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Five fair six-sided dice are rolled. The probability that the sum of the result being 14 is ?

My solution : I started it with this following method

$$\sum_{d_{1}=1}^{6}\sum_{d_{2}=1}^{6}\sum_{d_{3}=1}^{6}\sum_{d_{4}=1}^{6}\sum_{d_{5}=1}^{6} \left[z\right]^{14}z^{d_{1}+d_{2}+d_{3}+d_{4}+d_{5}}$$

And I obtained 85/972 as the solution , Is that right ?

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  • $\begingroup$ take a look here for an "alternative" way to count the valid cases. $\endgroup$ – Masacroso Jul 15 '17 at 5:21
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    $\begingroup$ Write $\left[\,{z^{14}}\,\right]$ instead of $\require{cancel}\cancel{\left[\,{z}\,\right]^{\color{#f00}{14}}}$. $\endgroup$ – Felix Marin Jul 15 '17 at 6:36
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From stars and bars the number of $n$-tuples of natural numbers summing to $k$ is given by $\binom {k-1}{n-1} $

The number of 5-tuples of natural numbers summing to 14 is given by $\binom {13}4 = 715 $

some of these will contain a number greater than six. To get the number of 5-tuples that correspond to 5 rolls of a 6 sided die you need to subtract these from 715. It is fortunate that no 5-tuple summing to 14 can contain more than one element greater than 6.

  • those that contain a 7 must also have 4 other natural numbers summing to 7,
  • there are $5\binom 63$ of these
  • and $5\binom 53$ will contain an 8

Total 5-tuples summing to 14 for which no element is greater then 6 is given by

$$ N = \binom {13}4-5\binom 63 -5\binom 53 -5\binom 43 -5\binom 33$$ where the last term comes from the 5-tuples containing one 10 and four 4's

$$ = 715 - 100-50-20-5= 540$$

Each 5-tuple has a probability of $(\frac 16)^5=\frac 1{7776}$

so the probability of summing to 14 is $\frac{540}{7776}= \frac 5{72}$

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  • $\begingroup$ Thank you i just correcting my Answer even with different method I obtained same answer , thanks again youve been really helpful $\endgroup$ – Znik Dzulqarnain Jul 15 '17 at 6:33
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There are $6^5$ possible outcomes when five six-sided dice are rolled.

The number of outcomes in which the outcomes sum to $14$ is equal to the number of solutions of the equation $$x_1 + x_2 + x_3 + x_4 + x_5 = 14 \tag{1}$$ in the positive integers subject to the constraints that $x_k \leq 6$ for $1 \leq k \leq 5$. The number of solutions of the equation $$x_1 + x_2 + x_3 + \cdots + x_k = n$$ in the positive integers is given by the formula $$\binom{n - 1}{k - 1}$$ Hence, equation 1 has $$\binom{14 - 1}{5 - 1} = \binom{13}{4}$$ solutions.

From these, we must exclude those solutions in which one or more of the variables exceeds $6$. Note that at most one variable may exceed $6$ since $2 \cdot 7 + 3 \cdot 1 = 17 > 14$.

Suppose $x_1 \geq 7$. Let $x_1' = x_1 - 6$. Then $x_1'$ is a positive integer. Substituting $x_1' + 6$ for $x_1$ in equation 1 yields \begin{align*} x_1' + 6 + x_2 + x_3 + x_4 + x_5 & = 14\\ x_1' + x_2 + x_3 + x_4 + x_5 + x_6 & = 8 \tag{2} \end{align*} Equation 2 is an equation in the positive integers with $$\binom{8 - 1}{5 - 1} = \binom{7}{4}$$ solutions.

By symmetry, there are an equal number of solutions for each variable that could exceed $6$. Hence, the number of solutions of equation 1 in which one of the five variables exceeds $6$ is $$\binom{5}{1}\binom{7}{4}$$ Hence, the number of solutions of equation 1 that satisfy the constraints is $$\binom{13}{4} - \binom{5}{1}\binom{7}{4} = 715 - 5 \cdot 35 = 540$$ Therefore, the desired probability is $$\frac{\binom{13}{4} - \binom{5}{1}\binom{7}{4}}{6^5} = \frac{540}{7776} = \frac{5}{72}$$

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