Conditional Probability exercise with dice

[Rampart]

Hey there community, I have a question on an exercise. Actually it is a general question based on it. Here is the exercise:

We throw 3 dice. If we know that the sum of these 3 is 10, then what is the probability of at least one of them being 3?

Well now, this exercise is very simple. I mean I can always find all the couples such as( 1,5,4),(5,4,1)...
then I can find which of these couples contain at least one "3" and then we are done.
But for an unknown reason, I don't " like" this solution. I mean, is this really the clever way? How about if I had 1000 dice and i knew that the sum of those were 3864 for example, would I have to count all the couples to answer this?

So my question is, is there a possible way to answer this without having to count each and every couple that are out there? Because if there is, it would be nice to start figuring this out, with your help of course(if needed). Anyway, that's all. Thanks in advance for your time!

 

[PeroK]

Sometimes you can find a shortcut, but in general you often have to count the number of ways. There is, however, an interesting way to count in the dice problem:

The numbers of ways of getting a total of $t$ from the roll of $n$ dice is the coefficient of $x^t$ in

$(x + x^2 + x^3 + x^4 + x^5 + x^6)^n$

And, if you want to get that total without a 3, then that's the coefficient of $x^t$ in:

$(x + x^2 + x^4 + x^5 + x^6)^n$

This gives you a useful mathematical way to count, but it's still counting essentially.

 

[Rampart]

It is indeed a very interesting approach. And now I can try some things as n is getting higher. Thank you again sir.

 

[PeroK]

It is indeed a very interesting approach. And now I can try some things as n is getting higher. Thank you again sir.

That's not something I know very much about, I'm sorry to say. You can do a lot by writing a computer program if you want to get a numerical answer for a specific type of problem. Let the computer do the counting.

 

[haruspex]

You could get quite an accurate figure if you were first to find the probability distribution of the 1 to 6 outcomes given the total.
Edit:
To do that, we can use the conditional probability rule, P(A|B)P(B)=P(B|A)P(A), where
A is the event that die 1 (say) shows x
B is the event that the sum of all 1000 dice is y.
We wish to find P(A|B), and we can estimate the other three terms by approximating the sum of 1000 dice as a normal distribution.