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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).

Thanks!

## Comments

http://www.wolframalpha.com/input/?i=sum((floor(k/4)+floor((X-k)/4))*(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.

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.

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.

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.

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.

Thanks!

The more players, the more chances of not all the points generating "scenes".

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.

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 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.

That may be good enough.

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.

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.

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.)