Probability of Dice

Question

Q) If 20 fair dice are rolled, then the probability that the sum obtained is between 30 and 40 is 0.325. True or False?

A) So far I have that $ \mu = \frac{7}{2}$ and $ \sigma^2 = \frac{35}{12} $. So for 20 rolls $ \mu = 20 \times \frac{7}{2} $ and $\sigma^2 = 20 \times \frac{35}{12}$. So using the continuity correction we have $ \frac{29.5-\mu}{\sigma}\leq z \leq \frac{40.5-\mu}{\sigma}$. However, this gives the probability equal to $0.0001$ using the normal distribution tables. I don't know if what I have done is right and the statement is false or if I have made a mistake somewhere.

Thanks for the help!

Answer 1

The statement is false. Arthur's argument is a proof, and your calculation is fairly persuasive

The true figure is closer to $0.00003$.

You can find it with the recursion $\displaystyle P_6(s,d) = \sum_{i=1}^6 \dfrac{P_6(s-i,d-1)}{6}$ starting at $P_6(0,0)=1$

which for $d=20$ gives the following probabilities for different sums

 s  Prob
30  5.430E-09
31  1.471E-08
32  3.763E-08
33  9.141E-08
34  2.118E-07
35  4.699E-07
36  1.001E-06
37  2.055E-06
38  4.071E-06
39  7.801E-06
40  1.449E-05

adding up to about 3.025E-05