# 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 – stretch Aug 14 at 2:26
• Please do Not Post an Image of the formulas. Use Latex or Format This table as code – miracle173 Aug 14 at 2:35

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, $$3$$).

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 $$\frac 16+ \frac{1}{18}=\frac 29$$.

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.

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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.

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.