We shuffle a 52-card French deck, then draw cards one by one from the deck until we find a black ace. On which draw is it most likely for the first black ace to appear?
If we draw the first black ace at the $k$-th position $(k=1,2, \ldots, 51)$, then the second black ace must be in the remaining $(52-k)$ cards, and it can be anywhere in this pack. Thus, considering all possible - and equally likely due to shuffling - placements of the two black aces, there are $(52-k)$ cases where the first black ace is drawn at the $k$-th position. The number of these possibilities is the largest when $k=1$, making the first draw the most likely for the first black ace to appear.

Remarks. 1. The positions of the two black aces can be chosen in $\binom{52}{2}$ ways from the 52-card deck, so the probability that the first black ace is found at the $k$-th position is $(52-k) /\binom{52}{2}$. If $k=1$, this probability is $\frac{1}{26} \sim 0.038$.

2. The papers generally used the "number of favorable events/total number of events" formula to write the probabilities under investigation, but rarely clarified which are the elementary events whose certain, equally likely outcomes they summed up. In the communicated solution, we considered the placement of the two indistinguishable black aces. The other end is if any two cards are distinguished. Then an elementary event is one of the 52! equally likely orders of the 52 cards, and the number of such orders where the first black ace is at the $k$-th position is $2 \cdot(52-k) \cdot 50$ !.
3. Similarly, it can be clarified that the occurrence of the second black ace at the last position is the most probable.