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 $k=1$.
You are watching: How many red aces are in a deck of cards
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.
Thanks for contributing solution to sdrta.net Stack Exchange!Please be certain to answer the question. Carry out details and also share your research!
But avoid …Asking because that help, clarification, or responding to other answers.Making statements based on opinion; earlier them increase with references or personal experience.
Use sdrta.netJax to layout equations. sdrta.netJax reference.
See more: How Do I Love My Family In Italian ? How To Say I Love You In Italian
To discover more, check out our tips on writing an excellent answers.
post Your prize Discard
Not the answer you're feather for? Browse other questions tagged probability or questioning your very own question.
Is the probability of picking an Ace together the 2nd card the a deck the exact same if the deck is shuffled after ~ the first card?
illustration 2 cards native a shuffled deck, what is the probability that the an initial is a spade and the second is an ace?
site design / logo © 2021 stack Exchange Inc; user contributions license is granted under cc by-sa. Rev2021.11.5.40661