Probability of rolling two dice and getting a sum of 7

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.

Probability means Possibility. It states how likely an event is about to happen. The probability of an event can exist only between 0 and 1 where 0 indicates that event is not going to happen i.e. Impossibility and 1 indicates that it is going to happen for sure i.e. Certainty. 

The higher or lesser the probability of an event, the more likely it is that the event will occur or not respectively. 

For example – An unbiased coin is tossed once. So the total number of outcomes can be 2 only i.e. either “heads” or “tails”. The probability of both outcomes is equal i.e. 50% or 1/2.
So, the probability of an event is Favorable outcomes/Total number of outcomes. It is denoted with the parenthesis i.e. P(Event).

P(Event) = N(Favorable Outcomes) / N (Total Outcomes)

Note: If the probability of occurring of an event A is 1/3 then the probability of not occurring of event A is 1-P(A) i.e. 1- (1/3) = 2/3

What is Sample Space?

All the possible outcomes of an event are called Sample spaces.

Examples-

1. A six faced dice is rolled once. So, total outcomes can be 6 and 
   Sample space will be [1, 2, 3, 4, 5, 6]
2. An unbiased coin is tossed, So, total outcomes can be 2 and 
   Sample space will be [Head, Tail]
3. If two dice are rolled together then total outcomes will be 36 and 
   Sample space will be [ (1, 1) (1, 2) (1, 3) (1, 4) (1, 5) (1, 6)     (2, 1) (2, 2) (2, 3) (2, 4) (2, 5) (2, 6)     (3, 1) (3, 2) (3, 3) (3, 4) (3, 5) (3, 6)     (4, 1) (4, 2) (4, 3) (4, 4) (4, 5) (4, 6)     (5, 1) (5, 2) (5, 3) (5, 4) (5, 5) (5, 6)   

     (6, 1) (6, 2) (6, 3) (6, 4) (6, 5) (6, 6) ]

Types of Events

Independent Events: If two events (A and B) are independent then their probability will be 
P(A and B) = P (A ∩ B) = P(A).P(B) i.e. P(A) * P(B)

Example: If two coins are flipped, then the chance of both being tails is 1/2 * 1/2 = 1/4

Mutually exclusive events:

• If event A and event B can’t occur simultaneously, then they are called mutually exclusive events.
• If two events are mutually exclusive, then the probability of both occurring is denoted as P (A ∩ B) and 
  P (A and B) = P (A ∩ B) = 0
• If two events are mutually exclusive, then the probability of either occurring is denoted as P (A ∪ B) P (A or B) = P (A ∪ B)                   = P (A) + P (B) − P (A ∩ B)                   = P (A) + P (B) − 0    

                 = P (A) + P (B)

Example: The chance of rolling a 2 or 3 on a six-faced die is P (2 or 3) = P (2) + P (3) = 1/6 + 1/6 = 1/3

Not Mutually exclusive events: If the events are not mutually exclusive then

P (A or B) = P (A ∪ B) = P (A) + P (B) − P (A and B)

What is Conditional Probability?

For the probability of some event A, the occurrence of some other event B is given. It is written as P (A ∣ B)

P (A ∣ B) = P (A ∩ B) / P (B)

Example- In a bag of 3 black balls and 2 yellow balls (5 balls in total), the probability of taking a black ball is 3/5, and to take a second ball, the probability of it being either a black ball or a yellow ball depends on the previously taken out ball. Since, if a black ball was taken, then the probability of picking a black ball again would be 1/4, since only 2 black and 2 yellow balls would have been remaining, if a yellow ball was taken previously, the probability of taking a black ball will be 3/4.

Solution:

When two dice are rolled together then total outcomes are 36 and Sample space is [ (1, 1) (1, 2) (1, 3) (1, 4) (1, 5) (1, 6)     (2, 1) (2, 2) (2, 3) (2, 4) (2, 5) (2, 6)      (3, 1) (3, 2) (3, 3) (3, 4) (3, 5) (3, 6)      (4, 1) (4, 2) (4, 3) (4, 4) (4, 5) (4, 6)      (5, 1) (5, 2) (5, 3) (5, 4) (5, 5) (5, 6)   

   (6, 1) (6, 2) (6, 3) (6, 4) (6, 5) (6, 6) ]

So, pairs with sum 7 are (1, 6) (2, 5) (3, 4) (4, 3) (5, 2) (6, 1) i.e. total 6 pairs

Total outcomes = 36
Favorable outcomes = 6

Probability of getting the sum of 7 = Favorable outcomes / Total outcomes

                                                      = 6 / 36 = 1/6

So, P(sum of 7) = 1/6.

Similar Questions

Question 1: What is the probability of getting 1 on both dice?

Solution:

When two dice are rolled together then total outcomes are 36 and Sample space is [ (1,1) (1,2) (1,3) (1,4) (1,5) (1,6)     (2,1) (2,2) (2,3) (2,4) (2,5) (2,6)      (3,1) (3,2) (3,3) (3,4) (3,5) (3,6)      (4,1) (4,2) (4,3) (4,4) (4,5) (4,6)      (5,1) (5,2) (5,3) (5,4) (5,5) (5,6)      

(6,1) (6,2) (6,3) (6,4) (6,5) (6,6) ]

So, pairs with both 1’s are (1,1) i.e. only 1 pair

Total outcomes = 36
Favorable outcomes = 6

Probability of getting pair of 1 = Favorable outcomes / Total outcomes 
                                                 = 1 / 36 

So, P(1,1) = 1/36.

Question 2: What is the probability of getting the sum of 4?

Solution:

When two dice are rolled together then total outcomes are 36 and Sample space is 

[ (1,1) (1,2) (1,3) (1,4) (1,5) (1,6)     (2,1) (2,2) (2,3) (2,4) (2,5) (2,6)      (3,1) (3,2) (3,3) (3,4) (3,5) (3,6)     (4,1) (4,2) (4,3) (4,4) (4,5) (4,6)     (5,1) (5,2) (5,3) (5,4) (5,5) (5,6)      

(6,1) (6,2) (6,3) (6,4) (6,5) (6,6) ]

So, pairs with sum 4 are (1,3) (2,2) (3,1)  i.e. total 3 pairs

Total outcomes = 36
Favorable outcomes = 3

Probability of getting the sum of 4 = Favorable outcomes / Total outcomes
                                                       = 3/36 = 1/12

So, P(sum of 3) = 1/12.

Question 3: What is the probability of getting the sum of 5?

Solution:

When two dice are rolled together then total outcomes are 36 and Sample space is [ (1,1) (1,2) (1,3) (1,4) (1,5) (1,6)     (2,1) (2,2) (2,3) (2,4) (2,5) (2,6)      (3,1) (3,2) (3,3) (3,4) (3,5) (3,6)      (4,1) (4,2) (4,3) (4,4) (4,5) (4,6)      (5,1) (5,2) (5,3) (5,4) (5,5) (5,6)      

(6,1) (6,2) (6,3) (6,4) (6,5) (6,6) ]

So, pairs with sum 5 are (1,4) (2,3) (3,2) (4,1) i.e. total 4 pairs

Total outcomes = 36
Favorable outcomes = 4

Probability of getting the sum of 5 = Favorable outcomes / Total outcomes
                                                       = 4 / 36 = 1/9

So, P(5) = 1/9.

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