Maths problem: fatigue vs critical hits

[lch]

lch said:
Now with two dice, I'm not sure but I think it goes like this: You want a combination of two rolls that add up to X, so X = Y+(X-Y) and 0 < Y < X. Your probability is given as follows: For every Y from 1 to X-1, multiply P(Y) and P(X-Y) as given above. Add all of them together, divide by X-1. That's your probability for two rolls. You can simplify it into a closed form, which could be kinda tricky because of the modulo representation for Y and X-Y. Maple helps.

Okay, the second part was pretty bogus. It shows that I never really learned probability calculus. Another try:

X = 6a + b, 0 <= b < 6. Then the probability that a DRN gives X or higher on a roll is P(">= X") = (7-b)(1/6)^(a+1).
The probability that it equals X is P("= X") = (1/6)^(a+1).

But measuring a two-die roll isn't as easy. Here's a small python program that I used to check cleveland's calculations:

Code:

---

#!/usr/bin/env python



def p_1(x):

	if x < 1:

		return 1

	b = x % 6

	a = (x-b)/6

	return (7-b)*pow((float(1)/6), a+1)



def p_2(x):

	if x < 1:

		return 0

	b = x % 6

	a = (x-b)/6

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



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



def doubledrn_2(x):

	s = p_1(x)

	for y in range(1,x-1):

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

	return s



for x in range(2,18):

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

---

And here's the output of it: 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 : 0.119599

>= 14 : 0.079475

>= 15 : 0.046296

>= 16 : 0.020062

>= 17 : 0.000772

---

So as far as I can tell, cleveland's table is right. The difference in later numbers probably comes from rounding precision being a ***** again, as my code probably isn't numerically stable. If I'd calculate greater values, the numbers even turn negative! I'd be interested what formulas cleveland used to be able to trick the rounding precision.