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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.

enter image description here

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    $\begingroup$ 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 $\endgroup$
    – stretch
    Commented Aug 14, 2020 at 2:26
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    $\begingroup$ Please do Not Post an Image of the formulas. Use Latex or Format This table as code $\endgroup$
    – miracle173
    Commented Aug 14, 2020 at 2:35

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

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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, $3$).
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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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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.

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    $\begingroup$ 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! $\endgroup$ Commented Nov 18, 2020 at 1:13
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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.

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