Theoretical Probability of the Sums when Rolling a Fair Die 6 Times

Question

You roll a fair die 6 times. What is the theoretical probability for the sum of each combination? For example, rolling a 1,1,1,1,1,1 has a sum of 6. Or what are all the possible combinations? I know there's 46,656. Is there any way to find these without having to do them by hand?

Answer 1

You can use generating functions here.

Suppose we represent a single die with the function $x + x^2+x^3+x^4+x^5+x^6$

The different exponents here represent the different possible outcomes of rolling the die.

Now, if you want the different possible outcomes of rolling two dice, just multiply this function by itself:

$$(x + x^2+x^3+x^4+x^5+x^6)\cdot(x + x^2+x^3+x^4+x^5+x^6)$$

$$= x^2+2x^3+3x^4+4x^5+5x^6+6x^7+5x^8+4x^9+3x^{10}+2x^{11}+x^{12}$$

Again, the different exponents here represent the possible outcoems, so they range from $2$ to $12$

But what's really neat, is that the coefficients represent the number of ways to get that outcome.

For example, the term $5x^6$ indicates that there are $5$ ways to get an outcome of $6$

Why does this work? Well, we get one term $x^6$ by multiplying $x$ from the first term and $x^5$ from the second (i.e. by throwing a $1$ with the first die, and a $5$ with the second), but we also got a term $x^6$ by multiplying $x^2$ from the first term and $x^4$ from the second (i.e. by throwing a $2$ with the first die, and a $4$ with the second), etc.

OK, so for $6$ dice, we just need to find the different coefficients and exponents for the following function:

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

Here is where a tool like WolframAlpha comes in:

Scroll down to the expanded form, and you'll find:

$$x^{36} + 6 x^{3}5 + 21 x^{34} + 56 x^{33} + 126 x^{32} + 252 x^{31} + 456 x^{30} + 756 x^{29} + 1161 x^{28} + 1666 x^{27}$$

$$ + 2247 x^{26} + 2856 x^{25} + 3431 x^{24} + 3906 x^{23} + 4221 x^{22} + 4332 x^{21} + 4221 x^{20} + 3906 x^{19}$$

$$ + 3431 x^{18} + 2856 x^{17} + 2247 x^{16} + 1666 x^{15} + 1161 x^{14} + 756 x^{13} + 456 x^{12} + 252 x^{11} + 126 x^{10}$$

$$ + 56 x^9 + 21 x^8 + 6 x^7 + x^6$$

So, for example, given the term $2856x^{17}$ we know that there are $2856$ ways for the $6$ dice to add up to $17$, meaning that the probability of getting a sum of $17$ is $\frac{2856}{46656}$

Answer 2

The number of ways you can make a sum $n$ from rolling $m$ die can be found with Pascal's Triangle. The pattern exists, though I don't know how to derive it. The diagonals of Pascal's Triangle are such: $$1,1,1,1,1,1...$$ $$1,2,3,4,5,6...$$ $$1,3,6,10,15,21...$$ $$1,4,10,20,35,56...$$ $$1,5,15,35,70,126...$$

When rolling 1 die, there's one way each to make the sums of 1, 2, 3, 4, 5, and 6.
When rolling 2 die, there's 1 way to make a sum of 2, 2 ways to make 3, 3 to make 4, 4 to 5, 5 to 6, 6 to 7 (being the average, we start decreasing the number of ways), 5 to 8, 4 to 9, etc.
When rolling 3, there's 1 to make 3, 3 to 4, 6 to 5, 10 to 6, etc.

Continue this pattern in both directions, and you can find $n$ for $m=6$!

Answer 3

If you want the distribution of the sum $S$ of faces appearing in six rolls of a fair die, you can get a reasonably accurate approximate distribution by simulation. (Illustrated below.)

For an exact analytical solution, consider using convolutions.

For an approximate analytical solution, you can let $D_i$ be the result on the $i$th roll. Find $E(D_i)$ and $Var(D_i).$ From there you can get $E(S)$ and $Var(S)$ because $S = \sum_{i=1}^6 D_i.$ Finally, from the Central Limit Theorem, you can get a pretty good approximation of $S$ by using the normal distribution with mean $E(S)$ and variance $Var(S).$ Using that approximation, you could also get reasonable approximations to such probabilities as $P(S = 25) = P(24.5 < S \le 25.5) \approx 0.06$ and $P(S > 25) = P(X > 25.5)$ $= 1 - P(S \le 25.5) \approx 0.14.$ (Computations in R statistical software.)

mu = 21; sg = sqrt(35/2)
diff(pnorm(c(24.5,25.5), mu, sg));  1 - pnorm(25.5, mu, sg)
[1] 0.06036241
[1] 0.1410294

I will leave the math for getting $E(S)$ and $Var(S)$ and finding the normal approximation to you. (From the first sentence of your question, I guess that may have been the assigned problem.)

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Below is a simulation of the distribution of $S$ based on 100,000 six-roll experiments.

set.seed(1102)
s = replicate(10^5, sum(sample(1:6, 6, rep=T)))
mean(s);  sd(s);  mean(s > 25);  mean(s == 25)
[1] 20.99436  # aprx E(S) = 6(7/2) = 21
[1] 4.181945  # aprx SD(S)
[1] 0.06134   # aprx P(S = 25)
[1] 0.14457   # aprx P(S > 25)

The histogram shows the simulated distribution of $S$ along with the density function (red) of the approximating normal distribution.