If we attribute a number to each card of a deck, what would be the probability of having an odd sum? Suppose you have a 52 card deck. We attribute $10$ points to the Ace, $5$ points to the King, $3$ points to the Queen, $1$ point to the Jack, and $0$ to the rest. If we simultaneously pick two cards from the deck, what is the probability of having a sum of points odd?
My first approach was first to calculate the number of issues possible: $\binom{52}{2}$.
Then, to see what issues would give an odd sum:
Let $A$ denote Ace, $K$ King, $Q$ queen, $J$ Jack, and $O$ others.
The issues that satisfy our need are: $$B =\left \{ (A,K);(A,Q);(A,J);(K,O);(Q,O);(J;O) \right \} $$
Now all is left to calculate is the cardinal of $B$.
As there are $4$ Aces and $4$ Kings, we have 16 of $(A,K)$ pairs. And so on for the $(A,Q);(A,J)$.
Now for each $(\cdot; O)$ we have 4 cards times 9, thus giving in total:
$$|B|= 4*(4 * 4) + 3*(4*9)= 156 $$
So the probability would be $$\frac{|B|}{|\Omega|}=\frac{156}{\binom{52}{2}}=\frac{156}{1326} \approx0.12 $$
Yet the correction show that the probability is equal to $\frac{12}{25}$. Where is my mistake?

The exact points per card seem like a distraction.  You have $40$ even cards, and $12$ odd....that's all that matters.  To get an odd sum, one draw must be odd and the other even.
the probability of drawing first an odd and then an even is $\frac {12}{52}\times \frac {40}{51}=\frac {40}{221}$.  Of course the probability of drawing them in the opposite order is the same.  Thus your answer is $$\boxed {\frac {80}{221}\approx .362}$$
