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 May 26, 2021

#1**+2 **

Each time you draw a card you have a 4/52 chance of an ace (or 1/13 chance)

(1/13)^{4 }=___________

ElectricPavlov May 26, 2021

edited by
Guest
May 26, 2021

edited by Guest May 26, 2021

edited by Guest May 26, 2021

#2**0 **

*.....what is the probability that I will end up with all four Aces?*

By this, do you mean

...probability that I will have pulled an ace each time?

OR

...probability that I will have pulled the ace of each suit?

Guest May 26, 2021

#3**0 **

Good point.....questions are often unclear and subject to interpretation by the answerer.

If you want to get EACH of the aces it would be a little more difficult :

(1/52)^{4 }

Guest May 26, 2021

#4**0 **

I think it would be 4/52 for the first ace, which could be any of the four aces

and then 3/52 for an undrawn ace, one of the remaining three

and then 2/52 for one of the two remaining undrawn aces

and then 1/52 for the last ace that hadn't been previously drawn

for the total probability multiply them together

4 3 2 1 24 **3**

–– x –– x –– x –– = –––––––– = **––––––––**

52 52 52 52 7,311,616 **913,952**

Guest May 27, 2021

edited by
Guest
May 27, 2021