Given $3$ dice, which number is the most likely to appear Given $3$ dice, what is the value of the sum of the number of the $3$ dice most likely to appear? I know that by symmetry, there would always be $2$ different values with the same probability to appear, however I don't know which pairs appear with the highest chance.
 A: Note that there are a total of $6 \times 6 \times 6 = 216$ options. The possible sums are from $3$ to $18$.
Also note that the distribution has to be symmetric since if a roll gives $x,y,z$ adding to $n$, then $7-x,7-y,7-z$ adds to $21-n$.
Hence, the number of ways of getting $n$ is same as the number of ways of getting $21-n$. We would hence expect the maximum to occur for $10$ and $11$.
Number of ways to get $3$ is $1$ i.e. $\dbinom{2}{2}$, which is the same as the number of ways to get $21-3=18$.
Number of ways to get $4$ is $3$ i.e. $\dbinom{3}{2}$, which is the same as the number of ways to get $21-4=17$.
Number of ways to get $5$ is $6$ i.e. $\dbinom{4}{2}$, which is the same as the number of ways to get $21-5=16$.
Number of ways to get $6$ is $10$ i.e. $\dbinom{5}{2}$, which is the same as the number of ways to get $21-6=15$.
Number of ways to get $7$ is $15$ i.e. $\dbinom{6}{2}$, which is the same as the number of ways to get $21-7=14$.
Number of ways to get $8$ is $21$ i.e. $\dbinom{7}{2}$, which is the same as the number of ways to get $21-8=13$.
Number of ways to get $9$ is $25$ i.e. $\dbinom{8}{2}-3$, which is the same as the number of ways to get $21-9=12$.
Number of ways to get $10$ is $27$ i.e. $\dbinom{9}{2} - 3 - 6$, which is the same as the number of ways to get $21-10=11$.
Now by symmetry you can get the number of ways to get the sum between $11$ and $18$.
Hence, the distribution peaks at $10$ and $11$ as expected. 
As a sanity check, we have $2 \left( 1 + 3 + 6 +10 +15 + 21 + 25 + 27\right) = 216$.
Below is the distribution of the number of times a number occurs as a sum of three dices, where the $X$-axis the sum of the three dice and $Y$-axis is the number of times the sum occurs.

A: You can find the answer easily as follows. For the sum of two dice, it is easy to see that $7$ is the most likely outcome ($6$ out of $36$ possibilities), and the probability decreases linearly with the distance from $7$: for the outcome $n$ there are $6-|n-7|$ possibilities, provided this is not negative (so $1\leq n\leq 13$).
Now for a given outcome $m$ for three dice, you must add up for the $6$ possible values for the third dice the number of possibilities for the proper outcome for the two other dice to end up at $m$. In formula the number of possibilities for $m$ is given by
$$
  \sum_{i=1}^66-|(m-i)-7|
$$
So to maximize this you want to minimize the sum of the disctances of $m-i$ from $7$. Since these form a consecutive range, this sum of distances is minimised when the range is either $[-3,-2,-1,0,1,2]$ or $[-2,-1,0,1,2,3]$, so if $m=10$ or $m=11$. The number of possibilities for these outcomes is $27$.
This analysis works only for three dice, but it is not hard to show using induction and similar arguments that for $k$ dice the distribution is symmetric around $\frac72k$, and (if $k\geq2$) the number of outcomes giving $m$ decreases strictly with the distance from  $\frac72k$.
