So if we have a regular deck of $52$ cards that is shuffled, then what is the probability of drawing the first red ace on the $k$-th card? And which k would maximize the probability?
I wasn't entirely sure of computing the probability.
Here are my thoughts:
Have $52$ cards, thinking about the $k$-th card. There are two red aces, so the probability would be $\frac{2}{52-(k+1)}$?
I am entirely lost on maximizing $k$. Would anyone be able to help me out? Or guide me on maximizing $k$? Is my probability correct?
Best Answer
In a well-shuffled deck, the positions of the two red aces are two randomly chosen numbers from $\{1,2,\dots, 52\}$. There are ${52\choose 2}$ such pairs of numbers.
If the first red ace is at position $k$, then the other red ace must belong to the set $\{k+1,\dots, 52\}$ and there are $52-k$ such possibilities.
Therefore the probability that the first red ace is in position $k$ is the ratio $${52-k\over {52\choose 2}},$$ which is maximized by taking $k$ as small as possible.