Roll a a die three times

Question

Draw probability mass function for the following random variables.

I have tried $P(X=1)=(6/6)^3-(5/6)^3$ which gives me the minimum value rather than the maximum value.

For $P(X=6)=1/6^3$.

Answer 1

Let $x_1$, $x_2$ and $x_3$ be the three rolls. Then for $1\leq k\leq 6$, $$P(\max(x_1,x_2,x_3)\leq k)=P(x_1\leq k)\cdot P(x_2\leq k) \cdot P(x_3\leq k)=\left(\frac{k}{6}\right)^3.$$ Hence $$\begin{align}P(\max(x_1,x_2,x_3)=k)&=P(\max(x_1,x_2,x_3)\leq k)-P(\max(x_1,x_2,x_3)\leq k-1)\\&=\left(\frac{k}{6}\right)^3-\left(\frac{k-1}{6}\right)^3=\frac{k^3-(k-1)^3}{6^3}.\end{align}$$

Answer 2

Hint:

For the maximum to be $n$

• you want all $3$ dice to be less than or equal to $n$

• but you do not want all $3$ dice to be strictly less than $n$

Answer 3

To compute the probability, we should add the probability of all cases that have a maximum of $m \in \{1,2,3,4,5,6\}$:

• Either all three dice $d_1$, $d_2$ and $d_3$ are equal to $m$;

• Or two of them are equal to $m$ and one is strictly less than $m$, there are 3 cases, because either $d_1$, $d_2$ or $d_3$ could be the smallest result;

• Or one of them is equal to $m$ and the two others are strictly less than $m$, there are 3 cases, because either $d_1$, $d_2$ or $d_3$ could be the largest result.

Computing this, results in:

$$P(X = m) = P(d_1 = m) \times P(d_2 = m) \times P(d_3 = m)$$ $$ \quad \quad \quad \quad \quad \quad + 3 \times P(d_1 = m) \times P(d_2 = m) \times P(d_3 < m)$$ $$ \quad \quad \quad \quad \quad \quad + 3 \times P(d_1 = m) \times P(d_2 < m) \times P(d_3 < m) .$$

Hence, $$ P(X = m) = 1/6^3 + 3 \times (m-1)/6^3 + 3 \times (m-1)^2/6^3$$ $$ = (3m^2 - 3m +1) / 216 .\quad \quad \quad \quad$$

Of course that I considered that the dice rolls are independent, so I multiplied the probability of the outcome of each dice.