A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 4 years ago
Try this fun question: A standard 52 poker deck is distributed among 4 persons. What are the chances of one of them getting all four 'A's?
anonymous
 4 years ago
Try this fun question: A standard 52 poker deck is distributed among 4 persons. What are the chances of one of them getting all four 'A's?

This Question is Closed

terenzreignz
 4 years ago
Best ResponseYou've already chosen the best response.0If a specific person is taken among the four, the chances of him/her getting all 4 aces, this is given by this expression: \[\frac{\left(\begin{matrix}48 \\ 9\end{matrix}\right)}{\left(\begin{matrix}52 \\13\end{matrix}\right)}\] Multiply this by 4, and you have the probability that one of them gets all 4 aces.

terenzreignz
 4 years ago
Best ResponseYou've already chosen the best response.0Hypergeometric distribution, the probability of k successes out of n draws from a finite population of size N containing m successes without replacement. (Wikipedia) Its probability density function is given by \[\frac{\left(\begin{matrix}m \\ k\end{matrix}\right)\left(\begin{matrix}Nm \\ nk\end{matrix}\right)}{\left(\begin{matrix}N \\ n\end{matrix}\right)}\] There are 52 cards in total, so N = 52 Each player is dealt 13 cards, so n = 13 If a "success" is to be defined as drawing an ace, then there are 4 such successes, therefore m = 4 The desirable outcome is for the player to draw 4 aces (successes), then k = 4 But \[\left(\begin{matrix}4 \\ 4\end{matrix}\right)=1\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0The first answer still has something missing out of it...something about the four.

terenzreignz
 4 years ago
Best ResponseYou've already chosen the best response.0I'm lost :) I'm not really that good in Statistics... Please enlighten me :D

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0lol I didn't use statistics. It'll complicated the solution considerably. Though you are invited to try permutations and combinatorics.

terenzreignz
 4 years ago
Best ResponseYou've already chosen the best response.0So what was lacking in my answer?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0um, quite a lot. specifically the point that the cards of one person can be shuffled around without disturbing his sample space.

terenzreignz
 4 years ago
Best ResponseYou've already chosen the best response.0Well, I'm completely lost :) I can do no further, I'm sorry :/

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Is it (4* 49C13 * 36C13 * 23C13)/(52C13 * 39C13 * 26C13)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Don't worry about it. My teacher spent an hour just trying to crack it. Almost went to statistic,too, before returning to standard permutations...

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Sorry, saurav, no...but you did come close...though I'm curious where did you get 49 from? 523=49cards?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Oh Sorry I guess I made a mistake there

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0(4* 48C13 * 35C13 * 22C13)/(52C13 * 39C13 * 26C13)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ah, You've come very very close to it. Just has an extra term, though.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Sorry, change the question to a specific person. Saurva, your answer was right. It was just a mistranslation on my part! lol

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0LOL yup! I translated this out of chinese...so please pardon my mistakes...a specific person gets all four As. What are the chances?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0(48C13 * 35C13 * 22C13)/(4*52C13 * 39C13 * 26C13)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Hmm...why the need to divide by four?.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Oh sorry....... I forgot to put 4 in the numerator.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0(4*48C13 * 35C13 * 22C13)/(4*52C13 * 39C13 * 26C13)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yup. This is the right answer! nice! though would an analysis of a tree diagram might simplify the whole thing nicely?

Zarkon
 4 years ago
Best ResponseYou've already chosen the best response.0you can also use the multinomial coefficient \[\frac{{48\choose 9,13,13,13}}{{52\choose 13,13,13,13}}\] \[=\frac{\frac{48!}{9!13!13!13!}}{\frac{52!}{13!13!13!13!}}\] \[=\frac{48!13!}{52!9!}=\frac{11}{4165}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Lol this was what I derived from my tree! wow. So, there was a name for that.

Zarkon
 4 years ago
Best ResponseYou've already chosen the best response.0yep it does... http://en.wikipedia.org/wiki/Multinomial_theorem#Multinomial_coefficients :)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Wow! Lol So all this time I was deriving something that has been derived. Thanks!
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.