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    Barbarian in the Playground
     
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    Default Random Encounter Chance (maths/probability question)

    As a DM, I do enjoy giving my party the occasional Random Encounter. I have made tables of 100 encounters for city/road/campsite, and if it's time for an encounter then I roll on those tables.

    So in order to determine IF an encounter takes place, I have come up with the following method:

    I roll 2 d20s. One green, one blue. If the green one rolls higher than the blue one, an encounter will happen. If the green one is lower, or if there's a tie, then nothing happens.

    Now, my question is what the odds are in this particular situation. I'm particularly keen on knowing, because I don't want to have the odds skewed heavily to one side. Problem is, while I do remember some probability calculations from secondary school, I just can't figure out how to deal with the fact that one has to be higher than the other. I'm stumped.

    Could anyone figure out what the % chance is to get a Random Encounter when determined this way?
    Last edited by Maelynn; 2019-03-28 at 05:52 AM.
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    Default Re: Random Encounter Chance (maths/probability question)

    There are 400 total possible dice rolls (20^2) since the dice are distinguishable.

    Given the value on the blue die, here are the number of ways you can get an encounter:

    Blue : Rolls that trigger an encounter
    1 : 19
    2 : 18
    3 : 17
    ...
    20 : 0

    This gives the general pattern x = sum(20-n) from n = 1 to n = 20 and a decimal probability of x/400.

    x evaluates to 190 (for this case), so the overall probability of an encounter is 47.5%.

    One thing to worry about as well is not only the frequency per roll, but the expected number of encounters per session or per in-game day. This is the product of the per-roll frequency (0.475) and the average number of rolls per unit time. Even with a low per-roll frequency, if you're rolling a lot you're going to have lots of encounters. Which may or may not be good, depending on your desires.
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    Default Re: Random Encounter Chance (maths/probability question)

    Most of your explanation went over my head... I guess my knowledge is even worse than I had estimated.

    Thanks for taking the time to do the calculations for me. :)

    Quote Originally Posted by PhoenixPhyre View Post
    One thing to worry about as well is not only the frequency per roll, but the expected number of encounters per session or per in-game day. This is the product of the per-roll frequency (0.475) and the average number of rolls per unit time. Even with a low per-roll frequency, if you're rolling a lot you're going to have lots of encounters. Which may or may not be good, depending on your desires.
    Good point, and it's the most important reason why I wanted to know the odds. I do want the party to have the occasional RE, but not too often. So if this method were to give a high probability, then I would've looked into a different one. Probably a slightly more boring one, like 'roll a 1 on a d8'. I do favour rolling to see if an RE happens, because I like the idea of 'fate' deciding rather than the DM. Or attribute it to Tymora, haha.

    I suppose a roughly 50-50 chance is a tad on the high side, but personally I feel it's not enough to make it skewed. I might decide to put a few measures in place - like forego the roll if they had an encounter the previous time. That would depend on how it goes. So far the amount of RE's has been balanced.
    Last edited by Maelynn; 2019-03-28 at 08:42 AM.
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    Default Re: Random Encounter Chance (maths/probability question)

    Quote Originally Posted by Maelynn View Post
    Most of your explanation went over my head... I guess my knowledge is even worse than I had estimated.

    Thanks for taking the time to do the calculations for me. :)
    Maybe I can explain the same concept better?

    So we know that the probability of rolling any given number on a d6 is 1/6, right? Because 1 is the number of rolls that 'event' can happen on, and 6 is the number of possible rolls total. With 2d20, the number of possible rolls is 20 x 20 = 400 (for every roll on one die, there's a possible outcome for each of all the rolls on the other). So our probability is ?/400. That is the ?, is our question.

    We can divide the possible outcomes into subsets. So we look at the blue die. If the blue die rolls a 1, how many ways does the event (green die rolling higher) happen? Well, the green die rolling a 2 or higher, so 19 ways. If the blue die rolls a 2, we have 18 ways (3 and higher on the green), and so on, all the way up to asking if the blue die rolls a 20 (zero ways our event can happen, because nothing is higher). We add all these together, we get:

    19+18+17+16+15+14+13+12+11+10+9+8+7+6+5+4+3+2+1+0 = 190

    So our probability is 190/400.
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    Quote Originally Posted by frogglesmash View Post
    I guess I'll amend my original statement and instead say that Pathfinder is close enough to 3.5 to spark an argument about how close it actually is.

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    Default Re: Random Encounter Chance (maths/probability question)

    Thank you very much, The Kool, for the simplistic layman version. I understand perfectly now. ^^
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    Default Re: Random Encounter Chance (maths/probability question)

    If you want to use your two different dice compare higher/lower method, but to adjust the frequency of random encounters lower, you could change the kinds of dice you roll. This is because you have no encounter if the numbers are the same, and if you roll dice with fewer different sides you'll get the same thing more often.

