What is the probability that the total of the two dice will add up to 7 or 11?
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Solution: Show
Sample space of rolling of 2 dice is as below, S = {(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)} n(S) = 36 Let A = sum of numbers is 7 = { (1, 6)(2, 5)(3, 4) (4, 3) (5, 2) (6, 1)} n(A) = 6 P (Sum of numbers is 7) = n(A) / n(S) = 6/36 = 1/6 Therefore, the probability of rolling two dice and getting a sum of 7 is 1/6. What is the probability of rolling two dice and getting a sum of 7Summary: The probability of rolling two dice and getting a sum of 7 is 1/6. You can draw up a possibility space and count how many of the outcomes meet the requirements. #{: (color(white)(0)," "ul1" "," "ul2" "," "ul3" "," "ul4" "," "ul5" "," "ul6" "),(1|," "2" "," "3" "," "4" "," "5" "," "6" "," "color(red)(7)" "),(2|," "3" "," "4" "," "5" "," "6" "," "color(red)(7)" "," "8" "),(3|," "4" "," "5" "," "6" "," "color(red)(7)" "," "8" "," "9" "),(4|," "5" "," "6" "," "color(red)(7)" "," "8" "," "9" "," "10" "),(5|," "6" "," "color(red)(7)" "," "8" "," "9" "," "10" "," "color(red)(11)" "),(6|," "color(red)(7)" "," "8" "," "9" "," "10" "," "color(red)(11)" "," "color(red)(12)" ") :}#
In general, the problem of restricted partitions is quite difficult. I'll frame the problem in a more general setting:
This problem can be de-worded as:
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. What is the probability of rolling a 7 or 11?Let us try to calculate the probability of winning. We can use the probabilities we calculated on the previous page. The probability of winning on the first roll is the probability of rolling 7 or 11, which is 1/6 plus 1/18, which equals to 2/9.
What is the probability of 2 dice adding up to 7?Detailed Solution. ∴ The probability of getting sum as 7 when two dice are thrown is 1/6.
How many ways are there to roll a sum of 7 or 11 on two dice?A simple but typical problem of this type: if we roll two dice, how many ways are there to get either 7 or 11? Since there are 6 ways to get 7 and two ways to get 11, the answer is 6+2=8.
What is the probability of getting a sum of either 11 or greater than 7 by throwing two dice?The probability is 25% .
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