# Probability of two fair dice rolls having a total of $7$ or $11$?

What's the probability of getting a total of $$7$$ or $$11$$ when a pair of fair dice is tossed?

I already looked it up on the internet and my answer matched the same answer on a site. However, though I am confident that my solution is right, I am curious if there's a method in which I could compute this faster since the photo below shows how time consuming that kind of approach would be. Thanks in advance.

• You should be able to solve problems like that without paper in a few seconds. First, you know there are 36 different possible throws: 2 die each with 6 faces, gives 36 possible throws. There are only 2 ways to get 11: 6-5 and 5-6, along with 6 ways to get 7: 1-6, 6-1, 5-2, 2-5, 3-4, 4-3. 8 total ways to 7 or 11: 8/36 = 2/9 = 0.22222 Commented Aug 14, 2020 at 2:26
• Please do Not Post an Image of the formulas. Use Latex or Format This table as code Commented Aug 14, 2020 at 2:35

To calculate the chance of rolling a $$7$$, roll the dice one at a time. Notice that it doesn't matter what the first roll is. Whatever it is, there's one possible roll of the second die that gives you a $$7$$. So the chance of rolling a $$7$$ has to be $$\frac 16$$.

To calculate the chance of rolling an $$11$$, roll the dice one at a time. If the first roll is $$4$$ or less, you have no chance. The first roll will be $$5$$ or more, keeping you in the ball game, with probability $$\frac 13$$. If you're still in the ball game, your chance of getting the second roll you need for an $$11$$ is again $$\frac 16$$, so the total chance that you will roll an $$11$$ is $$\frac 13 \cdot \frac 16 = \frac{1}{18}$$.

Adding these two independent probabilities, the chance of rolling either a $$7$$ or $$11$$ is $$\frac 16+ \frac{1}{18}=\frac 29$$.

For $$7$$, see that the first roll doesn't matter. Why? If we roll anything from $$1$$ to $$6$$, then the second roll can always get a sum of $$7$$. The second dice has probability $$\frac{1}{6}$$ that it matches with the first roll.

Then, for $$11$$, I like to think of it as the probability of rolling a $$3$$. It's much easier. Why? Try inverting all the numbers in your die table you had in the image. Instead of $$1, 2, 3, 4, 5, 6$$, go $$6, 5, 4, 3, 2, 1$$. You should see that $$11$$ and $$3$$ overlap. From here, just calculate that there are $$2$$ ways to roll a $$3$$: either $$1, 2$$ or $$2, 1$$. So it's $$\frac{2}{36} = \frac{1}{18}$$.

Key takeaways:

• $$7$$ is always $$\frac{1}{6}$$ probability
• When asked to find probability of a larger number (like $$11$$), find the smaller counterpart (in this case, $$3$$).

Gotta love stars and bars method.

The number of positive integer solutions to $$a_{1}+a_{2}=7$$ is $$\binom{7-1}{2-1}=6$$. Therefore the probability of getting $$7$$ from two dice is $$\frac{6}{36}=\frac{1}{6}$$.

For $$11$$ or any number higher than $$7$$, we cannot proceed exactly like this, since $$1+10=11$$ is also a solution for example, and we know that each roll cannot produce higher number than $$6$$. So we modify the equation a little to be $$7-a_{1}+7-a_{2}=11$$ where each $$a$$ is less than 7. This is equivalent to finding the number of positive integers solution to $$a_{1}+a_{2}=3$$, which is $$\binom{3-1}{2-1}=2$$. Therefore, the probability of getting $$11$$ from two dice is $$\frac{2}{36}=\frac{1}{18}$$

Try to experiment with different numbers, calculate manually and using other methods, then compare the result.

In general, the problem of restricted partitions is quite difficult. I'll frame the problem in a more general setting:

Suppose we have $$n$$ dice, having $$k$$ faces numbered accordingly. How many ways are there to roll some positive integer $$m$$?

This problem can be de-worded as:

How many solutions are there to the equation $$\sum_{i=1}^n x_i=m$$ With the condition that $$x_i\in \mathbb{N}_{\leq k}~\forall i\in\{1,...,k\}.$$

The solution to this problem is not so simple. In small cases, like $$n=2, k=6, m=7$$, this can be easily checked with a table; a so called brute force approach. But for larger values of $$n,k$$ this is simply not feasible. Based on this post I think in general the solution to this problem is the coefficient of $$x^m$$ in the multinomial expansion of $$\left(\sum_{j=1}^k x^j\right)^n=x^n\left(\frac{1-x^k}{1-x}\right)^n$$ In fact, let us define the multinomial coefficient: $$\mathrm{C}(n,(r_1,...,r_k))=\frac{n!}{\prod_{j=1}^k r_j!}$$ And state that $$\left(\sum_{j=1}^k x_j\right)^n=\sum_{(r_1,...,r_k)\in S}\mathrm{C}(n,(r_1,...,r_k))\prod_{t=1}^k {x_t}^{r_t}$$ Where $$S$$ is the set of solutions to the equation $$\sum_{j=1}^k r_j=n$$ With the restriction that $$r_j\in \mathbb{N}~\forall j\in\{1,...,k\}.$$ However, herein lies the problem: In order to compute the number of ways to roll $$m$$ with $$n$$ $$k$$ sided die, which is a problem of computing restricted partitions of the number $$m$$, we need to find the coefficient of $$x^m$$ in a multinomial expansion. But, in order to compute this multinomial expansion, we need to compute restricted partitions of $$n$$. As you can see the problem is a bit circular. But, $$n$$ is usually smaller than $$m$$, so it might speed up the computation process a little. But at the end of the day some amount of brute-force grunt work will be required.

• Lol, I feel like this answer is great but maybe too complicated. For somebody that finds dice rolls kind of complicated, they probably wouldn't know summations/products or partitions; good answer otherwise! Commented Nov 18, 2020 at 1:13

Welcome to the Math Stack Exchange.

There sure is a quicker way; you just have to quickly enumerate the possibilities for each by treating the roll of each die as independent events.

There are six possible ways to get 7 - one for each outcome of the first die - and two possible ways to get 11 - one each in the event that the first die is 5 or 6 - meaning you have eight total possibilities . There are $$6^2=36$$ possibilities for how the two dice could roll, so you have a $$\frac{8}{36}=\frac{2}{9}$$ chance of rolling either one.