What"s the probability of acquiring a complete of $7$ or $11$ as soon as a pair of same dice is tossed?

I currently looked it increase on the internet and my answer matched the exact same answer ~ above a site. However, though ns am confident that my systems is right, ns am curious if there"s a an approach in i m sorry I can compute this faster since the photo below shows how time spend that kind of approach would be. Many thanks in advance.

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For $7$, view that the an initial roll doesn"t matter. Why? If we roll anything indigenous $1$ to $6$, climate the second roll can constantly get a sum of $7$. The second dice has actually probability $\frac16$ the it matches v the very first roll.

Then, for $11$, I like to think the it together the probability of rojo a $3$. It"s much easier. Why? try inverting every the number in your die table you had in the image. Rather of $1, 2, 3, 4, 5, 6$, go $6, 5, 4, 3, 2, 1$. You have to see that $11$ and $3$ overlap. Native here, simply calculate the there room $2$ ways to roll a $3$: either $1, 2$ or $2, 1$. So it"s $\frac236 = \frac118$.

Key takeaways:

$7$ is constantly $\frac16$ probabilityWhen inquiry to find probability the a bigger number (like $11$), uncover the smaller equivalent (in this case, $3$).
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reply Aug 14 "20 at 2:33
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FruDeFruDe
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To calculation the chance of rojo a $7$, role the dice one in ~ a time. Notification that it doesn"t matter what the an initial roll is. Every little thing it is, there"s one feasible roll the the second die that gives you a $7$. So the opportunity of roll a $7$ needs to be $\frac 16$.

To calculate the possibility of rojo an $11$, roll the dice one in ~ a time. If the very first roll is $4$ or less, you have actually no chance. The very first roll will certainly be $5$ or more, maintaining you in the ball game, v probability $\frac 13$. If you"re quiet in the sphere game, your chance of obtaining the 2nd roll you require for one $11$ is again $\frac 16$, for this reason the total chance the you will certainly roll an $11$ is $\frac 13 \cdot \frac 16 = \frac118$.

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


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answer Aug 14 "20 in ~ 2:21
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Robert ShoreRobert coast
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Gotta love stars and also bars method.

The variety of positive integer solutions to $a_1+a_2=7$ is $\binom7-12-1=6$. Thus the probability of obtaining $7$ from 2 dice is $\frac636=\frac16$.

For $11$ or any kind of number greater than $7$, we cannot proceed exactly like this, because $1+10=11$ is likewise a solution for example, and we know that each roll cannot produce higher number than $6$. So we modify the equation a small to it is in $7-a_1+7-a_2=11$ where each $a$ is less than 7. This is equivalent to detect the variety of positive integers solution to $a_1+a_2=3$, which is $\binom3-12-1=2$. Therefore, the probability of acquiring $11$ from 2 dice is $\frac236=\frac118$

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


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reply Aug 14 "20 in ~ 2:33
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Rezha Adrian TanuharjaRezha Adrian Tanuharja
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Welcome come the rebab.net ridge Exchange.

There sure is a much faster way; you just have to quickly enumerate the possibilities because that each by treating the role of each dice as independent events.

There room six feasible ways to get 7 - one because that each result of the an initial die - and also two possible ways to obtain 11 - one each in the occasion that the first die is 5 or 6 - definition you have eight full possibilities . There are $6^2=36$ possibilities for just how the 2 dice could roll, therefore you have a $\frac836=\frac29$ chance of rolling either one.

See more: How To Do Tricks In Skate 3, How To Do Basic Flips In Skate 3


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reply Aug 14 "20 in ~ 2:26
Stephen GoreeStephen Goree
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In general, the problem of restricted partitions is fairly difficult. I"ll structure the trouble in a much more general setting:

Suppose we have $n$ dice, having $k$ faces numbered accordingly. How numerous ways space there to roll some confident integer $m$?

This problem can be de-worded as:

How many solutions room there come the equation$$\sum_i=1^n x_i=m$$With the problem that $x_i\in \rebab.netbbN_\leq k~\forall i\in\1,...,k\.$

The equipment to this difficulty is no so simple. In small cases, like $n=2, k=6, m=7$, this can be conveniently checked with a table; a so dubbed brute pressure approach. Yet for larger values that $n,k$ this is merely not feasible. Based upon this short article I think in general the solution to this problem is the coefficient of $x^m$ in the multinomial growth of$$\left(\sum_j=1^k x^j\right)^n=x^n\left(\frac1-x^k1-x\right)^n$$In fact, permit us define the multinomial coefficient:$$\rebab.netrmC(n,(r_1,...,r_k))=\fracn!\prod_j=1^k r_j!$$And state that$$\left(\sum_j=1^k x_j\right)^n=\sum_(r_1,...,r_k)\in S\rebab.netrmC(n,(r_1,...,r_k))\prod_t=1^k x_t^r_t$$Where $S$ is the collection of services to the equation$$\sum_j=1^k r_j=n$$With the restriction the $r_j\in \rebab.netbbN~\forall j\in\1,...,k\.$ However, herein lies the problem: In order come compute the number of ways to role $m$ v $n$ $k$ sided die, i beg your pardon is a difficulty of computing minimal partitions the the number $m$, we need to find the coefficient the $x^m$ in a multinomial expansion. But, in order come compute this multinomial expansion, we should compute minimal partitions of $n$. As you deserve to see the problem is a little bit circular. But, $n$ is usually smaller sized than $m$, therefore it can speed increase the computation procedure a little. However at the end of the day some amount that brute-force grunt occupational will it is in required.