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Three ordinary dice are thrown and the points obtained are added up. What is the probability of obtaining 10 points?

I tried to build by extension, the set of favorable results:

{

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

}

They are 18, except the 2 repeated.

Is there a way to do the calculation for these types of exercises, without having to count one by one?

  • 1
    Generating function $(x+x^2+x^3+x^4+x^5+x^6)^3$, find coefficient of $x^{10}$. Alternatively, inclusion-exclusion on stars-and-bars counting integral solutions to $\begin{cases}x_1+x_2+x_3=10\\1\leq x_i\leq 6\end{cases}$ – 2017-02-03
  • 2
    Also, you are missing many of the favorable results. you don't include for example $(1,5,4),(1,4,5)$ among others. There should in fact be $27$ favorable outcomes out of the $6^3$ possible. – 2017-02-03
  • 1
    I also get $27/6^3 = .125,$ and checked it by simulation. Also, the sum on three dice is 'pretty close' to normal, and a normal approx with continuity correction gives 0.127. – 2017-02-04

1 Answers 1

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Positive integral solutions of $x_1+x_2+x_3=10$
$=\dbinom{10-1}{3-1}$

Positive integral solutions of $x_1+x_2+x_3=10$ where $x_1\gt6$
$=\dbinom{4-1}{3-1}$
(this does not include $x_2\gt6$ and/or $x_3\gt6$ because if $x_1\gt6,x_2,x_3\lt6$

Positive integral solutions of $x_1+x_2+x_3=10$ where $x_2\gt6$
$=\dbinom{4-1}{3-1}$

Positive integral solutions of $x_1+x_2+x_3=10$ where $x_3\gt6$
$=\dbinom{4-1}{3-1}$

Therefore, required count

$=\dbinom{10-1}{3-1}-3\times \dbinom{4-1}{3-1}=27$

  • 0
    I do not understand where the numbers come from. Why do you evaluate when $ x_i $ is greater than 6? – 2017-02-04
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    to evaluate $x_1 \gt 6$, first we distribute $6$ to $x_1$ so that it will be always above $6.$ after distributing $6$ to $x_1,$ $10$ becomes $4.$ so it is equivalent to number of positive integral solutions to $x_1+x_2+x_3=4$ and therefore, it is $\dbinom{4-1}{3-1}$ – 2017-02-04
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    We did not see the method you just used. Try to understand it on my own but I could not (I do not even know what this method is called in Spanish). I thought the $ x_i $ could only take values from 1 to 6, so I do not understand why you have to evaluate $ x_i> 6 $ as in the system proposed @JMoravitz – 2017-02-04
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    @CarlosFrostte here we apply [Inclusion-Exclusion Principle](https://en.wikipedia.org/wiki/Inclusion%E2%80%93exclusion_principle) on the three events $A_1=\{x_1~\text{is too big}\}$, similarly $A_2,A_3$. We have $|A_1\cup A_2\cup A_3|=|A_1|+|A_2|+|A_3|-|A_1\cap A_2|-|A_1\cap A_3|-|A_2\cap A_3|+|A_1\cap A_2\cap A_3|$ is the size of the set of solutions which violates at least one of the upper bound conditions. Subtracting away from the amount of solutions where we don't care about upperbounds gives us our final answer. – 2017-02-04
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    Here, $|A_i|=\binom{3}{2}$ and $|A_i\cap A_j|=0$ and $|A_1\cap A_2\cap A_3|=0$, so calculations were minimal. – 2017-02-04
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    Would it be okay to say that $\begin{cases}x_1+...+x_k=n\\a\leq x_i\leq b\end{cases}=\dbinom{n-1}{k-1}-k\times \dbinom{n-b-1}{k-1}$ ? – 2017-02-04