    For example, if you used 2d4, here is the complete sample space:

    Green 1 Green 2 Green 3 Green 4
    Blue 1 Green and Blue the same Green higher. Encounter! Green higher. Encounter! Green higher. Encounter!
    Blue 2 Blue higher. No encounter. Green and Blue the same Green higher. Encounter! Green higher. Encounter!
    Blue 3 Blue higher. No encounter. Blue higher. No encounter. Green and Blue the same Green higher. Encounter!
    Blue 4 Blue higher. No encounter. Blue higher. No encounter. Blue higher. No encounter. Green and Blue the same

    Since there are 16 outcomes and 6 of them are "encounter", you would have an encounter probability of 6/16 or 0.375. (The "green is lower" outcome is also 6/16, and the "both are the same" outcome is 4/16.)

    Compare that to the same thing with 2d6:

    Green 1 Green 2 Green 3 Green 4 Green 5 Green 6
    Blue 1 Green and Blue the same Green higher. Encounter! Green higher. Encounter! Green higher. Encounter! Green higher. Encounter! Green higher. Encounter!
    Blue 2 Blue higher. No encounter. Green and Blue the same Green higher. Encounter! Green higher. Encounter! Green higher. Encounter! Green higher. Encounter!
    Blue 3 Blue higher. No encounter. Blue higher. No encounter. Green and Blue the same Green higher. Encounter! Green higher. Encounter! Green higher. Encounter!
    Blue 4 Blue higher. No encounter. Blue higher. No encounter. Blue higher. No encounter. Green and Blue the same Green higher. Encounter! Green higher. Encounter!
    Blue 5 Blue higher. No encounter. Blue higher. No encounter. Blue higher. No encounter. Blue higher. No encounter. Green and Blue the same Green higher. Encounter!
    Blue 6 Blue higher. No encounter. Blue higher. No encounter. Blue higher. No encounter. Blue higher. No encounter. Blue higher. No encounter. Green and Blue the same

    There are now 36 outcomes and 15 of them are "encounter", so the probability of an encounter is 15/36 or about 0.416667. Because the "diagonal" of "both dice are the same" takes up less and less proportional space as the number of die faces increases, this "what to do when there's a tie" part of your system has less and less impact as you use dice with more ways to not tie.

    (Math nerd comment: this is because the number of ways to tie when rolling two dice and looking at which is higher increases linearly with the number of die faces, and the number of total possibilities increases by a power of 2 because you're rolling two dice. So, for 2d4, there are 4 ways to tie and 42=16 total possibilities, but for 2d20 there are 20 ways to tie and 202=400 total possibilities, so the ties are really swamped by the non-ties.)

    In general, if you're trying to compare two events (such as rolling two dice) and you're having a trouble getting a feel for how common each possible outcome is, making a table like that can be a good way to get a handle on it. There are more efficient ways to figure out the probabilities since making tables is time-consuming, but if you need to get a clearer picture of what is going on it can help you visualize things and see patterns.

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    Default Re: Random Encounter Chance (maths/probability question)

    Alternately, if you want to be flexible and have varied probability of encounter based on the situation, make yourself a table of probabilities. "Okay, on this road it's fairly well traveled and protected, so there's a 12% chance of an encounter each night and only a 7% chance of encounter during each day." Then grab those faithful underused d% dice. Roll that number or lower, congrats you have an encounter. PCs flaunting wealth? Increase the odds. PCs being intimidating? Decrease the odds. The distinct advantage to this method is the blatant transparency. You can directly see and control the probability of having an encounter, without having to do maths.
    Last edited by The Kool; 2019-03-28 at 03:22 PM.
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    Quote Originally Posted by frogglesmash View Post
    I guess I'll amend my original statement and instead say that Pathfinder is close enough to 3.5 to spark an argument about how close it actually is.

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    Default Re: Random Encounter Chance (maths/probability question)

    Quote Originally Posted by Algeh View Post
    If you want to use your two different dice compare higher/lower method, but to adjust the frequency of random encounters lower, you could change the kinds of dice you roll. This is because you have no encounter if the numbers are the same, and if you roll dice with fewer different sides you'll get the same thing more often.
    Or you could mix die types to add rows of blue or columns of green. The number of ties and smaller die winning are the same.
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    In general for a N*M throw, with N<M
    You have N*M total possible outcomes
    N draws (along a diagonal)
    Given you throw 2 die. You could probably get some more from the two die.
    You have N-1 cases where the smaller die is 1 higher (a routine encounter)
    N-2 cases where the die is 2 higher
    ...
    1 cases where the die is N-1 higher (a special encounter)
    [The total number of these wins is N*(N-1)/2 or N*(N+1)/2 if you count draws]
    The other cases total is the difference or equivalently N*(M-(N+1)/2)

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    Default Re: Random Encounter Chance (maths/probability question)

    Why do it so complicated? Just decide on the chance you want for a random encounter and roll a d10. Ex.; 30 % chance - 1-3 gives an encounter. Use a d20 if you want finer 5 % increments.

