Math help! Most likely number of times passing threshold

edited January 2016 in Game Design Help
I have a supervillain game where players earn points in various ways: some for being assigned tasks, some for volunteering for tasks, and many as fanmail.

There are two types of points: Story points and Fan points. If you get 4 FPs, you get a bonus scene for your villain. If you get 4 SPs, you get to direct a bonus scene for someone else's villain.

I have found that 3 bonus scenes is a good number per villain; 4 is okay too; 2 is okay, but not as good as 4; 5 is too many; 1 is too few. If the group wants a quicker game, then 4 is too many.

I have come up with various ways of putting points in circulation which I estimate at anywhere from 10 to 19 per player. However, when I look at those, I realize that I don't know how many bonus scenes to expect because of remainders.

Let's say the average player earns 14 total points. That's 3.5 scenes. But it isn't, right? We shouldn't actually expect that the most typical outcome (median or mode; I'm curious about both) for your villain is 2 bonus scenes you control and 1 or 2 bonus scenes someone else controls (or the reverse), right? Because with 14 points, you CAN'T get 4 scenes. In fact, if you have 11 SPs and 3 FPs (or the reverse), you could wind up with only 2!

Can anyone give me a formula that I can plug my expected points into that will turn out likely numbers of scenes?

Re: the tweaks I'm considering for what points to put into circulation, they include SPs, generic points that could become FPs or SPs at equal rates, and generic points that are likely to become FPs 2 to 3 times as often as SPs (just based on how people tend to award fanmail).



  • @Deliverator I know you haz the mad math skills...
  • edited January 2016
    Suppose you have X points, and each is SP with independent probability p. Already different than your actual game but let's go. :) Plug in expected points e.g. X=14 and probability a point is a SP e.g. p=0.5, get average scenes:*(choose(X,k)+*+p^k+*+(1-p)^(X-k)),+k=0..X)+at+X=14,+p=0.5

    Probability a point is a SP can be estimated as...

    A = SPs put into circulation
    B = generic becoming FP or SP at equal rate, assuming players can't strategically convert
    C = generic points becoming FPs 2-3x as often as SPs

    p = (A + B/2 + C/3.5) / (A + B + C)

    Though presumably different players will tend to get different types of points.
  • And actually the average total number of scenes for a player with X points is very close to (X-3)/4, so you can just use that. :D
  • I'd like to help, but I'm not sure I'm 100% clear on the problem. So maybe just clarifying it will help.

    Are you saying that X points are "in circulation", meaning "X points get awarded during the total run of a game"? Or do you mean something else? It could be interpreted a few different ways.

    After all, when the points get awarded also affects the math (if, for instance, half are distributed during the first phase play, and the second half during the second).

    One thing you can do by yourself is to just take the "remainders" (small numbers, after all) and write out all the possible combinations, to get a sense of the possible outcomes. (It's not that hard, since there aren't that many possible combinations.)

    For instance, how many different ways can you divide 5 points among 3 players?

    0,0,5 x3
    0,1,4 x6 (014 104 140 041 410 401)
    0,2,3 x6
    1,2,2 x3

    So, with five points in circulation, you have 18 possible distributions.

    Out of those:

    Half (9/18) mean no scenes at all.
    Half (9/18) mean one scene happens.

    It's not a solution to your problem (yet), but we can see that adding 5 points results in half a scene (50% possibility of one extra scene).

    This won't hold if players already have points to begin with; we'd have to estimate further. But if your method of distribution is fairly equilateral, you can very roughly estimate that 12 points + 5 points, for a total of 17, gives you 3 or 4 scenes total. Two scenes is unlikely, but possible (3,3,4,5), and five is clearly impossible.

    You might be able to get a decent estimate this way (possibly good enough for a playtest), until a proper calculation can be made by someone more math-savvy than me.

  • You rang? :)

    Simplest solution, it seems to me, would be to allow both rounding and horse-trading: players can give each other FPs and SPs on a 1-to-1 basis in order to make even groups of 4.

    Hmm, actually. Try this. There are only three possible remainders, right? 1, 2, or 3. If you have a remainder of 1 in one of your pools, it converts into the other type of point. (If you have remainders of 1 in both pools, you choose which converts into which.) If you have 2 or 3 remainder points, then you just round up.

    Not enough games use rounding correctly, in my opinion. They're all like "always round down" and I'm like motherfucker it's like 9/10 what the hell.
  • edited January 2016
    Or alternatively (perhaps too complex, but worth considering):

    Can you make up for being a point or two short with some kind of compromise?

