Maths problem: fatigue vs critical hits

[DonCorazon]

And now in English?

[cleveland]

DonCorazon said:
And now in English?

English? I didn't do very well in English...

[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.

[Carkaton]

Very useful, thanks.