Which is the probability to obtain one pair in poker? I'm interested in the case when I decide which is the pair.
I know the probability for one pair is: $$\frac{C^{4}_{13}\cdot4\cdot C^{2}_{4}\cdot4^3}{C^{5}_{52}}$$ but I don't know which is the probability if I choose the pair to be composed by two Kings(K).
Thanks :)
The probability of one specific pair and no other is $\frac 1{13}$ of the chance of any one pair, assuming by one pair you mean only one pair. I don't see how to derive your formula easily. If you want two kings and any three other non-kings, it is $\frac {C_4^2 \cdot C_{48}^3}{C_{52}^5}$ where the numerator is choosing which two kings and which three other cards.
Added: vadim123 has shown the derivation of your formula, which makes sure there are no other pairs. To get kings specifically, you can just divide by $13$. Otherwise, you can do $\frac {C_4^2 C_{12}^3\cdot 4^3}{C_{52}^5}$, picking the kings, three other ranks, then the cards within those ranks. Since $4C_{13}^4=C_{12}^3$ this checks