Mode of 3 Dice

Method 1:

Clearly $6$ will be the mode iff there are $2$ $6's$ or if there are $3$ $6's$

There is only $1$ configuration where we have $3$ $6's$ (i.e. $6,6,6$ )

There are $\binom{3}{2}=3$ ways of arranging the $2$ $6's$ and $5$ choices for the other number giving $3 \times 5=15$ configuations.

There is a total of $6 \times 6 \times 6=216$ possible combinations of $3$ dice and $15+1=16$ where $6$ is the mode so the probability is:

$\dfrac{16}{216}=\dfrac{2}{27} \implies x=2,y=27 \implies y-x=27-2=\color{#20A900}{\boxed{\boxed{25}}}$

Method 2:

For the numbers to not have a mode, all three must be different. There are $6 \times 5 \times 4=120$ ways doing this.

This means the number of configurations where the numbers have a mode (not necessarily $6$ ) is $6 \times 6 \times 6 -120=96$

As the dice are fair, each of the $6$ faces are equally likely to be the mode so the number of combinations where $6$ is the mode is $\dfrac{96}{6}=16$ giving the probability​ as:

$\dfrac{16}{216}=\dfrac{2}{27} \implies x=2,y=27 \implies y-x=27-2=\color{#20A900}{\boxed{\boxed{25}}}$

This question can be solved easily with the Binomial Distribution formula.

Note that the three throws of the fair dice can be expressed as a binomial experiment where success represents a $6$ rolled and failure being a number other than $6$ rolled.

Let $p$ represent the probability of success, and $q$ be the probability of failure. We have:

$n=3,\;p=\dfrac{1}{6},\;q=1-\dfrac{1}{6} = \dfrac{5}{6}$

For $6$ to be the mode, we need to have at least $2$ rolls of $6$ . Therefore, the probability of the mode being $6$ is

$\text{P}(X \geq 2)\\ =\text{P}(X = 2) + \text{P}(X=3)\\ =\displaystyle\binom{3}{2}\left(\dfrac{1}{6}\right)^2\left(\dfrac{5}{6}\right)^1+\binom{3}{3}\left(\dfrac{1}{6}\right)^3\left(\dfrac{5}{6}\right)^0\\ =3\left(\dfrac{1}{36}\right)\left(\dfrac{5}{6}\right)+ 1\left(\dfrac{1}{216}\right)(1)\\ =\dfrac{15}{216} + \dfrac{1}{216}\\ =\dfrac{16}{216}\\ =\dfrac{2}{27}$

$x=2,\;y=27,\;y-x=27-2=\boxed{25}$