Maths problem: fatigue vs critical hits

[cleveland]

lch said:
If I'd calculate greater values, the numbers even turn negative!

Negative probabilities? Sounds like my odds the last time I was in Vegas.

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.

2) Constructed a vector B(j) of the probabilities a second drn would be less than or equal to j. So:
B(1)=A(1)
B(2)=A(1)+A(2)
B(3)=A(1)+A(2)+A(3)
Etc.

3) Constructed a matrix M such that M(i,j) = A(i)*B(j). Therefore, each M(i,j) = P(X<=i+j | drn1=i).

4) Summing the / diagonals of this matrix (i.e. all M(i,j) whose i+j are identical) gives the probability that the 2d6oe will be less than or equal to i+j.

So for example, lets take the case when we want the 2d6oe <= 3. This is only possible 2 ways: if drn1=1 & drn2<=2, or if drn1=2 & drn2=1.

We therefore have:
A(1)=1/6
A(2)=1/6

B(1)=1/6
B(2)=2/6

M(1,2)=(1/6)*(1/6)
M(2,1)=(1/6)*(2/6)

So M(1,2)+M(2,1)= 0.083

Of course, this summation becomes larger for larger X=i+j.
When X=7, you'll need to add M(1,6)+M(2,5)+M(3,4)+...+M(6,1).

Phew.