homepage.divms.uiowa.edu/~mbognar/applets/bin.html

After 12 rolls on the table your team has a 50% chance of having an escape per the binomial calculator. Their chances are really greater than that because of rerolls. They have 100% chance of exiting after 16 rolls because you'll have seen every encounter in the table.

I can't do this in my head, but you can calculate the improved probability per roll by taking (number is ways to roll 2,3 or 20, which is 4l/(total number of allowed results), which will require a bit of mathematical footwork.

[bayjorntreeroot]

It sounds like this might be getting tedious for your group, especially since they don't seem to have a way to influence what happens other than to roll randomly, especially since the encounters don't push them closer to their goal. I would generally allow wisdom modifiers to help guide them, add bonuses for each completed encounter, or some other similar way that they can improve the overall situation. Lacking that, "fumbling for several sessions" sounds... not fun. Assuming they don't know what the rolls mean I would just make whatever next session's roll lead to the local tribe encounter. (Unless it's a 20).

[Telok]

Well, it depends on what you roll. Most dice rollers can't do what you need.

2-20 range weighted towards 11, success on 2, 3, 20, rerolling previously rolled numbers. Chances straight on the first try are 4/96

So it depends. If you whack out a 4 the first roll your success chance goes to 4/93 (if the in-head math is right). But if your first roll is 11 with it's... 10? die combos you drop to 4/86 after the first roll.

I'm sure this is calculable, but my personal approach would be a quick programming script to run the die results across an array. I'm busy today or I'd do it now. If nighttime allows and there's no decent answer then I'll see about a quick & dirty script answer.

Edit: total butt pull here, guesstimate by laying out your roll matrix, count off as if you rolled each large chance number in sequence (11, 10, 12, 9, 13, etc.), stopping when you get down to a 4/8 or less, and take that number of rolls as an answer. Now I'm curious and if free time will check how close that gets, likely in the next 3-5 days.

[Telok]

Ok, there's a 4% success rate for each of the first about 5 rolls, a 5% success rate for rolls 6-9, a 6% success rate for 11 & 12, 7% for roll 14, 8% for rolls 15 & 16, then another 7% for roll number 17. Your 50/50 is about 10 or 11, depending on how you like rounding. The 75% point is roll 14

TLDR: not much different from looking for a 20 on a d20. Suggest: drop the d12 to a d6 & just to get 8%-10% range, midpoint at roll 6, max 11 rolls.

[Sorinth]

I don't think you've helped yourself here, a simple d20 roll would have given your players a much better chance of getting out. And I'd argue a Survival check(s) would've been even better since they have a bit more control on the outcome. You could for example have required 3 successful DC 20 checks to find the pass where a natural 20 in either of the first 2 checks be meeting the helpful tribes. That way the players should at least feel like they are making progress, where the presumably hidden d8+d12 gives them no sense that they are making headway.

And there's also the argument that so long as they stick with a direction they should get out of the mountain range eventually regardless of finding the proper mountain pass or helpful tribe.


In cases like these I find it better to just set a number of miles they have to travel and just use the travel rules more or less as is. When picking the number you don't need to use the exact distance on the map because when travelling through the mountains you will often end up following a winding trail that leads to somewhere impassible forcing you to double back and take a different route. This method works well with the getting lost rules where it costs 1d6 hours of travel time, and the travel pace has a direct impact too.

[Witty Username]

Okay but they don't need to roll a 20, they need to roll a 20, a 2 or a 3. Any of those of those three results gets them through the mountain

Also I don't think you're accounting for the no duplicates in your maths.


That is easy, add the chance for 2 and 20 together, which gets you 2.08%, and then 3 which has 2 possible results which is equal to 2 and 20 together. So 4.16%

As for no duplicates the math goes something like
4/((8×12) - X) where X is the number of previous rolls.

The worst case is after 93 rolls, as that will clear all fail results then a pass if you need big O value.

Using a d20 seems prudent 20 as pass 1 and 2 as safety, it also cuts mass rolling.

Or use 20 cards from a deck, as that will speed up the random.

[Boci]

Had a session on Monday, they finally made it out of the mountains. I asked them if they enjoyed it and they said they did, as each encounter was fun, so it didn't matter as much that actual progress was hard to track. Likely helped that in one of the encounters they found 3 magic items.

As encounters starting being used, and so where removed from the pool, I did drop the dice size. 2d6 at 11 encounters, and 2d6 at 7 encounters. I don't know if this changed the numbers, or just sped things up. It likely had some effect on the probability, but not too much (I think). In any case the players needed 14 encounters to get through the mountains, skipping 5. So it sounds like they did get a bit unlikely, but not that much.


And there's also the argument that so long as they stick with a direction they should get out of the mountain range eventually regardless of finding the proper mountain pass or helpful tribe.

In other terrain maybe, but not really in mountains with no roads, bridges, or any serious magic.