 Cold blood ld Probabilities - Warhammer 40K Fantasy

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# Thread: Cold blood ld Probabilities

1. ## Cold blood ld Probabilities

There has been some talk in the VC forum recently about lizardman ld and how it relates to normal tests. I said i would give a conclusive report and here it is. Make sure the tables are read carefully. I drafted this in word first and not all of it copied over as i'd have liked but i made a few changes and it should read ok now. Be warned that it does take a certain knowledge of mathematics to read Probabilities with tossing two and three fair dice

Questions:

1) Say I roll two dice, and I want the sum of the two numbers to be less than or equal to five. The probability of this isn't too difficult to work out.

2) Now say I roll three dice and I pick the lowest two scores, what is the probability of these two adding up to less than or equal to five? And what is this probability equivalent to on two dice? i.e. Is getting less than or equal to seven on two dice equal in probability to getting less than or equal to five on three dice when you pick the lowest two?

3) How do these relationships change as I change the desired number?

The probabilities are not difficult to work out for independent fair dice. Let X be the score on a single die. The distribution of the score is given by the probability function
P(X = x) = 1/6, for x = 1, 2, 3, 4, 5, 6.
If we have two dice, the joint probability function is
P(X1 = x1, X2 = x2) = 1/62, for x1 = 1, 2, 3, 4, 5, 6 and x2 = 1, 2, 3, 4, 5, 6.
Then the probability function of the sum Y = X1 + X2 is easily found to be as shown in the table below:

Table 1: Probability function of the sum of the scores on two dice
y:2 3 4 5 6 7 8 9 10 11 12
P(Y = y)*6^2:1 2 3 4 5 6 5 4 3 2 1 respectively!

Thus, the answer to question 1 is given by P(Y <= 5) =(1 + 2 + 3 + 4)/36 = 10/36.

Now consider three dice.
The joint probability function for the scores on three dice is given by
P(X1 = x1, X2 = x2, X3 = x3) = 1/63,
for x1 = 1, 2, 3, 4, 5, 6, x2 = 1, 2, 3, 4, 5, 6, and x2 = 1, 2, 3, 4, 5, 6.
Now suppose that we order the three scores (X1, X2 and X3) such that the ordered values are represented by X(1), X(2) and X(3) where equal values are allowed, i.e. X(1) <= X(2) <= X(3).
Now, the joint probability function of the ordered values needs careful handling because of the possible equalities. The joint probability function is given by the following expressions where equalities and strict inequalities must be carefully observed.

P(X(1) = x1, X(2) = x2, X(3) = x3) = 6(1/6)3, for x1 < x2 < x3,
= 3(1/6)3, for x1 = x2 < x3,
= 3(1/6)3, for x1 < x2 = x3,
= (1/6)3, for x1 = x2 = x3,

The probability function for the sum of the lowest two, Y =X(1) + X(2), is now given in the following table (you need to carefully add up all the possible probabilities from the joint density function â an interesting exercise in itself):

Table 2: Probability function of the sum of the two lowest scores on three dice
y:2 3 4 5 6 7 8 9 10 11 12
P(Y = y)*6^3:16 27 34 36 34 27 19 12 7 3 1 respectively

Thus, P(X(1) + X(2) <= 5) = (16 + 27 + 34 + 36)/63 = 113/216.
The second part of question 2 was âIs getting less than or equal to seven on two dice equal in probability to getting less than or equal to five on three dice when you pick the lowest two?â
We see from Table 1 that for two dice
P(X1+ X2 <= 7) = (1 + 2 + 3 + 4 + 5+6)/62 = 21/36 = 126/216.
So these two probabilities are not the same.
You can answer question three for other values by reference to Tables 1 and 2.

So there we have it!

Ciao

Stonehambey  Reply With Quote

2.

3. what!?! :wacko: :wacko: :wacko:  Reply With Quote

4. Well, I understood it after reading it twice, but it looks good. Good work on that . Could you just add what the actual odds are for each ld value though, I think this would help alot of people who don't really get the math of the ld probability.  Reply With Quote

5.

6. im only 13! (and not that good at math's)  Reply With Quote

7. Originally Posted by ^Shadow-Stalker
im only 13! (and not that good at math's)
Llike i said before, a good understanding of maths is required to read this and fully understand it.

For those of you who have asked me about summarising the probabilities, there are two tables in the report (not very clear i know) from which you can work out the probabilities of each ld two or three dice. Some simple addition is required but i'm sure that isn't out of anyone's reach! Ciao

Stonehambey  Reply With Quote

8. wow! and you calculated that (tell me are you studying degree level maths???)
i'll be going to uni in september to study electronics, if i look over it multiple times i'll understand it:rolleyes:
good work(Y) but, i think the odds of passing the leadership will be easier for Mr. Ninja of understanding the principles.

-Kai-Itza-  Reply With Quote

9. Lol, yeah i am studying degree level maths:yes: I'm approaching the end of my second year, only one more to go!

Then i'm gonna have to get a job which requires a bit more than standing behind a till Ciao

Stonehambey  Reply With Quote

10. Originally Posted by Kai-Itza
good work(Y) but, i think the odds of passing the leadership will be easier for Mr. Ninja of understanding the principles.

-Kai-Itza-
whos mr. ninja, and what does he understand?

EDIT: oh, i hadnt read the 2.3k list yet... so its ME...  Reply With Quote

11. Well I'm not gunna lie to ya, I hate stats, so I'm not even gunna try to understand that. Though if I actually did do work in class I should've been able to understand the majority. So what I'm asking is if some1 will write up a clear table of the calculated probabilities for each ld. test with cold-blooded. I understand the method is there but it would be much easier for myself and others if it were calculated.  Reply With Quote

12. fellas, we're looking at a genious, amazing really, a kid asked if i was once, i think he was serious:rolleyes:

-Kai-Itza-  Reply With Quote