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.

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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 $frac16$ 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 $frac236 = frac118$.

Key takeaways:

$7$ is always $frac16$ probabilityWhen asked to find probability of a larger number (like $11$), find the smaller counterpart (in this case, $3$).
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answered Aug 14 "20 at 2:33
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FruDeFruDe
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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 $frac 13 cdot frac 16 = frac118$.

Adding these two independent probabilities, the chance of rolling either a $7$ or $11$ is $frac 16+ frac118=frac 29$.


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answered Aug 14 "20 at 2:21
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Robert ShoreRobert Shore
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Gotta love stars and bars method.

The number of positive integer solutions to $a_1+a_2=7$ is $inom7-12-1=6$. Therefore the probability of getting $7$ from two dice is $frac636=frac16$.

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 $inom3-12-1=2$. Therefore, the probability of getting $11$ from two dice is $frac236=frac118$

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


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answered Aug 14 "20 at 2:33
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Rezha Adrian TanuharjaRezha Adrian Tanuharja
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Welcome to the les-grizzlys-catalans.org 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 $frac836=frac29$ chance of rolling either one.

See more: What Are Things That Come In A Dozen Cool Things In Our Online Collection For 12


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answered Aug 14 "20 at 2:26
Stephen GoreeStephen Goree
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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 $x_iin les-grizzlys-catalans.orgbbN_leq k~forall iin1,...,k.$

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$$left(sum_j=1^k x^j ight)^n=x^nleft(frac1-x^k1-x ight)^n$$In fact, let us define the multinomial coefficient:$$les-grizzlys-catalans.orgrmC(n,(r_1,...,r_k))=fracn!prod_j=1^k r_j!$$And state that$$left(sum_j=1^k x_j ight)^n=sum_(r_1,...,r_k)in Sles-grizzlys-catalans.orgrmC(n,(r_1,...,r_k))prod_t=1^k x_t^r_t$$Where $S$ is the set of solutions to the equation$$sum_j=1^k r_j=n$$With the restriction that $r_jin les-grizzlys-catalans.orgbbN~forall jin1,...,k.$ 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.