Is rolling a dice a Gauss distribution?

Question

I'm in an argument with a friend over rolling a dice several times, for example rolling 5 times. His argument is, that is far more difficult to roll out 1-1-1-1-1 than any other combination (for example 1-5-2-4-3) as the results of the rolls distribute in a Gauss manner so the edges are somehow less probable. His arguments have to do with Bayes theory. My intuition tells just the opposite since the dices have no memmory and one roll isn't conditioned by the latter, so any combination is equally probable. I know that dices are a very recurrent topic here but I haven't found an straight answer to my question...

In the final terms it seems is a discussion of the Bayessian against frequentist approaches to probabilistics. We're not going to solve it today... I abandon you since we have a duel at twelve o'clock in the church yard.

EDIT As I'm seeing many comments and answers here I would like to point out that the discussion is about combinations WITHOUT PERMUTATIONS, that is 12345 is different of 54321. Furthermore, I said 5 rolls to put a concrete example, but what we were discussing was for a large number of rolls (take large as the number you want...)

Answer 1

We can calculate $$\left(\sum_{i=1}^6 x_i\right)^5$$

We can expand this with the Multinomial theorem. The coefficients before each monomial is the number of configurations to get the occurrences of the corresponding exponents. For example, the term $x_1^2x_2^2x_3$ will tell us that there are as many combinations of 2 ones, 2 twos and one three as the coefficient before it.

For the general case this coefficient will be $$\frac{n!}{\prod_{i=1}^n e_i!}$$ where $e_i$ are the exponents, i.e. the individual occurrences.

In the extreme case if all exponents are one it becomes all possible permutations: $n!$ which is $5! = 120$ configurations.

And if one exponent is n and all other 0:

$$\frac{n!}{n!0!0!\cdots} = \frac{n!}{n!} = 1$$

Answer 2

Your friend is right, is far more difficult to roll out $11111$ than any other sequence. Let's see why.

Formalism

Let $X$ be number of times a $1$ appears in $5$ independent rolls of a die. Then, $X \sim \text{Binom}(5, \frac{1}{6})$, and

$$\text{Prob. of sequence 11111} = P(X=5) = \left(\frac{1}{6} \right)^5$$

$$\text{Prob. of any other sequence} = 1-P(X=5) = 1-\left(\frac{1}{6} \right)^5$$

Intuition

Any other combination different than $11111$ is more probable because in general that any other combination has more options per roll. Even in the more restrictive case of any other combination where a $1$ does not appear, you would have $5$ possibilities ($2$ to $6$) per roll that would make that event occur. In turn, the sequence $11111$ only has one possibility per roll ($1$).

The Gaussian approximation

If you define a random variable $X$ as above over the experiment of successive and independent rolls of a die, then that binomial random variable with parameters $n$ (number of rolls) and $p$ (probability of a success) can be approximated through a Gaussian distribution with mean $np$ and variance $np(1-p)$ when $n$ is large and $p$ is not too close to neither $1$ or $0$. Do the values of $n$ and $p$ in your problem satisfy that requirement? Let's see how the binomial PMF looks like for those values:

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Not so good, right? You can find here some rules of thumb to decide when the normal approximation could be good and here a proof about the Gaussian approximation of the binomial distribution.

If in the description of the problem, you make $n$ as large as you want, then, with $p=\frac{1}{6}$, the Gaussian approximation is good. Let's see:

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About why this approach works only when $12345$ is different than $54321$

Let's see now why the definition of $X$ as above indeed works only when we consider that $12345$ is different of $54321$. To do that, let's consider a toy example. Suppose you roll three times a $2$-sided die and let $X$ be the number of times a $1$ appears. So, for example, $X=2$ when any of the following sequences occur

$$121\quad 112\quad 211$$

In this way $P(X = 2) = 3\left(\frac{1}{2}\right)^3$, but let's write this in a more interesting way

$$P(X = 2) = \binom{3}{2}\left(\frac{1}{2}\right)^2\left(\frac{1}{2}\right)$$

...and what is this? Nothing more that a binomial probability! In this way, we can see that using binomial probabilities, we are considering that $121$, $211$ and $112$ are all different sequences but with the same probability! and that's why we multiply the probability term $\displaystyle \left(\frac{1}{2}\right)^2\left(\frac{1}{2}\right)$ by $\displaystyle \binom{3}{2}$.