I wasn"t totally sure of computer the probability. Below are mine thoughts:

Have $52$ cards, thinking around the $k$-th card. There space two red aces, so the probability would certainly be $frac252-(k+1)$?

I am completely lost top top maximizing $k$. Would certainly anyone it is in able to help me out? Or overview me ~ above maximizing $k$? Is mine probability correct?

In a well-shuffled deck, the location of the 2 red aces space two randomly favored numbers native $1,2,dots, 52$. There room $52choose 2$ such pairs that numbers.

If the very first red ace is at position $k$, climate the other red ace should belong to the set $k+1,dots, 52$ and also there space $52-k$ together possibilities.

Therefore the probability that the very first red ace is in place $k$ is the ratio $$52-kover 52choose 2,$$ which is maximized by taking $k$ as tiny as possible.

There room two red aces in the deck, diamond ace and heart ace. The probability the the very first card is no red ace is $frac52-252$. The probability the the second one is not the red ace, provided than the an initial one was not is $frac51-251$. The probability the the very first $k-1$ room not red ace and also the $k$-th one is for this reason equals:$$ left(prod_m=1^k-1 frac52-1-m52-m+1 ight) cdot frac252-(k-1) = left( frac(52-(k-1))(52-k)(52-1+1)(52-2+1) ight) cdot frac252-(k-1) = frac2(52-k)52 (52-1)$$Thus the probability is maximal because that $k=1$.

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Another way of gaining the very same answer:

Consider a heat of 52 empty slots, and also put the **ordered** cards in randomly selected slots. The very first two cards come be inserted are the two red aces (why not?). The opportunities of the first red ace walking in the k-th slot out of the 52 empty ones is $frac152$, and also after that the chance of the 2nd going in any kind of slot **after** that, the end of the 51 remaining, is $frac52-k51$

Or, the first red ace might go in any kind of of the "after" positions, probability $frac52-k52$, and the 2nd red ace can go in the k-th slot the end of the 51 remaining, with probability $frac151$.

Multiplying the end and including the two results provides the very same answer as above.

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