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Question: Six dice are thrown independently. What is the probability of getting sum 10?

Attempt: There are total $6^6$ many possible cases. A number 6 can't appears in the sum so that total become 10.

1+1+1+1+1+5 there 6 way to happen this 2+2+2+2+2+2

Is there any better approach to solve this problem?

N. F. Taussig
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    You can use the stars and bars method to figure out how many ways six integers can add up to $4 = 10 - 6$ – Matti P. Apr 10 '18 at 12:20

3 Answers3

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We have that $$ \left\{ \matrix{ 1 \le x_{\,k} \le 6 \hfill \cr x_{\,1} + x_{\,2} + \cdots + x_{\,6} = 10 \hfill \cr} \right.\quad \Rightarrow \quad \left\{ \matrix{ 0 \le y_{\,k} \left( { \le 5} \right) \hfill \cr y_{\,1} + y_{\,2} + \cdots + y_{\,6} = 4 \hfill \cr} \right. $$ and we can omit the upper bound on the admissible range for the $y$'s , since it is greater than the sum.

Thus the number of solutions is equal to the number of weak compositions of $4$ into $6$ parts which is $$\binom{4+6-1}{6-1} = \binom{9}{5}$$

For a more general approach you may refer to this related post.

G Cab
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Let $x_k$ be the result of the $k$th throw. Then the number of favorable cases is the number of solutions of the equation $$x_1 + x_2 + x_3 + x_4 + x_5 + x_6 = 10$$ in the positive integers subject to the restrictions that $x_k \leq 6$ for $1 \leq k \leq 6$. However, none of the variables can exceed $6$ since $1 + 1 + 1 + 1 + 1 + 6 = 11 > 10$ is already too large. A particular solution of the equation corresponds to the placement of five addition signs in the nine spaces between successive ones in a row of ten ones. $$1 \square 1 \square 1 \square 1 \square 1 \square 1 \square 1 \square 1 \square 1 \square 1$$ For instance, if we choose the second, third, fifth, eighth, and ninth spaces, we obtain $$1 1 + 1 + 1 1 + 1 1 1 + 1 + 1$$ which corresponds to the solution $x_1 = 2$, $x_2 = 1$, $x_3 = 2$, $x_4 = 3$, $x_5 = x_6 = 1$. The number of such solutions is the number of ways we can choose which five of the nine spaces between successive ones in a row of ten ones to fill with an addition sign, which is $$\binom{9}{5}$$

N. F. Taussig
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There are only 5 combinations to consider:

A. 1,1,1,1,1,5
B. 1,1,1,1,2,4
C. 1,1,1,1,3,3
D. 1,1,1,2,2,3
E. 1,1,2,2,2,2

A will occur 6! / (5! * 1!) = 6 times
B will occur 6! / (4! * 1! * 1!) = 30 times
C will occur 6! / (4! * 2!) = 15 times
D will occur 6! / (3! * 2! *1! ) = 60 times
E will occur 6! / ( 2! * 4! ) = 15 times

6 + 30 + 15 + 60 + 15 = 126 so the probability is 126 / 6^6 = approx. 0.27 %

Brian Trial
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