    If you are missing one point, the player on your left gets to state one unpleasant detail for you to include, that kind of thing? If you're two points short, both the person on your left and the person on your right do?

    This allows you to "round up", as Matt wisely suggests, but still makes it *matter*, if that's something you want to matter.
  • edited January 2016
    I'm not trying to make up for being a point or two short, and I'm not trying to maximize the number of points that turn into scenes. I'm trying to take the point-generating processes which I have in place (which I like as is) and understand how many scenes they will likely produce, so that I can tweak them a little up or down if I want.

    My amateur math looks like "14 points = 3 or 4 scenes (50% chance of each)". I would prefer pro math that tells me a more accurate "14 points =". And, y'know "13 points =" and "15 points =", etc.

  • edited January 2016
    To calculate this, we need to know how many players there are, so we can figure various possible "remainders". Interesting problem!

    The more players, the more chances of not all the points generating "scenes".
  • 3 to 5 players. 2 scenes per player. Each scene, one player (V) has 1 point to give out, while the others have 2. There are also 2 auto points that other players earn for GMing and Distraction-authoring. My tweaks involve maybe taking V's point away and/or adding a 3rd auto point for a 2nd Distraction.

    I think all of those details are completely beside the point of my request here, though. I've already done the math on average points earned in each of these scenarios for each number of players, based on guesses about who tends to award points to V (in which case they become FP) vs others (in which case, SP). I have all the numbers. But my numbers don't factor in likelihoods of completed sets vs uncompleted sets.
  • edited January 2016
    Well, I only asked for the number of players...

    Do you have somewhat reliable numbers of minimum and maximum numbers of points handed out, as well? That will *almost* give you your answer.

    I actually wonder if a proper mathematical model would give you any better information than that. It might not...

  • I know exactly how many points are given out. I know the range of points any given players will likely receive (10-19). What I don't know is how many points, probabilistically, equals how many scenes, probabilistically.

    I figured there was a very simple answer to this, like "subtract .25" or "multiply by .75" to cover the issue of remainders. That's all I'm looking for here. But perhaps not.
  • edited January 2016
    ...okay, looks like if I just posit 18 points, go through all the FP/SP combos, count the number that result in 4 scenes (12), compare that rough avg scene outcome (3.7) to the simple division (18/4 = 4.5), then I have some ballpark idea of how much to discount my simple division by. Discounting 0.6 to 0.8 appears to cover the range pretty well.

    That may be good enough.
  • edited January 2016
    Players could be allowed to pool points together to make a shared scene. On the Joker's turn, he earns his third FP and asks Harley Quinn to invest 1 FP from her reserve to narrate a scene in which the two villains obtain something from Batman.
    The responsibility and character involvement should be, if possible, shared between all contributing players (e.g. the Joker is the main character of the scene, likely having a confrontation with Batman, and Harley Quinn is featured in the Joker's scene in a minor role).

    Leftover points at the end of the "regular" scenes could be assigned to other players who have more, spending them until no more than 3 points of each type are left in total.
    Example rule: sort the player by decreasing total points, then by decreasing highest number of leftover points in the two categories, then by other extremely important tie-breaking criteria.
    Then the top player does an extra scene, of the type that corresponds to the category in which he has the most points, paid with all his leftover points and some or all of the points of the appropriate category of the bottom player or players. Sort the players again and repeat until all the players together don't have enough points in any category.
  • David, here's the simple way to discount: Start with number of points. For example, 18. Subtract the average wasted points for each type of scene, which is the average of {0,1,2,3} or 1.5. Two types of scenes, so subtract twice, or in total subtract 3. Now we have 15. Divide by 4, getting 3.75. The number of scenes will be 3 (25%) or 4 (75%).

    Alternatively and equivalently, do the simple division and subtract 0.75.

    (X-3)/4, integer part is base number of scenes, fractional part is probability you'll get +1 scene.
  • edited January 2016
    That matches our shorthand calculations (from last night), thanks!

    Nice. (Although I'm not 100% sure about the probability part... the math makes sense, but doesn't match my brute force methods for some reason.)
  • The probability's not exact, but it is the best simple approximation you can get, I believe.
  • Glad to hear! My number was roughly 30-31% for the previous example (compared to your 25%), so I was curious what was going on.
  • Thanks, both of you!
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