    Rolling multiple dice just obfuscate what you are actually doing, and is just unneccesary IMO. Deciding on the rolling scheme first and afterwards figuring out the probabilites seems backwards.

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    Default Re: Random Encounter Chance (maths/probability question)

    Quote Originally Posted by Pelle View Post
    Why do it so complicated?
    Because I can. I like coming up with ways to do things differently, because it gives me pleasure to create something myself - whether it's a quest, a magic item, or a method for rolling random encounters.

    Quote Originally Posted by Pelle View Post
    Rolling multiple dice just obfuscate what you are actually doing, and is just unneccesary IMO. Deciding on the rolling scheme first and afterwards figuring out the probabilites seems backwards.
    'Unnecessary' and 'backwards'. Thank you for being so derogatory about something I created.

    I came up with this method during a session (I was playing with 2 d20s in my hand) and I liked it. It's different from the boring d%. However, after having used it a few times I now wanted to know if it's a somewhat reliable method to determine odds (and doesn't turn out to have a 75% chance of an encounter happening). It's called tweaking your initial idea, adjusting it where necessary. Part of a creative process.
    Last edited by Maelynn; 2019-03-29 at 07:34 AM.
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    Default Re: Random Encounter Chance (maths/probability question)

    Negative, fair enough, but derogatory? Come on. Anyways, it was more aimed at the other posters expanding on it than the initial post.

    Multiple dice and complex procedures is fine if it achieves something special though, like a non-linear progression of the probabilites etc...

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    Default Re: Random Encounter Chance (maths/probability question)

    "Which die rolls higher" is (mostly) just a fancy way to flip a coin. (Yeah, it's not quite 50% because of the ties....)
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    Default Re: Random Encounter Chance (maths/probability question)

    Quote Originally Posted by Maelynn View Post
    Because I can. I like coming up with ways to do things differently, because it gives me pleasure to create something myself - whether it's a quest, a magic item, or a method for rolling random encounters.

    ...

    I came up with this method during a session (I was playing with 2 d20s in my hand) and I liked it. It's different from the boring d%. However, after having used it a few times I now wanted to know if it's a somewhat reliable method to determine odds (and doesn't turn out to have a 75% chance of an encounter happening). It's called tweaking your initial idea, adjusting it where necessary. Part of a creative process.
    That's fair. I'm a bit the same way myself. I like to fiddle with various dice. Go to anydice and play around, it'll give you probabilities and you can easily check a variety of options against each other. I remember one time wondering if it would be beneficial to replace the standard d20 roll with d8+d12... I think I determined I liked the probabilities on d4+d16 better but who has a d16 lying around? I think I have a d14 somewhere though...

    Oh, here's an idea! If you want the probability to be slim, grab a pair of dice and roll them together. If they match, encounter. If you want slimmer chances, roll bigger dice. The neat thing about this is that if you use a pair of dice the exact same size, the probability of them matching is exactly the same as, say, the probability of rolling a 1 on one of them. (For example, for each number on the green die, only one number on the blue die triggers an encounter. So the probability, with d20s, is 20/400... or 1/20.) This makes it similarly transparent and easy to control, but gives a neat rolling mechanic. Chances go as high as 25% for a pair of d4s, or as low as 5% for a pair of d20s.
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    Quote Originally Posted by frogglesmash View Post
    I guess I'll amend my original statement and instead say that Pathfinder is close enough to 3.5 to spark an argument about how close it actually is.

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    Default Re: Random Encounter Chance (maths/probability question)

    The advantage of 2d20 is that it's basically an opposed roll of the type that already occurs throughout the game. You might as well take that idea and run with it, expanding it to cover (say) time pressure, similar to the time pool system. For example, you can add a stacking "danger bonus" to the green die (the danger roll) every time the party does something alarming, or just in very dangerous areas (and danger penalties in particularly safe areas). And while you're at it, you might want to give the party some impact on the rolls via the blue die (the protect-against-danger roll). For example, a party might get a safety bonus/danger penalty based on marching order, scouting abilities, or their reputation in the area.
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    Default Re: Random Encounter Chance (maths/probability question)

    I think a great method to check for encounters is to have one happen on the roll of a 1. If the current area is very dangerous, you roll a d4, which means you have a 1 in 4 chance (25%) for an encounter. If the current area is not very dangerous, you could roll a d8, which means you have a 1 in 8 chance (12.5%) of an encounter. If it is quite safe, you roll a d12, for a 1 in 12 chance (8%) of an encounter.

