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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!

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  • $\begingroup$ You said no calculators allowed. How do you obtain an answer from the normal graph? $\endgroup$
    – Dan
    Jul 29, 2022 at 8:24
  • $\begingroup$ oh, I didn't. I explained my approach to the interviewer $\endgroup$
    – Charlie
    Jul 29, 2022 at 8:28
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    $\begingroup$ I would rather go for $P(34.5<X<35.5)\sim P(-\frac1{10.8}<Z<\frac1{10.8})=2\Phi(\frac1{10.8})-1$. $\endgroup$
    – drhab
    Jul 29, 2022 at 8:40

2 Answers 2

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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.

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  • $\begingroup$ I think the interviewer was probably looking for the exact solution. :( Anyway, thanks! $\endgroup$
    – Charlie
    Jul 29, 2022 at 10:55
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    $\begingroup$ If that's the case I'm surprised the interviewer didn't tell you that he'd like you to calculate it exactly, since the approximation works quite well here and is certainly a reasonable approach. Perhaps he took issue with your explanation why the approximation should work. $\endgroup$
    – user159517
    Jul 29, 2022 at 11:48
  • $\begingroup$ I'm not sure that the formula for K and your computation of $K\cdot 6^{-10}$ count as 'no calculators allowed'. Of course it could be done by hand in principle but in practice 'no calculators' usually means something like 'within 5 minutes with pen and paper' not 'within a couple of hours'. $\endgroup$
    – quarague
    Jul 29, 2022 at 16:36
  • $\begingroup$ @quarague the OP could tell us more, but I don't think that the "no calculators allowed" requirement means that the final result needs to be a number in its decimal representation, but that a formula ready to be put into a calculator should be equally acceptable. After all the "end result" the OP gave was also in this form. In the interviews I've had, the ability to come up with the right answer / right approach was important to the interviewer, and the requirement of "no calculators" primarily served to prevent cheating. $\endgroup$
    – user159517
    Aug 1, 2022 at 7:22
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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.

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