@Xirema here's one for you rpg.stackexchange.com/questions/156379/…

I'm trying to figure out hot to express the probability of X in the system of 4d6 drop lowest reroll lower than 8.

Thus far, I've got a piecewise function that just flattens probability to 0 for results 1 - 7

@GcL I don't know how to write an AnyDice function to do this, but I could definitely feed this into my probability calculator.

I'm writing the anydice code. F'ing looping using only recursion reminds me of university

I started by trying to sort out the probability function expression but didn't get further than a piecewise function that was kind of crude.

Hmm, looks like I need to add a few new functions to my calculator for this.

FML... f'ing max recursion depth

Yeah, anydice just isn't the place for that.

Without an actual while loop, I can't get that to work.

Stats for Roll [4d6 Drop 1T]
╔═════════╤══════════╤══════════╤═══════════╗
║ Outcome │     Odds │   Trials │ Pass Odds ║
╟─────────┼──────────┼──────────┼───────────╢
║      +8 │   5.074% │   31/611 │  100.000% ║
║      +9 │   7.447% │     7/94 │   94.926% ║
║     +10 │   9.984% │   61/611 │   87.480% ║
║     +11 │  12.111% │   74/611 │   77.496% ║
║     +12 │  13.666% │ 167/1222 │   65.385% ║
║     +13 │  14.075% │   86/611 │   51.718% ║
║     +14 │  13.093% │   80/611 │   37.643% ║
║     +15 │  10.720% │ 131/1222 │   24.550% ║

Does that jive with your calculations?

It's surprisingly easy to calculate: you just generate a spread for 4d6 Drop 1, just remove any outcomes that are being rerolled, and then just normalize all the probabilities so they all add up to 1.

how do you do the scrap if more than two 15+

@goodguy5 Well, the actual rule is scrap if there's not at least two 15+. And it's the same method, just with a much much larger dataset.

I set my calculator to run about a minute ago, we'll see if it finishes in a reasonable span of time. Some of the steps took far less time than I was expecting them to.

But actually performing the truncation is (apparently) the slowest part of the process.

@Xirema my mistake, but the same in effort to implement

Although I kind of shot myself in the foot a bit; I could have implemented this as a O(N) function and I think it implemented it O(NlogN). So it's definitely taking longer than it needs to.

Oh, good. OutOfMemoryError.

Never mind then. XD

@goodguy5 I mean, that's what I tried to do, but I keep running out of heap space.

There's just too many outcomes.

............ Although....

Hmm.

I have an idea.

╔══════════════════════════╤══════════╤═══════════════════════════════════╗
║                  Outcome │     Odds │                            Trials ║
╟──────────────────────────┼──────────┼───────────────────────────────────╢
║     [15, 15, 8, 8, 8, 8] │   0.000% │      15848543881/6317769072690000 ║
║     [16, 15, 8, 8, 8, 8] │   0.000% │       5686118797/1579442268172500 ║
║     [17, 15, 8, 8, 8, 8] │   0.000% │         362943753/175493585352500 ║
║     [18, 15, 8, 8, 8, 8] │   0.000% │        846868757/1052961512115000 ║

Well APPARENTLY Stack Exchange didn't stop me from copying that giant block of text. =D

can you sort the lists and collapse them? so that 10,9,10 and 10,10,9 are the same?

I wonder how many rolls it takes, on average to get 6 stats with the colvillle method

@goodguy5 It's already done that, it reduced the whole thing down to only 5,236 unique sets of rolls.

@Xirema oh derp

Okay, a few more optimizations, and now it calculates this array in 0.4 seconds instead of 10 seconds. I'm happy with it now.