Likelihood of getting a certain sum after rolling several six-sided dice 
Do you believe your friend whom claims to achieve a sum of $400$ after
rolling $100$ six-sided dice?

I was asked this question in a data science interview, and I was wondering if someone can please explain if my reasoning understanding is correct now. I think the answer is no, I shouldn't believe them. By the normal approximation to the binomial distribution, I found that the probability of getting a sum of 400 which is less than 1/501, which is the "average" probability of the 501 outcomes.
Is my reasoning okay? Or is there something wrong with it.
 A: *

*The exact outcome of 400 isn't the relevant quantity for evaluating the plausibility of the claim; as you add more dice, even the most probable outcomes will become extremely improbable on their own, so it's not the probablility of that exact outcome that tells you about how consistent it is with the apparent circumstances (that some outcome is plausible on a large number of fair dice).


*The binomial doesn't seem to be directly relevant, so you probably wouldn't do yourself any favours mentioning it. Even if it was relevant it's not clear from your post how you got to 1/501 from the binomial. It sounds like you switched to treating the sum on 100 dice as a discrete uniform (which would suggest what are perhaps even worse misunderstandings than thinking it was binomial).
If my numerical convolution didn't lose too much accuracy from moving piles of sand, it looks like the probability for $\geq 400$ from direct calculation is about $0.00182$ (normal approximation with continuity correction is about $0.00187$).
However, in an interview I'd just compute the standard deviation (var = 3500/12 is a little under 300, so just to a rough approximation, sd is around 17),  and say "The total will be approximately normal. Now 400 is almost 3 sd's from the mean of 350. If the dice were fair a value at least this far from the mean would be pretty unlikely". Done.
Here's the right half of the pmf (with the far upper tail cut off), with the normal approximation to those probabilities dotted in red:

It's symmetric so I'm just showing half (allowing us to get a tiny bit more detail).
As we can see from the plot the normal approximation is going to work quite well. Convolutions of discrete uniforms settle in pretty quickly, apart from the far tail; 100 dice is plenty good enough to use a normal approximation for this sort of back of the envelope calculation.
If you don't know that the variance of the outcomes on a die with k sides is $\frac{k^2-1}{12}$ (so 35/12 for a six-sided die), you'd need to work it out, but it's not a long mental computation:
The sum of the first 6 squares is simple - $k(k+1)(2k+1)/6$ for $k=6$ gives $7\times 13=91$, so the variance is $91/6-(7/2)^2 = (182-147)/12 = 35/12$. (Though if I was doing this in my head, I'd work this slightly differently, in terms of squared deviations from the mean - $\frac16\cdot 2(\frac14+2\frac14+6\frac14)=35/12$.)
So then the sum of 100 dice has variance 3500/12 and you proceed by the earlier reasoning to the "nearly 3 sd's from the mean" (assuming you also know $17^2 =289$ and $18^2=324$, so the square root of something in the region of 290-300 is 17-and-a-bit; at worst you should definitely be able to say the sd is somewhere between 16 and 18 and still get to "roughly 3 sd's from the mean").
[The actual number of sd's is about 2.9 but we don't need to be that accurate to answer the question.]
Just with some basic number facts, the CLT, and knowing some basic things about the normal distribution you should be able to give a good answer quite quickly.
It doesn't much matter whether the probability calculation should be one or two tailed (both are quite small), but I'd probably mention that if you do two tailed you want to double $P(T\geq 400)$, but "it's still a very small number".

How is any of this relevant to this kind of job? I think it speaks to several things, but IMO, perhaps the most important is your ability to perform rapid ballpark reasonableness checks on your computations. If you have no idea how to approach these sorts of calculations, you're left accepting whatever output you got, which means at any given level of competence you're letting through a lot more errors than if you know how to say "wait, this makes no sense, let's check that calculation again".
Being able to identify a suitable quick approximation or bound and evaluate it is an essential everyday skill for anyone doing a lot of computation. People make errors all the time - of many kinds - but if you can catch a lot of them before it matters, the fact that you will make errors will hurt your employer much less - you weed a lot of them out.
