Ok, borrowing this from that last post about Grimwild, because this has been tingling my math brain.
I love looking at how the math of tabletop mechanics works out. Ever since I sat down with some graph paper in high school and histogrammed the difference between 3d6 and 3d6 drop 1.
So the first thing I thought of when I saw this was "well, given a pool of size N, how long would it take it to get to 0, on average?"
Each die rolls either high (4-6), or low (1-3). After you roll a pool, you throw away all your low dice. If all your dice roll high, you roll again with the same pool size. If all your dice roll low, you're done.
An individual die has a 50 percent chance of rolling high. It always gets used in at least one pool roll, and whether it survives to another depends on what value it rolls. However, if it does roll high, we are now in the same situation we started in: we have one die. So we can express e, the expected number of pool rolls an individual die would survive, using a recurrence relation: e = 1 + 0.5 e. Using some algebra gives us e = 2.
If we have a pool of size N, then we might be tempted to say we should also expect it to survive 2 pool rolls. After all, each individual die is expected to survive that long, and no die affects any other.
But the truth is more fun than that.
Let E(N) be the expected number of pool rolls we get out of a pool of size N. We know we don't get any more rolls if we have zero dice, so E(0) = 0. There's only one way to roll all highs, but there's N ways to roll one low and N-1 highs, since that could be any one of the N dice, and N(N-1)/2 ways of rolling two lows and N-2 highs, since that's how many different pairs of dice there are.
In general, there's N choose H ways of rolling H highs and N-H lows. That means the probability of any particular combination of highs and lows is (N choose L)(probability of H dice all rolling highs)(probability of N-H dice all rolling lows). The probability of K dice all rolling the same value is 0.5 ** K, so the probability of rolling exactly H highs with N dice works out to (N choose H)(0.5 ** H)(0.5 ** (N-H)) which simplifies to just (N choose H)(0.5 ** N).
Now we can get a recurrence relation for E(N) in terms of the probability of our next pool having H dice and the expected number of rolls that pool has, E(H).
E(N) = 1 + Σ (N choose H) (0.5 ** N) E(H), 0 ≤ H ≤ N
This is great! But slightly unsatisfying as if we look at the right hand side closely, we can see that we're defining E(N) in terms of E(N).
E(N) = 1 + (0.5 ** N) (E(N) + E(0) + Σ (N choose H) E(H), 1 ≤ H < N)
Our friend algebra steps in again, and helpfully simplifies this to
E(N) = (2**N + Σ (N choose H) E(H), 1 ≤ H < N) / (2**N - 1)
And... this is where I'm stuck for now. I'd like to have a closed form for E(N), but I haven't figured it out yet.
But, it is good enough to tell me how many pool rolls the 4d/short, 6d/mid, and 8d/long pools should last:
E(1) = (2) / (2 - 1) = 2 / 1 = 2 E(2) = (4 + (2 choose 1) E(1)) / (4 - 1) = (4 + 2 x 2) / 3 = 8/3 E(3) = (8 + (3 choose 1) E(1) + (3 choose 2) E(2)) / (8 - 1) = (8 + 3 x 2 + 3 x 8/3) / 7 = 22/7 E(4) = (16 + (4 choose 1) E(1) + (4 choose 2) E(2) + (4 choose 3) E(3)) / (16 - 1) = (16 + 4 x 2 + 6 x 8/3 + 4 x 22/7) / 15 = 368/105 E(5) = (32 + (5 choose 1) E(1) + (5 choose 2) E(2) + (5 choose 3) E(3) + (5 choose 4) E(4)) / (32 - 1) = (32 + 5 x 2 + 10 x 8/3 + 10 x 22/7 + 5 x 368/105) / 31 = 2470/651 E(6) = (64 + (6 choose 1) E(1) + (6 choose 2) E(2) + (6 choose 3) E(3) + (6 choose 4) E(4) + (6 choose 5) E(5)) / (64 - 1) = (64 + 6 x 2 + 15 x 8/3 + 20 x 22/7 + 15 x 368/105 + 6 x 2460/651) / 63 = 7880/1953 E(7) = (128 + (7 choose 1) E(1) + (7 choose 2) E(2) + (7 choose 3) E(3) + (7 choose 4) E(4) + (7 choose 5) E(5) + (7 choose 6) E(6)) / (128 - 1) = (128 + 7 x 2 + 21 x 8/3 + 35 x 22/7 + 35 x 368/105 + 21 x 2460/651 + 7 x 7880/1953) / 127 = 150266/35433 E(8) = (256 + (8 choose 1) E(1) + (8 choose 2) E(2) + (8 choose 3) E(3) + (8 choose 4) E(4) + (8 choose 5) E(5) + (8 choose 6) E(6) + (8 choose 7) E(7)) / (256 - 1) = (256 + 8 x 2 + 28 x 8/3 + 56 x 22/7 + 70 x 368/105 + 56 x 2460/651 + 28 x 7880/1953 + 8 x 150266/35433) / 255 = 13303613/3011805
Which means that on average, we can expect a 4d pool to last E(4) = 368/105 ≈ 3.50 pool rolls, a 6d pool to last E(6) = 7880/1953 ≈ 4.03 pool rolls, and a 8d pool to last E(8) = 13303613/3011805 ≈ 4.42 pool rolls.
Now I really want to histogram them to see what the distributions look like, but it's late and I should go to bed.
