How can I calculate the average result of rolling a given number of N sided dice?

The specific question in this case is what is the average summed result of the rolls on 18 ten sided dice, but I'd like to know how to calculate the average total of any number of dice of any number of sides. I've considered that it could be as simple as the average throw times the number of throws, but I have a hard time believing it could really be that simple.|||Since the dice has ten sides, the dice is numbered 1 through 10. The sum of the numbers that appear on each dice is 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 = 55. Since there are ten sides, the average number that one can roll is 55/10 = 5.5. Therefore, the required expected sum after rolling 18 such die is 18(5.5) = 99.





I hope this helps!|||I would first like to point out that as far as fair dice are concerned (fair meaning that each face has the same probability of occurring), the only possible numbers of sides are 4,6,8,12, and 20, because these are the only possible regular polyhedra (often called "platonic solids"). Now, if we ignore that conceptual difficulty, we can move on to probability theory.





The expectation of a sum of random variables is the sum of the expectations of those random variables. Essentially, this means that you are right with regards to that last statement. Assuming the faces are numbered 1 ... n on each die, and that the dice are fair, the average throw of a single die will be (1+n)/2. The average of a sum of N throws will then be (N/2)(1+n). In the case of 18 ten-sided dice, we have (18/2)(1+10) = 99.|||You are correct.





The average result can be calculated by taking the average throw of one die and multiplying it by the total number of dice.


Let us think why this is true. What is the average throw of one normal six sided die?


(1+2+3+4+5+6) / 6 = 3.5


Now what happens when we roll two dice?


Each die has an average value of 3.5. So, when you add them together you get a combined average value of 7. This in fact is the most common roll with two dice!





It may be even clearer with a simpler example. Imagine a coin that we assign the value 1 to heads and the value 0 to tails. What is the average value of flipping this coin? It is obviously 0.5.


Now imagine flipping two of these coins? Let us calculate the average value of this by looking at all possible scenarios.


1) Heads and Heads -- 1+1 = 2


2) Heads and Tails -- 1+0 = 1


3) Tails and Heads -- 0+1 = 1


4) Tails and Tails -- 0+0 = 0





Now averaging all of these possibilities gives us (2+1+1+0) / 4 = 1


This is the same answer we would get by multiplying the average value of one coin by the number of coins!





This long method of looking at all possible scenarios can be repeated for the six and ten sided dice but that would take a lot of room, but it would show the same answer as multiplying the average value of one die by the total number of die.