Probability finishing on a given turn two independent draw I have two decks of card. I draw from each deck independently. I need to find a specific card from each individual deck. What is the odd that on turn $n$, I have found the card in each deck (this is not an assignment)?
From what I understand, the probability to finish on the first turn is $P(1) = 1/52^{2}$. On the second turn, $P(2) = 2*1/52*1/51 + 1/51^2$ where the first term is the odds that the card has been found in one of the deck on the first turn multiplied with the odds to find it on the second turn and the second term, that of finding the two cards on the same turn.
I'd like to express the probability in an easy way to calculate. If I had to expand this for the $n^{th}$ turn, things would get messy. How would one proceed to do so?

I am not quite sure what your $P(2)$ is supposed to be,
but if it is the probability that you have seen both individual cards by the second turn drawing from both decks but not both by the first turn and you are drawing without replacement,
then it should be $\left(\frac{2}{52}\right)^2 - \left(\frac{1}{52}\right)^2$.
If so, then more generally you get $P(n)= \dfrac{2n-1}{2704}$ for $n \in \{1,2,\ldots,52\}$, which is a discrete triangular distribution.
