What"s the probability of getting a full of $7$ or $11$ once a pair of fair dice is tossed?

I already looked it up on the internet and also my answer suitable the very same answer ~ above a site. However, though ns am confident the my systems is right, i am curious if there"s a an approach in i beg your pardon I could compute this faster since the photo below shows just how time spend that type of method would be. Thanks in advance.

You are watching: What is the probability of getting a total of seven when you roll two dice

*


*

*

For $7$, view that the first roll doesn"t matter. Why? If we roll anything from $1$ come $6$, then the 2nd roll can always get a sum of $7$. The second dice has probability $frac16$ the it matches with the an initial roll.

Then, for $11$, I choose to think of it together the probability of roll a $3$. It"s much easier. Why? shot inverting all the number in your dice table you had actually in the image. Instead of $1, 2, 3, 4, 5, 6$, walk $6, 5, 4, 3, 2, 1$. You need to see that $11$ and also $3$ overlap. Indigenous here, just calculate that there space $2$ means to role a $3$: one of two people $1, 2$ or $2, 1$. Therefore it"s $frac236 = frac118$.

Key takeaways:

$7$ is constantly $frac16$ probabilityWhen asked to find probability that a larger number (like $11$), uncover the smaller counterpart (in this case, $3$).
re-superstructure
mention
monitor
answered Aug 14 "20 in ~ 2:33
*

FruDeFruDe
1,79411 yellow badge44 silver- badges2020 bronze title
$endgroup$
include a comment |
3
$egingroup$
To calculation the possibility of rolling a $7$, role the dice one at a time. Notification that it doesn"t matter what the an initial roll is. Everything it is, there"s one possible roll the the 2nd die that gives you a $7$. For this reason the possibility of rolling a $7$ needs to be $frac 16$.

To calculate the chance of rolling an $11$, roll the dice one at a time. If the an initial roll is $4$ or less, you have no chance. The very first roll will be $5$ or more, maintaining you in the ball game, through probability $frac 13$. If you"re quiet in the ball game, your opportunity of acquiring the 2nd roll you need for one $11$ is again $frac 16$, therefore the complete chance the you will roll one $11$ is $frac 13 cdot frac 16 = frac118$.

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


re-superstructure
point out
monitor
answered Aug 14 "20 at 2:21
*

Robert ShoreRobert coast
15.3k22 yellow badges1111 silver badges3838 bronze title
$endgroup$
add a comment |
1
$egingroup$
Gotta love stars and bars method.

The variety of positive integer services to $a_1+a_2=7$ is $inom7-12-1=6$. As such the probability of acquiring $7$ from 2 dice is $frac636=frac16$.

For $11$ or any kind of number greater than $7$, us cannot proceed exactly like this, since $1+10=11$ is likewise a solution for example, and also we understand that each roll cannot produce greater number 보다 $6$. So us modify the equation a tiny to be $7-a_1+7-a_2=11$ wherein each $a$ is much less than 7. This is identical to finding the variety of positive integers equipment to $a_1+a_2=3$, which is $inom3-12-1=2$. Therefore, the probability of getting $11$ from 2 dice is $frac236=frac118$

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


share
cite
follow
reply Aug 14 "20 in ~ 2:33
*

Rezha Adrian TanuharjaRezha Adrian Tanuharja
8,07155 silver badges2121 bronze badges
$endgroup$
add a comment |
0
$egingroup$
Welcome to the smashville247.netematics Stack Exchange.

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

There room six possible ways to get 7 - one because that each outcome of the very first die - and two feasible ways to gain 11 - one every in the event that the first die is 5 or 6 - meaning you have actually eight complete possibilities . There room $6^2=36$ possibilities for just how the 2 dice can roll, therefore you have a $frac836=frac29$ chance of rolling one of two people one.

See more: What Happens When You Remove An Electron From An Atom ? Can You Remove Protons From An Atom


re-publishing
point out
follow
answer Aug 14 "20 at 2:26
Stephen GoreeStephen Goree
38022 silver badges1010 bronze badges
$endgroup$
add a comment |
0
$egingroup$
In general, the problem of restricted partitions is rather difficult. I"ll frame the problem in a more general setting:

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

This problem can it is in de-worded as:

How numerous solutions space there to the equation$$sum_i=1^n x_i=m$$With the problem that $x_iin smashville247.netbbN_leq k~forall iin1,...,k.$

The solution to this difficulty is not so simple. In little cases, like $n=2, k=6, m=7$, this have the right to be quickly checked v a table; a so referred to as brute pressure approach. Yet for larger values the $n,k$ this is just not feasible. Based on this short article I think in general the equipment to this difficulty is the coefficient the $x^m$ 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 specify the multinomial coefficient:$$smashville247.netrmC(n,(r_1,...,r_k))=fracn!prod_j=1^k r_j!$$And state that$$left(sum_j=1^k x_j ight)^n=sum_(r_1,...,r_k)in Ssmashville247.netrmC(n,(r_1,...,r_k))prod_t=1^k x_t^r_t$$Where $S$ is the collection of services to the equation$$sum_j=1^k r_j=n$$With the restriction that $r_jin smashville247.netbbN~forall jin1,...,k.$ However, herein lies the problem: In order come compute the variety of ways to roll $m$ through $n$ $k$ face die, i beg your pardon is a trouble of computing limited partitions of the number $m$, we require to discover the coefficient the $x^m$ in a multinomial expansion. But, in order come compute this multinomial expansion, we have to compute minimal partitions the $n$. Together you deserve to see the difficulty is a little bit circular. But, $n$ is usually smaller sized than $m$, for this reason it could speed increase the computation process a little. But at the end of the day part amount of brute-force grunt job-related will be required.