From a pack of 52 cards, we draw one by one, without replacement. If f(n) is the probability that an Ace will appear at the $$n^{th}$$ turn, then

Probability that ace will appear in 1st turn $$f\left(1\right)=\dfrac{4}{52}=\dfrac{1}{13}$$

Probability that ace will appear in 2nd turn $$f\left(2\right)=\dfrac{48}{52}\times \dfrac{4}{51}=\dfrac{1}{13}\times \dfrac{48}{51}$$

Probability that ace will appear in 3rd turn $$f\left(3\right)=\dfrac{48}{52}\times \dfrac{47}{51}\times \dfrac{4}{50}=\dfrac{1}{13}\times \dfrac{48}{51} \times \dfrac{47}{50}$$

$$f\left(1\right)=\dfrac{1}{13}$$

$$f\left(2\right)=\dfrac{1}{13}\times \dfrac{48}{51}=f(1)\times (number<1)$$ => $$f(2)<f(1)$$

$$f\left(3\right)==\dfrac{1}{13}\times \dfrac{48}{51} \times \dfrac{47}{50}=f(2)\times (number<1)$$ => $$f(3)<f(2)$$

$$\dfrac{1}{13} > f(2) > f(3)$$