Is rolling a dice a Gauss distribution? 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...)
 A: Keep in mind that since dice events are independent and each die is a fair die (no side is more likely than the other), rolling one die multiple times is equivalent to rolling multiple dice all at once.
So, whether rolling a die $n$ times, or $n$ dice at once, each permutation is as likely as the next. For five six-sided dice there are $6^5$ possible outcomes. Of those $7776$ possibilities only one is five 1's. That is roughly a 0.0129% chance of rolling five 1's.
Any other sequence has the same probability. Five 2's, five 3's, five 4's, five 5's, all 6's, 1-2-3-4-5-6, 6-5-4-3-2-1, 1-3-2-4-5-6, 1-4-2-3-5-6, etc. all have a 0.0129% chance.
But if only the combination matters (any order -- like a poker hand) and not the permutation (a specific order -- like a combination lock) you should see why $n$ of any kind is so much rarer. There are multiple ways to roll some combinations.
That is why the odds in Craps are distributed the way they are. Sevens are the most common dice combination. Snake eyes and boxcars are the least common.
Rolling dice is a discrete distribution, while the normal distribution, AKA the Gaussian distribution, is continuous by definition. The distribution is technically binomial, which approximates the normal distribution as n gets large.
So while your friend is right that dice COMBINATION probabilities approximate the Gaussian distribution, you are also correct that each PERMUTATION of dice are equally as likely. It is hard to think of a real life example where dice permutations are used. If anyone knows of any I would be interested to learn about them.
A: 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 $\displaystyle \binom{3}{2}$.
A: 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:
$$\frac{n!}{n!0!0!\cdots} = \frac{n!}{n!} = 1$$
