# Thread: Probability problem with 4 Dice

1. ## Probability problem with 4 Dice

I'm stuck with the following problem: If I roll 4 regular six-sided dice what is the probability that the total will be greater than or equal to 20? I can't work out how to calculate this short of writing out all the combinations of throws that would yield a sum of at least 20. Ideally I would like to derive a formula.

2. You could convolutions twice, which would be a bit easier (still messy though!)

The first convolution gets the distribution of the sum of a pair of dice.

Then do convolutions of that distribution to see the distribution of two [I]pairs of[I] dice.

http://www.stat.science.cmu.ac.th/~w...0_132_P001.pdf

3. Originally Posted by StaryNight
I'm stuck with the following problem: If I roll 4 regular six-sided dice what is the probability that the total will be greater than or equal to 20? I can't work out how to calculate this short of writing out all the combinations of throws that would yield a sum of at least 20. Ideally I would like to derive a formula.
You will see the expansion of $\displaystyle \left( {\sum\limits_{k =1 }^6 {x^k } } \right)^4$.
The coefficients of the $\displaystyle x^n$ tell you how many ways one can roll a sum of n with four dice.
Now how can you use that to work this question?

4. Originally Posted by Plato
You will see the expansion of $\displaystyle \left( {\sum\limits_{k =1 }^6 {x^k } } \right)^4$.
The coefficients of the $\displaystyle x^n$ tell you how many ways one can roll a sum of n with four dice.
Now how can you use that to work this question?

that is the most interesting bit of probability/statistics ive ever seen

5. Originally Posted by Plato
You will see the expansion of $\displaystyle \left( {\sum\limits_{k =1 }^6 {x^k } } \right)^4$.
The coefficients of the $\displaystyle x^n$ tell you how many ways one can roll a sum of n with four dice.
Now how can you use that to work this question?
I can sum the number of ways of getting 20,21,22,23,24. But why does the formula you gave work?

6. Originally Posted by StaryNight
But why does the formula you gave work?
That is known as a generating function. It is a standard technique used in general counting problems such as this problem. It is a very difficult problem to do by hand.

I fail to understand why you are expected to solve a problem that you do not have the tools to use. Maybe that is a conversation you can have with your education authority.

7. We've covered probability generating functions, but these were used to calculate E(X), E(X^2) etc.. I think this is different.

8. Originally Posted by StaryNight
We've covered probability generating functions, but these were used to calculate E(X), E(X^2) etc.. I think this is different.
It is somewhat different.