Probability to get a certain number or less by throwing two 6-faced die

[aperson]

TL;DR

I'm trying to write the general formula to obtain the probability of getting a certain number(e.g. all the combinations to obtain 4) or less by throwing two six-faced dices.

In a board game, I need to reach a certain place in a board divided into boxes, I move by throwing 2 six-faced dices.
If my goal is 4 boxes away I need to obtain at least a combination of numbers that sums to 4, but any higher number is also a favorable outcome.
I want to calculate the probability of reaching a precise distance with a single dice throw.

This is my procedure:
two dices have 36 possible combinations, and the likeliness of a numerical outcome Is measured by the number of combinations that sums to it, divided by all possible combinations.
e.g. of all 36 possible combinations, there are 3 that sum to four (1 and 3, 3 and 1, 2 and 2) so the probability of having 4 as an outcome is equal to 3/36.
But to reach a place in the game board that is 4 boxes away I can also roll 5, 6, 7, 8, 9, 10, 11, or 12 so I calculate the number of all possible combinations (36) and subtract the number of combinations(1) that give 2 as an outcome and the number of combinations(2) that give 3 as an outcome:
36 - (1+2) =
36 - 3 = 33
so the possibility of reaching a place 4 boxes apart with a single throw of dice is 33/36.

This is what I have obtained:
2 boxes apart: any combination 36/36
3 boxes apart: any combination but one 35/36
4: 33/36
5: 30/36
6: 26/36
7: 21/36
8: 15/36
9: 10/36
10: 6/36
11: 3/36
12: 1/36

Is this correct?
I don't have much of an education on probability and I would like to write a general formula to embody this procedure but I don't know where to start, could you help me?

 

[Hornbein]

Yes that is correct and the best approach. I will quibble that "combinations" are when order does not matter, where you are not distinguishing the two die, where (1,6) is considered the same as (6,1). You are counting them as different, so these are called "permutations."

 

[FactChecker]

I will quibble that "combinations" are when order does not matter, where you are not distinguishing the two die, where (1,6) is considered the same as (6,1). You are counting them as different, so these are called "permutations."

Ha! I would not call this "quibbling". The first probability lesson that @aperson should be aware of is that the difference between "combinations" and "permutations" is the cause of many nightmares for a first-year student of probability. :-)

 

[Hornbein]

Ha! I would not call this "quibbling". The first probability lesson that @aperson should be aware of is that the difference between "combinations" and "permutations" is the source of many nightmares for a first-year student of probability. :-)

True enough, but he got everything right except the terminology, which is arbitrary. It is useful for referring to the literature.

Looking back at college it seems to me that it was mostly the teaching of terminology, which you will need later to understand the literature. The students are exposed to concepts so at least they know they exist. In short, it gets you to square one. The rest is up to you.

 

[aperson]

Thank you very much for the answers!
Here is an attempt at a formula what do you think?

 

[Hornbein]

Thank you very much for the answers!
Here is an attempt at a formula what do you think?View attachment 326692

Yes that is the way to do it.

 

[Hornbein]

The basic technique is to avoid anything that is procedural, anything that takes the process into account. That confuses things. You are correctly avoid that by doing what is called "a counting argument."