# How do I determine the average number of successes, variance, and standard deviation, for a given set of die rolls.

Assuming that I roll 10 dice, and am going to call a 5 or 6 a 'success' on any single die, how can I determine the average number of successes, the variance, and the standard deviation for the associated bell curve?

If I use $N$ as the number of dice used, and $P$ as the probability of getting a success on a single die roll, I believe the average is simply $NP$, or in this case, $10×\frac26=3.33$ successes on average.

What I'd like to know now, is what is my range of possible successes at 1 and 2 standard deviations. Beyond that, I'd like to understand how to perform this calculation assuming I change either $N$ or $P$.

• In my opintion 10 dice (random variables) are not enough to approximate the distribution by the normal distribution. The thumb rule is $n\cdot p\cdot q > 9$. Think about that. – callculus Dec 6 '17 at 1:32

The distribution is binomial with p=1/3. Use the standard form to get the first and second moment and therefore the variance and standard deviation.

As mentioned by callculus, normal approximation would not be a good method in this case. Instead I will use the binomial distribution.

The expected value can be obtained in two ways:

\begin{align*} E(X) &=\sum_{k=0}^ n {n\choose{k}} k\cdot p^k\left(1-p\right)^{n-k}\\\\ &=\sum_{k=0}^{10} {10\choose{k}} k\cdot \left(\frac{1}{3}\right)^k\left(1-\left(\frac{1}{3}\right)\right)^{10-k} \\\\ &= \frac{10}{3}\\\\ &\approx 3.333 \end{align*}

You could also just do $np$ to get $$E(X)=10\cdot\frac{1}{3}=\frac{10}{3}\approx 3.333$$

The variance can be obtained in two ways:

\begin{align*} Var(X) &=E(X^2)-E(X)^2 \\\\ &=\sum_{k=0}^{10} {n\choose{k}} \cdot k^2 \cdot p^k\left(1-p\right)^{n-k} - \left(\sum_{n=0}^{10}{n\choose{k}} \cdot k \cdot p^k\left(1-p\right)^{n-k}\right)^2 \\\\ &=\sum_{k=0}^{10} {10\choose{k}}\cdot k^2 \cdot \left(\frac{1}{3}\right)^k\left(1-\left(\frac{1}{3}\right)\right)^{10-k} - \left(\sum_{n=0}^{10} {10\choose{k}}\cdot k \cdot \left(\frac{1}{3}\right)^k\left(1-\left(\frac{1}{3}\right)\right)^{10-k}\right)^2 \\\\ &= \frac{40}{3} - \left({\frac{10}{3}}\right)^2\\\\ &= \frac{20}{9}\\\\ &\approx 2.222 \end{align*}

You could also just do $npq$ to get

$$Var(X)=10\cdot\frac{1}{3}\cdot\frac{2}{3}=\frac{20}{9}\approx 2.222$$

The standard deviation is just

$$S.D.=\sqrt{Var(X)}=\sqrt{\frac{20}{9}}\approx 1.490711985$$

• @Calculus and Remy - Thank you both very much. I had come up with something similar to what Remy posted, and it seemed to be working fine, but when I tried the equation with 100 dice, it fell apart. I wasn't considering the NPQ>9 thumb rule. Thanks again! – JJ Bone Dec 6 '17 at 12:39