anonymous 3 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?

1. terenzreignz

If 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.

2. terenzreignz

Hypergeometric 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 $\frac{\left(\begin{matrix}m \\ k\end{matrix}\right)\left(\begin{matrix}N-m \\ n-k\end{matrix}\right)}{\left(\begin{matrix}N \\ n\end{matrix}\right)}$ 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 $\left(\begin{matrix}4 \\ 4\end{matrix}\right)=1$

3. anonymous

The first answer still has something missing out of it...something about the four.

4. terenzreignz

I'm lost :) I'm not really that good in Statistics... Please enlighten me :D

5. anonymous

lol I didn't use statistics. It'll complicated the solution considerably. Though you are invited to try permutations and combinatorics.

6. terenzreignz

So what was lacking in my answer?

7. anonymous

um, quite a lot. specifically the point that the cards of one person can be shuffled around without disturbing his sample space.

8. terenzreignz

Well, I'm completely lost :) I can do no further, I'm sorry :/

9. anonymous

Is it (4* 49C13 * 36C13 * 23C13)/(52C13 * 39C13 * 26C13)

10. anonymous

Don't worry about it. My teacher spent an hour just trying to crack it. Almost went to statistic,too, before returning to standard permutations...

11. anonymous

Sorry, saurav, no...but you did come close...though I'm curious where did you get 49 from? 52-3=49cards?

12. anonymous

Oh Sorry I guess I made a mistake there

13. anonymous

(4* 48C13 * 35C13 * 22C13)/(52C13 * 39C13 * 26C13)

14. anonymous

52-4=48

15. anonymous

ah, You've come very very close to it. Just has an extra term, though.

16. anonymous

Sorry, change the question to a specific person. Saurva, your answer was right. It was just a mistranslation on my part! lol

17. anonymous

Is it new question?

18. anonymous

LOL yup! I translated this out of chinese...so please pardon my mistakes...a specific person gets all four As. What are the chances?

19. anonymous

(48C13 * 35C13 * 22C13)/(4*52C13 * 39C13 * 26C13)

20. anonymous

Hmm...why the need to divide by four?.

21. anonymous

Oh sorry....... I forgot to put 4 in the numerator.

22. anonymous

(4*48C13 * 35C13 * 22C13)/(4*52C13 * 39C13 * 26C13)

23. anonymous

u got it?

24. anonymous

yup. This is the right answer! nice! though would an analysis of a tree diagram might simplify the whole thing nicely?

25. Zarkon

you can also use the multinomial coefficient $\frac{{48\choose 9,13,13,13}}{{52\choose 13,13,13,13}}$ $=\frac{\frac{48!}{9!13!13!13!}}{\frac{52!}{13!13!13!13!}}$ $=\frac{48!13!}{52!9!}=\frac{11}{4165}$

26. anonymous

Lol this was what I derived from my tree! wow. So, there was a name for that.

27. Zarkon
28. anonymous

Wow! Lol So all this time I was deriving something that has been derived. Thanks!