Suppose you have $2$ dices of $20$ faces each. I want to calculate the probability that both faces have a value which is a multiple of $5$, given that I know that one of the two has always a value which is a multiple of $5$. I am tempted to say that the answer is $2*\frac{4}{20}$ because either the first has value multiple of $5$ and then I just need the second dice to have a value multiple of $5$, plus the symmetric case. However, I tried to also use Bayes formula and I get a wildly different result. Let $A$ be the event that both dices have a value which is multiple of $5$. Let $B_i$ be the event that the $i-th$ dice has value multiple of $5$. Then, I want to calculate: $\mathbb{P}(A \vert B_1 \cup B_2) = \frac{\mathbb{P}(A \cap ( B_1 \cup B_2) )}{\mathbb{P}(B_1 \cup B_2)} = \frac{\mathbb{P}(A)}{\mathbb{P}(B_1) + \mathbb{P}(B_2) - \mathbb{P}(B_1 \cap B_2)} = \frac{16}{1584}$
Needless to say, I am slightly confused. Can someone clear this up?
The requested probability is clearly $\frac{16}{144}=\frac{1}{9}$
Using Bayes theorem you get
$$\frac{\frac{4}{20}\cdot\frac{4}{20}}{1-\left( \frac{16}{20} \right)^2}=\frac{16}{144}$$
This result is right and you can visualize it in the following table
where you can count 144 possible cases (jellow cells) and 16 favourable cases (red cells)