# A person rolls five fair dice simultaneously. Find the probability that the total number of spots showing is less than or equal to $11$

I am having trouble with this problem. When they say spot I think they are essentially saying the sum, so its the probability that the sum of dice is $11$ or less.

I understand that there are $6^5$ combinations.

I found 6 ways that it can equal to $11$ $(2,3,2,2,2)(3,3,1,1,3),(4,4,1,1,1),(5,2,2,1,1),(6,2,1,1,1)(7,1,1,1,1)$, but I know there has to be an easier way than just counting. Is it like $\binom{6}{5}$? Thanks for the help.

• You can probably use combinations with repetitions. Maybe it's easier to think that you have dice that go from 0 to 5 and ask what is the probability that their sum is less or equal to 6. – Matti P. Feb 13 '18 at 6:41

You can solve an equivalent question which is slightly easier. Consider the following:

You have five different bins to which you need to distribute up to eleven balls. First you will put one ball at each bin, as Matti P suggested, since it is impossible to have a score lower than 1 on a dice. Now you have up to six balls left to distribute. So, we will use an additional bin - a sort of trash bin, if you will, which will take all the balls we won't put in the original five bins (corresponding with the dice). The amount of ways to distribute up to six balls in five bins is equivalent to distributing six balls to six bins, as in $nSr(6,6)=nCr(11,5)$. However, you still need to subtract the number of distributions which have more than a total six balls in a single bin (that one mean a dice got 7 or more). You already have five bins with one ball inside, and six balls left, so your only option is to choose 1 out of 5 bins then insert all 6 balls to that bin.

To sum up, you have $nSr(6,6)-5=nCr(11,5)-5=457$ ways to get less than 11. Now divide that by the total number of possible dice outcomes and you'll get the probability of 457 over 7776, i.e. 0.05877...

• Very nice. (+1) It has been a while since I have done 'balls-in-boxes'. Would you mind explaining $nSr$ notation? – BruceET Feb 13 '18 at 7:53
• Combinations with repetition. Perhaps you've seen it as $CC$, but I'm used to this notation. – P.Diddy Feb 13 '18 at 7:59

My guess is that you have been studying the Central Limit Theorem and that you are to assume that the total $T$ on five dice is approximately normal. Here is an outline of that method.

It is not difficult to show that the number $X$ on a single die has $E(X) = 7/2$ and $var(X) = 105/36.$

Thus $T$ has $\mu=E(T) = 17.5,\,\sigma^2=Var(T) = 14.58333,$ and $\sigma=SD(T) = 3.818813.$

Then, approximately, $T \sim \mathsf{Norm}(\mu, \sigma),$ and $$P(T \le 11) = P(T \le 11.5) = P\left(\frac{T-\mu}{\sigma} \le \frac{11.5-17.5}{3.819}\right) \approx P(Z \le -1.571) \approx 0.0581,$$ where $Z$ is standard normal and the probability can be evaluated using printed normal tables.

As a reality check on the normal approximation, I simulated a million 5-dice experiments. One run of the simulation gave $P(T \le 11) = 0.0589$ (and also results for $E(T)$ and $SD(T)$ that are in good agreement with known values).

t = replicate(10^6, sum(sample(1:6, 5, rep=T)))
mean(t <= 11);  mean(t);  sd(t)
## 0.058977  # aprx P(T <= 11)
## 17.50287  # aprx E(T) = 17.5
## 3.824495  # aprx SD(T) = 3.819


Below is a histogram of the simulated values of $T$ along with the approximating normal density curve. The fit is not perfect, but it is pretty good for values of $T$ below about 12, which is what we need.

Notes: Of course, an exact combinatorial solution for $P(T \le 11)$ would be best. You have listed possible ways for dice numbers to add to eleven, but you have not considered the various ways those numbers might be assigned to the individual dice, so $6/6^5$ is not the correct answer for $P(T = 11).$ Maybe one of our experts in combinatorics will find a way to get an exact solution. But I have seen similar problems before, where a normal approximation was intended.

We can solve the problem using generating function. \begin{align}\left(\frac16\sum_{i=1}^6x^6\right)^5=\frac{1}{6^5}\left( \frac{x(1-x^6)}{1-x}\right)^5 \end{align}

Our goal is to find the sum of coefficients of $x^5$ to $x^{11}$.

$$(1-x^6)^5 =1-5x^6+\text{higher order terms}$$

By negative binomial series $$(1-x)^{-5}=\sum_{k=0}^\infty \binom{4+k}{k}x^k$$

Hence,

\begin{align}\left(\frac16\sum_{i=1}^6x^6\right)^5=\frac{1}{6^5}x^5(1-5x^6+\text{higher order terms})\sum_{k=0}^\infty \binom{4+k}{k}x^k\end{align}

and the sum of coefficients is

$$\frac1{6^5}\left(\sum_{k=0}^{6}\binom{4+k}{k}-5\binom{4+0}{0} \right)=\frac1{6^5}\left(\sum_{k=0}^{6}\binom{4+k}{4}-5 \right)=\frac{462-5}{6^5}=\frac{457}{7776}$$

• Congratulations. Very nice. (+1) Your approximation 0.0588 is within the margin of error of my simulation, and summing five discrete uniforms does not quite give a good approximation to normal (especially near the mean). – BruceET Feb 13 '18 at 7:44