Approximating the result of probability So, here's a question asked in an interview:

10 dices are rolled simultaneously. What is the probability of getting the sum of numbers appearing on top of all the dice as 35?

If we calculate manually, I think it's pretty tedious with no calculators allowed. So, I thought of approximating the answer. We know that rolling $10$ dice, has total number of ways as $6^{10}$ which is ~$ 1e7$ which makes it ideal for using a normal distribution. So, mean of $10$ throws is $35$ and SD of 10 throws is ~$ 5.4$ and so $Z = \dfrac{(X-35)}{5.4}$ and so we want $P(\leq35) - P(\leq34)$ as my answer, $P(\leq35) = 1/2$ and the other one can be obtained from the Normal graph and hence, we have the answer.
But to my surprise, the interviewer didn't seem convinced! Is there something wrong with my method? Please point out. Thanks!
 A: Your approach is valid, however arguing that the "total number of outcomes" is large is not a good justification for using the central limit theorem - by that logic you should be able to approximate any continuous distribution, which has (uncountably) infinite possible outcomes by a normal distribution, which is not true.
The random variable whose distribution you want to find is $S_n = X_{1} + \dots + X_{n}$, where the $X_{i}$ are iid and uniformly distributed on $\{1,\dots,6\}$. Using the normal distribution as an approximation via the central limit theorem is justified when $n$ is large, but here $n=10$, so it's at least debatable whether that counts as large.
Also, in this case one can give the exact solution. We can enumerate all the possibilites for the sum being $35$ with integers $k_{1},\dots,k_{10}$ in $\{1,\dots,6\}$ such that
$k_{1} + \dots + k_{10} = 35$. Call this number $K$. Then $K$ is equal to the
coefficient of $X^{35}$ in the polynomial $(X + X^2 + \dots + X^{6})^{10}$,
which can be rewritten as $X^{10} (1 + X + \dots + X^{5})^{10}$. We are
therefore looking for the coefficient of $X^{25}$ in the polynomial $(1+X+\dots+X^{5})^{10}$, which is a stars and bars problem with upper bound and can be solved e.g. as in this answer. Applying the formula from this answer, we obtain
$$K = \sum_{q=0}^{4} (-1)^{q}\binom{10}{q}\binom{25 - 6q + 9}{9} = 4395456$$
The probability then comes out as $K\cdot 6^{-10} \approx 0.0727$.
The normal approximation you gave, centered around 35 (as @drhab suggests) gives $\mathbb{P}(34.5 < X < 35.5) \approx \mathbb{P}(-\frac{1}{10.8} < Z < \frac{1}{10.8}) \approx 0.0737$, so it's actually fairly good.
A: The answer can be approximated without a calculator.
Let $X=$ {sum of scores on $n$ dice} $-\text{ }3.5n$.
We are looking for $P(X=0)$.
$\text{E}(X)=0$
$\text{Var}(X)=\dfrac{35n}{12}$
Using the central limit theorem and the Maclaurin series for $e^x$:
$P(X=0)$
$=2\sqrt{\frac{12}{35n(2\pi)}}\int_0^{1/2}{e^{-\frac12\left(\frac{12x^2}{35n}\right)}}\text{d}x$
$=\sqrt{\frac{24}{35n\pi}}\int_0^{1/2}{\left(1-\frac{6x^2}{25n}+...\right)}\text{d}x$
$\approx\sqrt{\frac{6}{35n\pi}}$
If $n=10$ then $P(X=0)\approx \sqrt{\frac{6}{1100}}\approx \frac{1}{\sqrt{183}}\approx\frac{1}{13.5} \approx 0.07$.
This matches the actual value (found in another answer) to two decimal places.
