Maths problem: fatigue vs critical hits

[lch]

cleveland said:
Here's roughly what I did:
1) Constructed a vector A(i) of the explicit probabilities for each single drn outcome i. This is pretty straightforward:
A(1)=A(2)=...=A(5)=1/6
A(6)=A(7)=...=A(10)=(1/6)^2
Etc.

I think this is already where I fail, because the rest is pretty much the same. I was jumping back and forth between modulo 5 and modulo 6 just because the whole thing is offset by 1 and the gaps didn't work out. Let's correct this now...

Let X-1 = 5a + b, 0 <= b < 5. (notice the -1!!) Then the probability that a DRN gives exactly X is p(DRN = X) = (1/6)^(a+1).
The probability that it is X or greater is:


Now, we have two ways of measuring all the (x,y) in IN x IN which sum is equal to or greater than X: Either we take all the (x,y) with x+y < X and subtract the probability of them showing up simultaneously from 1, thus inverting the set, like you did if I understood right, or measure the following (with a loose mathematical notation): Code:

---

  (1, >= X-1)

  (2, >= X-2)

  (3, >= X-3)

   ... etc.

  (X-1, >= 1)

  (>= X, *)

---

which should be equal to {(x,y) | x+y >= X}. I implemented both versions in Python again: Code:

---

#!/usr/bin/env python



# p(DRN) >= x

def p_1(x):

	if x < 1:

		return 1

	b = (x-1) % 5

	a = (x-1) / 5

	return (6-b)*pow((float(1)/6), a+1)



# p(DRN) == x

def p_2(x):

	if x < 1:

		return 0

	b = (x-1) % 5

	a = (x-1) / 5

	return pow((float(1)/6), a+1)



# p(2d6oe) >= x

def doubledrn(x):

	M = [(a,b) for a in range(1,x-1) for b in range(1,x-a)]

	s = 0.0

	for (x_1, x_2) in M:

		s += p_2(x_1)*p_2(x_2)

	return 1-s



# p(2d6oe) >= x (alternate version)

def doubledrn_2(x):

	s = p_1(x)

	for y in range(1,x):

		s += p_2(y)*p_1(x-y)

	return s



for x in range(2,20):

	print ">= %2d : %f" % (x, doubledrn(x))

	print ">= %2d : %f" % (x, doubledrn_2(x))

---

and huzzah, they both give the same result, and it is: Code:

---

>=  2 : 1.000000

>=  3 : 0.972222

>=  4 : 0.916667

>=  5 : 0.833333

>=  6 : 0.722222

>=  7 : 0.583333

>=  8 : 0.462963

>=  9 : 0.361111

>= 10 : 0.277778

>= 11 : 0.212963

>= 12 : 0.166667

>= 13 : 0.127315

>= 14 : 0.094907

>= 15 : 0.069444

>= 16 : 0.050926

>= 17 : 0.039352

>= 18 : 0.029578

>= 19 : 0.021605

---

[lch]

lch said:
The probability that it is X or greater is: (...) (6-b)(1/6)^(a+1)

I just noticed that it is way easier to get the same result in a different way than I did. We already know that each element in a modulo "layer" shares the same probability, and that all the following layers together are cramped into something having the same probability, too. So you just need to count from the current element to the end and sum up.

[Pehmyt]

I don't know if you are still interested in this, but I was, as it says "maths problem" in the title For future reference, if nothing else, the actual probability distributions for 1d6oe and 2d6oe (=DRN in the manual) are:


and for the original question,

In all expressions X=5n+r, with 1<=r<=5 (note limits!).

The first one is more or less trivial, the second is proven by induction in n and the third follows from that by straightforward summation.

[Revolution]

I started reading all of this and thought of this picture...

[moderation]

So that is what kids are learning in school these days.

[Wick]

With only a little calculus long, long ago I can't play math, but I do enjoy watching it.

[lch]

Pehmyt said:

In all expressions X=5n+r, with 1<=r<=5 (note limits!).

With a little redesign, this translates into python code as follows: Code:

---

# p(2d6oe) >= x (Pehmyt's version)

def doubledrn_3(x):

	n = (x-1) / 5

	r = (x-1) % 5

	r += 1

	return float((62-15*r+r*r)*5*n+70+3*r-r*r)/(2*6**(n+2))

---

which produces the right output. Very nice! I think that if you lose the tricky quotient/remainder representation of yours and just work on x-1 like I did in my third attempt, your formula might look a little simpler than the current one.