If I choose four cards from a standard 52-card deck,with replacement, what is the probability that I will end up with all four Aces?

Guest Sep 25, 2021

#1**0 **

*If I choose four cards from a standard 52-card deck,with replacement, what is the probability that I will end up with all four Aces?*

I wish the problem hadn't specified the "with replacement" condition.

What does "with replacement" mean? Do you draw four cards, keep any aces, replace the non-aces

back into the deck, and just keep on drawing cards until you eventually end up with all four aces?

Do you draw four cards, and if all four aren't all aces, replace all four back into the deck, and draw again?

Do you keep going until you do finally get all four aces on the same draw?

Guest Sep 25, 2021

#2**0 **

I think it is 4/52 * 3/52 * 2/52 * 1/52 = 24 / 7311616 = 0.0000032824480936

ElectricPavlov Sep 25, 2021

#3**0 **

*If I choose four cards from a standard 52-card deck,with replacement, what is the probability that I will end up with all four Aces?*

**This question is poorly written**. The primary defect is the phrase “*end up with all four Aces,*” which is a colloquialism as used here. The capital “A” in “Aces” is nonstandard, and gives ambiguous emphases to the word aces. Use of the word “all” *could *imply different suits for the four aces, rather than some combination of aces where one or more is repeated.

This is a primer statistics question so a reasonable interpretation is:

If I choose four cards from a standard 52-card deck, with replacement, what is the probability that I will select *four aces of different suits*?

**The phrase “with replacement” is an explicit standard variation in the sampling method**. In this case, it means the card (no matter what it is) is replaced after it is drawn. It’s also important to note that the deck or selection process remains randomized after the card is replaced.

**A success in this experiment is** **four aces in four different suits**. Any other combination is a failure.

\(\dfrac{4}{52}*\dfrac{3}{52}*\dfrac{2}{52}*\dfrac{1}{52} =\dfrac{3}{913952} \)

This happens about (3) times per million attempts. It’s noteworthy that EP’s solution is correct and (mostly) matches this, which also happens about (3) times per million attempts.

IF...

**A success in this experiment is** **four aces without regard to the suit. **

\(\left(\dfrac{4}{52}\right)^4 = \dfrac{1}{28561}\)

This happens about (35) times per million attempts.

GA

Guest Sep 27, 2021