for this reason if we have actually a consistent deck the $52$ cards the is shuffled, climate what is the probability of illustration the very first red ace on the $k$-th card? and also which k would certainly maximize the probability?

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?


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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 $k=1$.

You are watching: How many red aces are in a deck of cards


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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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