Open Test..., A "mathematical" problem Options

[Prototype]

Due to the ridiculous level of variance in the open test I only use them for stress points pretty much, where the numbers getting ludicrous doesn't matter so much.

For stealth tests I have my players roll against base TN4 and add +1 to the Perception test TN for each success. Stops things getting too ludicrous (though 13 stealth dice and centering can really make it tough at times!)

[Gorath]

Okay, i have some ideas for those propabilities p(x,y) that someone (A) with x dice gets more or equal as someone (B) with y dice.

First i will calculate the probability p(1,1). We can assume that p(1,1)>0.5, because "tie goes to the attack" at stealth versus intelligence.

If x=1, the only success is for y=1.
If x=2, we need y=1,2
...
If x=5, y=1,2,3,4,5 is okay.
If x=6, we have a automatic success, if y is not 6. Else both roll 1d6 again.

==>

p(1,1) = 1/6*1/6 + 1/6*2/6 + 1/6*3/6 + 1/6*4/6 + 1/6*5/6 + 1/6*5/6 + 1/6*1/6*p(1,1)

Simplifing:

p(1,1)=20/36 + 1/36*p(1,1)

Solving for p(1,1):

p(1,1)=36/35*20/36=4/7 > 0.5

as we aspected before.

[Gorath]

Now i will develop a recursiv solution for p(2,1):

We must distinguish between some cases:

1. A has two 6s, B has no 6:
In this case A wins.

2. A has two 6s, B has one 6:
In this case we have the same situation as before, so p(2,1).

3. A has one 6, B has no 6:
A wins.

4. A has one 6, B has one 6:
We are in the situation p(1,1), because A can only roll 1 die.

5. A and B don't have any 6s:
In this case we must look, who gets teh higher result. The propability that A wins in this case is called R(2,1) and is calculated later...

If we stick all this informations together we get following equation:

p(2,1) =
B(2,2)*(1/6)^2*(5/6)^0 * B(1,0)*(1/6)^0*(5/6)^1 [1]
+ B(2,2)*(1/6)^2*(5/6)^0 * B(1,1)*(1/6)^1*(5/6)^0 * p(2,1) [2]
+ B(2,1)*(1/6)^1*(5/6)^1 * B(1,0)*(1/6)^0*(5/6)^1 [3]
+ B(2,1)*(1/6)^1*(5/6)^1 * B(1,1)*(1/6)^1*(5/6)^0 * p(1,1) [4]
+ B(2,0)*(1/6)^0*(5/6)^2 * B(1,0)*(1/6)^0*(5/6)^1 * R(2,1) [5]

If we simplify all those terms ( :D ) for x=2, y=1 we get:

p(2,1) = (1/6)^3 * ( 5 + p(2,1) + 50 + 10*p(1,1) + 5^3*R(2,1))

Now its time to give a formula for R(2,1). If you write down all combinations possible form (1,1,1) to (5,5,5) and look if one of the first to numbers is equal or higher then the last you get the possibility R(2,1).

If you check the old thread, you see, that we already had a solution to calculate how often 1,2,3,.. is the highest number in such a tuple.

k^n-(k-1)^n gives the number of tuples with the maximum number "k" when you roll "n" dice. I.e. if you roll two dice and want to know how many tuples have a maximum number of 2 you get: (1,2); (2,1) and (2,2). 2^2-1^2=3

So we can calculate R(2,1):

In 1^2-0^2=1 cases "1" is the maximum for A, with p=1/5 A wins ( if B has a 1 too).
In 2^2-1^2=3 cases "2" is the maximum for A, with p=2/5 A wins (if B has a 1 or 2).
...
In 5^2-4^2=9 cases "5" is the maximum for A, with p=5/5 A wins (because B can not have a 6).

So R(2,1)= 1/25*1/5 + 3/25*2/5 + 5/25*3/5 + 7/25*4/5 + 9/25*5/5
R(2,1) = (1/5)^3 * (1+6+15+28+45)= (1/5)^3*95

For our case we get:

p(2,1) = (1/6)^3 * (5 + p(2,1) + 50 + 10*p(1,1) + 95)

Substituting p(1,1)=4/7 and solving for p(2,1) we get:

p(2,1) = 1/(6^3-1) * (150+10*4/7) = 218/301 = 0.72425

I will post the general recursive formula to calculate all those possibilities later....

[Gorath]

Here is my recursive formula to calculate the propability that you beat with your x d6s an opponent with y d6s at an open test...

[gknoy]

Wow :) Very nice.

Would anyone else like to implement this (or input it into some Handy Math Calculation Tool (Mathmatica? Matlab?)), or shall I whip up some perl? :-) (which, of no one posts numbers, I might. Next week. :))

One favor, though -- when choosing variables and putting them in monospaced font, Please don't use a lower cased L -- l and 1 look a lot alike. Same for O and 0 ... just as using s and 5 and S is problematic when doing it by hand, a lot of the time. I had a professor that liked to use s, it was bloody annoying. ;)

[Gorath]

When i get some time, i will implify the formula and post some propabilities.Perhaps you could verify p(1,1) and p(2,1) by brute force, so we know the formula is right?

I change the index small "L" in the last sum to n...

[gknoy]

wow, cool =)

although, in this case, it might be easier to calculate it than to verify by brute force -- simply because the testing seems like it will take a bit of code to do. ;)

[Gorath]

Okay, here it is:

[phelious fogg]

I just wanted to say, I HATE STATISTICS. That said I feel better. Maybe Ill give some time to think of a good combinatorics proof for this over the weekend. Gah, why cant you people ask a math question on Graph Theory or something.. sheesh..

[Gorath]

I hate statistics, combinatory too. But someone has to do the dirty job :rotfl:

phelious fogg as you are good in math i would appreciate if you could check my given formulas.