Calculate the probability that the maximum is greater Two dice are rolled simultaneously. Calculate the probability that the maximum is greater than or equal to $3$.
Attempt: The answer is $\frac{8}{36}$. Why can't I say that since there are $4 \cdot 4$ ways to give data respecting the imposed conditions, then we have the probability of $\frac{16}{36}$. Am I missing something?
 A: You would calculate that the outcome of both dice are greater than or equal to 3. But then you would miss the outcome $(2,4)$, for instance. So you are looking for the combinations where at least one outcome is greater than or equal to 3. You can make a table and count the favorable outcomes.
Or you look for the complementary events. These are the combinations where both outcomes are less or equal to 2: $\max(1,1)=1 ;\max(1,2)=2, \max(2,1)=2, \max(2,2)=2$
Then the complementary probability is $\frac{4}{36}$. So the asked probability is $P\left(\max(X_1,X_2)\geq 3\right)=1-\frac{4}{36}=\frac{32}{36}=\frac{8}{9}$. This is the double of the given answer.
A: A formula for all $1\leq n \leq 6$:
Considering that the two dice are rolled independently and from definition of the maximum function we have,
$\qquad\begin{align}P(\max(X_1, X_2) \geq n) &= 1  - P(\max(X_1, X_2) \leq  n-1)\\&= 1 - P(X_1 \leq n-1, X_2 \leq n-1) \\&= 1 - P(X_1 \leq n-1) \,P(X_2 \leq n-1) \\&= 1 - \dfrac{n-1}{6}\, \dfrac{n-1}6\\&= 1- \dfrac {(n-1)^2}{36}\end{align}$
So for $n=3$,
$\qquad P(\max(X_1, X_2) \geq 3) = 1 - \dfrac{4}{36}$
