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Question: A dice is rolled 3 times. Let X denote the maximum of the three values rolled. What is the distribution of X (that is, P[X = x] for x = 1,2,3,4,6)? You can leave your final answer in terms of "x". [Hint: Try to first compute P[X ≤ x] for x = 1,2,3,4,5,6].

Solution: Let $X$ denote the maximum of the three values rolled. We are interested in $P(X = x)$, where $x = 1,2,3,4,5,6$. First, define $X_1, X_2, X_3$ to be the values rolled by the first, second, and third dice. These random variables are independent and identically distributed between 1 and 6 inclusive. Following the hint we first compute $P[X ≤ x]$ for $x = 1,2,3,4,5,6$:

$$P(X ≤ x) = P(X_1 ≤ x)P(X_2 ≤ x)P(X_3 ≤ x) = \left(\frac x6\right)\left(\frac x6\right)\left(\frac x6\right) = \left(\frac {x}{6}\right)^3$$

Then, observing that $P(X = x) = P(X ≤ x)−P(X ≤ x−1),$

$$P(X = x) = \left(\frac {x}{6}\right)^3 - \left(\frac {x-1}{6}\right)^3 = \left(\frac {3x^2-3x+1}{216}\right)^3$$

And using the above formula, we can find $P(X = x)$ for all values of $x$. My question is how did we come up with the equality in the first equation? Why is the following true?:

$$P(X ≤ x) = P(X_1 ≤ x)P(X_2 ≤ x)P(X_3 ≤ x)$$

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First note that the event $(X\leq x)=(X_1\leq x)\cap (X_2\leq x)\cap (X_3\leq x)$ which basically is a fact about numbers namely that $\max_{i=1}^n(x_i)\leq x$ iff $x_i\leq x$ for all $i$. Then use independence to write $$ P(X\leq x)=P((X_1\leq x)\cap (X_2\leq x)\cap (X_3\leq x))=P(X_1\leq x)P(X_2\leq x)P(X_3\leq x) $$

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If $x_1,x_2,x_3$ are real numbers then $\max \{x_1,x_2,x_3\} \leq x$ iff $x_i \leq x$ for each $i$. So $P(X \leq x)=P(X_1 \leq x,X_1 \leq x,X_1 \leq x)=P(X_1 \leq x) P(X_2 \leq x)P(X_3 \leq x)$ by independence.

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