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I am trying to figure out the odds of rolling x or higher on 2d6.

I know how to figure out the odds on 1d6 (7-X)/6)
I know how to find the odds on a twin linked d6 ((7-X)/6)+((1-((7-X)/6))*((7-X)/6))

What is the formula of rolling X or higher?

It's a bit tricky. For X between 7 and 12 inclusive, let Y = 13 - X. The formula is then (Y*(Y+1))/2 out of 36. For X less than 7, it's 36 - (X*(X-1))/2 out of 36.

An interesting implication of all of this is that you are twice as likely to fail an unmodified LD9 check than an unmodified LD10 check, and more than three times as likely to fail LD8 versus LD10.

MrEconomics wrote:It's a bit tricky. For X between 7 and 12 inclusive, let Y = 13 - X. The formula is then (Y*(Y+1))/2 out of 36. For X less than 7, it's 36 - (X*(X-1))/2 out of 36.

An interesting implication of all of this is that you are twice as likely to fail an unmodified LD9 check than an unmodified LD10 check, and more than three times as likely to fail LD8 versus LD10.

Thanks Mr.E. i now have a headache!

MrEconomics wrote:It's a bit tricky. For X between 7 and 12 inclusive, let Y = 13 - X. The formula is then (Y*(Y+1))/2 out of 36. For X less than 7, it's 36 - (X*(X-1))/2 out of 36.

An interesting implication of all of this is that you are twice as likely to fail an unmodified LD9 check than an unmodified LD10 check, and more than three times as likely to fail LD8 versus LD10.

Is there a way to solve with out a 2 part equation, cause I need it to consider values of 1 to 12 inclusive? (Yes I know >=1 and >=2 on 2d6 are both 100%) I tried to map out the odds on a table and got (unreduced):

1 = 36/36 with a variance of 0
2 = 36/36 with a variance of 1
3 = 35/36 with a variance of 2
4 = 33/36 with a variance of 3
5 = 30/36 with a variance of 4
6 = 26/36 with a variance of 5
7 = 21/36 with a variance of 6
8 = 15/36 with a variance of 5
9 = 10/36 with a variance of 4
10 = 6/36 with a variance of 3
11 = 3/36 with a variance of 2
12 = 1/36 with a variance of 1


So I understand why it folds at 7 but I don't know how to state that in a single formula.

It is like an upside down v if you graph it but off set on the x axis by 6 with a slope of 1, if only I remember what math equation produced the upside down V! Isn't it absolute value? y = -|x-7|+6 (Edit: forgot counting numbers start at 1 not 0) would produce the right graph for the variance between the values right?

Basically I need to find the equation for y=function(x) for the values of y[0, 0, 1, 3, 6, 10, 15, 21, 26, 30, 33, 35] to correspond with x[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12] ... I knew I shouldn't have skipped linear algebra to go out with that blonde girl...

If you take your formula as a function then you can take the integral of the function from one to x to come up with a formula that you can simply plug numbers in to.

I will work on this today (I am a bit rusty on my integrals) and will come up with a solution, hopefully, by this afternoon.

hmmm... while I dont have my graphing calc, I seem to remember a function where if you put in the data points from 2 to 12 for x and y in the graph it will spit back the formula for you, doing your work so much easier.

Anyone with a working graphing calc who can assist?

alarmingrick wrote:

MrEconomics wrote:It's a bit tricky. For X between 7 and 12 inclusive, let Y = 13 - X. The formula is then (Y*(Y+1))/2 out of 36. For X less than 7, it's 36 - (X*(X-1))/2 out of 36.

An interesting implication of all of this is that you are twice as likely to fail an unmodified LD9 check than an unmodified LD10 check, and more than three times as likely to fail LD8 versus LD10.

Thanks Mr.E. i now have a headache!

Maths is only difficult when you write it down.
Keep it in your head ;D

After spending some time with this it is generally easier to add the numbers from the table you posted rather than trying to come up with a formula.

Even if someone came up with one formula to use all the time chances are it would be more complicated than remembering:
1,2,3,4,5,6,5,4,3,2,1

=SUM(IF(MMULT(TRANSPOSE({1,1,1,1,1,1}),{1,2,3,4,5,6}) +MMULT(TRANSPOSE({1,2,3,4,5,6}),{1,1,1,1,1,1})-A1+1>0,1,0))/36

this will be in excel.

key in what you like in cell a1

Dazuni- I'm not getting correct values in excel using your formula...is there something I'm missing?


Automatically Appended Next Post:
Nevermind I got it to work, sorry for the previous post. Great work Dazuni

I took another look after sleeping some so I could come at it fresh, dose this look right?
Let Y represent the number of d6's
(((6*y+1)-x)/6^y)+((1-(((6*y+1)-x)/6))*(((6*y+1)-x)/6^y))

No need to do the math if someone else did it

http://www.thedarkfortress.co.uk/tech_reports/2_dice_rolls.htm

Also, I think you are all trying to find a formula for something that is best solved with an algorithm.

The chance of a single result is equal to the amount of different dice combinations for that result divided by the total amount of dice combinations.

So for the probability of rolling any result N with 2d6:
P(N) = (number of combinations)/36

If you want to find the chance X of rolling N or lower, you just add up all probabilites from 0 to N:
P(X) = P(0) + P(1) + P(2) + ... + P(N-1) + P(N)

If you want to find the chance Y of rolling N or higher, you add up the probabilites from N to the maximum possible result MAX, in this case 12:
P(Y) = P(N) + P(N+1) + ... + P(MAX-1) + P(MAX)

I can explain it a bit easier.

Make a colum on paper with 1 to 6 both up-down and left-right

Now add up the two numbers in the collems. You will get a list with 6 7's going from top left to bottom right (if I am explaining this badly look in the back of the rulebook under streagh vs thoughnes to wound chart and you will get the idea.)

You will get 36 numbers (1d6 times 1d6) and numbers ranging from 2 to 12 with 7 being the most common (Ocuring 7 times.) The number you are expecting to roll moast often is 7 (1 in 6 times) and numbers that are closer to 7 will be more regular then numbers that are far away.

Dazuni wrote:=SUM(IF(MMULT(TRANSPOSE({1,1,1,1,1,1}),{1,2,3,4,5,6}) +MMULT(TRANSPOSE({1,2,3,4,5,6}),{1,1,1,1,1,1})-A1+1>0,1,0))/36

this will be in excel.

key in what you like in cell a1

this formula will help you to do it in one go but it is only in excel.

it use vector to create 2 matrices, one with every row 1s,2s,...6s. one with every column 1s,2s,...6s.

one matrix plus another give you the 36 possible outcome. remember the two dices are independent.

the result minus the 2d6 number of your choice, then it get converted into boolean, i.e. success or failed.

a quick count and divided by 36 give you the probability.





another way that I can think of doing this is to pre-calculate the probability of each value of 2d6, much like the article linked. and just sum the cuminative probabilities.

or read up the table.

Or just memorise them after a couple of decades of math hammering games with 2D6.

This is just sad. If you cant instinctively get an approximative result there's something wrong. And if you need those kind of formulas to play warhammer it's just depressing, means you have no instinctive feel AND have made the game into a job.
Sorry for the harshness, but that's what the thread inspired me