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I'd like to get some hints of solving this problem.

Nobody seems to be answering to my comment which I have replied. Any feedback guys?

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    Assume that the dice are differently colored: say red, blue, and green. How many outcomes are there in the sample space? Now, either count manually, or use more advanced techniques to count how many outcomes in the sample space result in the sum of the three rolls being at most ten. Note: This is equivalent to counting how many non-negative integer solutions there are to $\begin{cases}x_1+x_2+x_3\leq 10\\ 0\leq x_i\leq 6\end{cases}$ – 2017-01-27

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One half.

It's the same as the probability of getting a sum of at least $11$, which is the other half of the symmetric distribution.

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    Is it possible to get a step by step explanation please? – 2017-01-27
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    There's one way to get $3$. There's also one way to get $18$, which is the symmetric case. There are three ways to get $4$, and three ways to get $17$. Continue up to the number of ways to get $10$ and $11$, which are the two center cases. Jorge's answer explains this observation a bit more explicitly than mine, but it still boils down to an argument that makes the answer evident. – 2017-01-30
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    Ok got it thank! – 2017-02-04
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You want to find the number of solution to $x_1+x_2+x_3=\leq 10$ where each variable is in the range $\{1,2,3\dots,6\}$.

Notice that you can pair the triples that sum $k$ with the triples that sum $21-k$ ( by sending $x_1,x_2,x_3$ to $(7-x_1,7-x_2,7-x_3)$.

This tells us that the probability they add $3,4,5,6,7,8,9$ or $10$ is the same as the probability they add $18,17,16,15,14,13,12$ or $11$. So the answer is $\frac{1}{2}$.

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    I think I'm going to need to go step by step for this question. I know the sample space is 120 possibilities ($6P3$) I just don't understand the next part, we need the number of possibilities that the sum of these three numbers is at most 10 in order to compute the probability. Now I kind of got lost at $21 - k$. Don't quite understand the concept of pairing the triples – 2017-01-27
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    $6^3=216$, not $120$. – 2017-01-29
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    @JKawa We don't calculate the triples directly, we just prove that the number of possibilities where sum is not greater than 10 and the number of those when it is greater are the same. It follows that the probability of each is $\frac12$ – 2017-01-29