In this answer, there is a section titled "Summing Dice". It describes how convolution of the discrete function that is $1$ for each integer from $1$ through $6$, and $0$ otherwise, yields the distribution for the sum of $n$ six-sided dice.
Rolling $50$ six-sided dice will yield an approximately Normal Distribution whose mean is $\mu=50\times\frac72$ and whose variance is $50\times\frac{35}{12}$; thus, a standard deviation of $\sigma=\sqrt{50\times\frac{35}{12}}$.
Mean and Variance of a Single Die
The mean of a single die whose faces vary from $1$ to $n$ is $\frac{n+1}{2}$. For $n=6$, this gives $\frac72$.
The variance of a single die whose faces vary from $1$ to $n$ is "the mean of the squares minus the square of the mean." The sum of the squares from $1$ to $n$ is $\frac{2n^3+3n^2+n}{6}$, so the mean is $\frac{2n^2+3n+1}{6}$. Subtracting $\frac{n^2+2n+1}{4}$ yields $\frac{n^2-1}{12}$. For $n=6$, this gives $\frac{35}{12}$.
ETA: OK, I think I've fixed the problem. Off-by-one error...
I think this can be done with generating functions. The generating function for a single die is given by
$$
F(z) = \frac{t-1}{d} + \frac{(d-t)z}{d} + \frac{zF(z)}{d}
$$
We can interpret this as follows: The probability that there are no hits on the one die is $\frac{t-1}{d}$, so $F(z)$ has that as the coefficient for $z^0 = 1$. The probability that there is one hit and the die doesn't "explode" (repeat) is $\frac{d-t}{d}$, so $F(z)$ has that as the coefficient for $z^1 = z$. In the remaining $\frac{1}{d}$ of the cases, the die explodes and the situation is exactly as it was at the start, except that there is one hit already to our credit, which is why we have $zF(z)$: the $F(z)$ takes us back to the beginning, so to speak, and the multiplication by $z$ takes care of the existing hit.
This expression can be solved for $F(z)$ via simple algebra to yield
$$
F(z) = \frac{t-1+(d-t)z}{d-z}
$$
whose $z^h$ coefficient gives the probability for $h$ hits. For example, for the simple case $n = 1, d = 20, t = 11$:
\begin{align}
F(z) & = \frac{10+9z}{20-z} \\
& = \frac{10+9z}{20} \left(1+\frac{z}{20}+\frac{z^2}{20^2}+\cdots\right) \\
& = \left( \frac{1}{2} + \frac{9}{20}z \right)
\left(1+\frac{z}{20}+\frac{z^2}{20^2}+\cdots\right) \\
\end{align}
and then we obtain the probability that there are $h$ hits from the $z^h$ coefficient of $F(z)$ as
$$
P(H = h) = \frac{1}{2\cdot20^h}+\frac{9}{20^h} = \frac{19}{2\cdot20^h}
\qquad h > 0
$$
with the special case
$$
P(H = 0) = \frac{1}{2}
$$
In general, we can obtain the expectation of the number of hits $\overline{H}$ as
$$
\overline{H} = F'(1) = \frac{d(d-t)+t-1}{(d-1)^2} = \frac{d+1-t}{d-1}
$$
Now, for $n$ dice, we have
$$
[F(z)]^n = \left[ \frac{t-1+(d-t)z}{d-z} \right]^n
$$
We can write this as $N(z)M(z)$, where
\begin{align}
N(z) & = [t-1+(d-t)z]^n \\
& = \sum_{k=0}^n \binom{n}{k} (t-1)^{n-k}(d-t)^kz^k
\end{align}
and
\begin{align}
M(z) & = \left(\frac{1}{d-z}\right)^n \\
& = \frac{1}{d^n} \left( 1+\frac{z}{d}+\frac{z^2}{d^2}+\cdots \right)^n \\
& = \sum_{j=0}^\infty \binom{n+j-1}{j} \frac{z^j}{d^{n+j}}
\end{align}
so we can obtain a closed form for $P(H = h)$ from the $z^h$ coefficient of $[F(z)]^n = N(z)M(z)$ as
\begin{align}
P(H = h) & = \sum_{k=0}^{\max\{h, n\}} \binom{n}{k} \binom{n+h-k-1}{h-k}
\frac{(t-1)^{n-k}(d-t)^k}{d^{n+h-k}} \\
& = \frac{(t-1)^n}{d^{n+h}}
\sum_{k=0}^{\max\{h, n\}} \binom{n}{k} \binom{n+h-k-1}{h-k}
\left[ \frac{d(d-t)}{t-1} \right]^k
\end{align}
For example, for $n = 1, d = 6, t = 5$ (the example in the OP), the above expression yields
$$
P(H = h) = \frac{5}{3 \cdot 6^h} \qquad h > 0
$$
with the special case
$$
P(H = 0) = \frac{2}{3}
$$
which coincides with the conclusions drawn in the comments to the OP.
The expectation for the number of hits could be obtained by evaluating $\frac{d}{dz} [F(z)]^n$ at $z = 1$, but owing to the linearity of expectation, it is obtained more straightforwardly as $n$ times the expected number of hits for one die, namely
$$
\overline{H} = \frac{n(d+1-t)}{d-1}
$$
I think this all checks out, but some independent verification (or disproof, as appropriate) would be nice.
Best Answer
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.