    The other important factor is how often you want to roll. I like every 10 minutes in a dungeon and every 6 hours outdoors.
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    Default Re: Random Encounter Chance (maths/probability question)

    Quote Originally Posted by Yora View Post
    I like every 10 minutes in a dungeon and every 6 hours outdoors.
    See, this I can get behind. Dungeons are far too often written with 'roll every hour', and I don't know about you but my PCs can totally blow through an entire dungeon in half that time.
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    Quote Originally Posted by frogglesmash View Post
    I guess I'll amend my original statement and instead say that Pathfinder is close enough to 3.5 to spark an argument about how close it actually is.

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    Default Re: Random Encounter Chance (maths/probability question)

    I recommend use of the following two tables for random encounters (unmodified d20).


    When the game is progressing smoothly, and conversations are short and/or productive:
    1-20: No encounter
    21+: Encounter

    When the game is bogged down, or players are arguing uselessly:
    1-20: Encounter
    21+: No encounter

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    Default Re: Random Encounter Chance (maths/probability question)

    Quote Originally Posted by Pelle View Post
    Negative, fair enough, but derogatory? Come on.
    It came across like that to me. Glad I was wrong and read more into it than you meant. :)

    Quote Originally Posted by The Kool View Post
    If you want the probability to be slim, grab a pair of dice and roll them together. If they match, encounter. If you want slimmer chances, roll bigger dice. [...] Chances go as high as 25% for a pair of d4s, or as low as 5% for a pair of d20s.
    Hm, that's also a nice one. Gives something more of a gambling feel. I also like how you can modify the chances by using different dice, hadn't thought of that yet - guess it works that way with my method as well.

    What I also like to do, as a little inside joke in our group, is have a player roll a die (any will do) and say that the 'odds are in their favour' - meaning that an odd number will get them the result they want. This started with a little abandoned shrine to Tymora they encountered as an RE, which had that line inscribed into the stone. After placing an offering, they got to roll a d6 for the outcome - the even numbers had a negative outcome, the odd numbers a positive one. They liked it, especially one player who had chosen her as his deity, so now if they need to 'flip a coin' for something I use this method.

    Quote Originally Posted by Yora View Post
    I like every 10 minutes in a dungeon and every 6 hours outdoors.
    Do you mean this as in-game hours? I think 6 hours would be too long in my campaign, because my players have their quests quite close to the city and go to places that require little travelling. I think right now the frequency is about 1 RE per 2-3 sessions. Depends also on how much of the session is spent on travelling, of course.

    What may also influence the decision to roll, is what kind of encounter you get. The lists I use are far more than just combat encounters - they can also be NPCs they run into or observe from a distance, or something that tells them something has recently happened (like a burnt cart with a dead merchant). I wanted encounters to give flavour as well, to make them aware there's other people on the roads they travel and the city they walk in. This could justify rolling more often, as only a small part is actual combat.
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    Default Re: Random Encounter Chance (maths/probability question)

    Quote Originally Posted by The Kool View Post
    Alternately, if you want to be flexible and have varied probability of encounter based on the situation, make yourself a table of probabilities. "Okay, on this road it's fairly well traveled and protected, so there's a 12% chance of an encounter each night and only a 7% chance of encounter during each day." Then grab those faithful underused d% dice. Roll that number or lower, congrats you have an encounter. PCs flaunting wealth? Increase the odds. PCs being intimidating? Decrease the odds. The distinct advantage to this method is the blatant transparency. You can directly see and control the probability of having an encounter, without having to do maths.
    While those are good percentages for the game, because we actually WANT random encounters, I have to point out that from a realistic perspective, if traveling one day on a "well traveled and protected" road still results in almost 1 chance in 5 of assault and possible death, it depicts a pretty grim world. I wonder how merchants survive on longer trips.
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    Default Re: Random Encounter Chance (maths/probability question)

    Quote Originally Posted by King of Nowere View Post
    While those are good percentages for the game, because we actually WANT random encounters, I have to point out that from a realistic perspective, if traveling one day on a "well traveled and protected" road still results in almost 1 chance in 5 of assault and possible death, it depicts a pretty grim world. I wonder how merchants survive on longer trips.
    The curiosity that is the Story Worth Telling tells us that interesting things are going to happen to our party more often, because the party we're interested in hearing about is the one all the things happen to. It need not be a realistic representation of the world. If you ascribe to that view, I find it doesn't strain suspension of disbelief.
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    Quote Originally Posted by frogglesmash View Post
    I guess I'll amend my original statement and instead say that Pathfinder is close enough to 3.5 to spark an argument about how close it actually is.

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