anonymous
  • anonymous
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?
Probability
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SOLVED
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katieb
  • katieb
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terenzreignz
  • 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.
terenzreignz
  • 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 \[\left(\begin{matrix}4 \\ 4\end{matrix}\right)=1\]
anonymous
  • anonymous
The first answer still has something missing out of it...something about the four.

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terenzreignz
  • terenzreignz
I'm lost :) I'm not really that good in Statistics... Please enlighten me :D
anonymous
  • anonymous
lol I didn't use statistics. It'll complicated the solution considerably. Though you are invited to try permutations and combinatorics.
terenzreignz
  • terenzreignz
So what was lacking in my answer?
anonymous
  • anonymous
um, quite a lot. specifically the point that the cards of one person can be shuffled around without disturbing his sample space.
terenzreignz
  • terenzreignz
Well, I'm completely lost :) I can do no further, I'm sorry :/
anonymous
  • anonymous
Is it (4* 49C13 * 36C13 * 23C13)/(52C13 * 39C13 * 26C13)
anonymous
  • 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...
anonymous
  • anonymous
Sorry, saurav, no...but you did come close...though I'm curious where did you get 49 from? 52-3=49cards?
anonymous
  • anonymous
Oh Sorry I guess I made a mistake there
anonymous
  • anonymous
(4* 48C13 * 35C13 * 22C13)/(52C13 * 39C13 * 26C13)
anonymous
  • anonymous
52-4=48
anonymous
  • anonymous
ah, You've come very very close to it. Just has an extra term, though.
anonymous
  • anonymous
Sorry, change the question to a specific person. Saurva, your answer was right. It was just a mistranslation on my part! lol
anonymous
  • anonymous
Is it new question?
anonymous
  • 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?
anonymous
  • anonymous
(48C13 * 35C13 * 22C13)/(4*52C13 * 39C13 * 26C13)
anonymous
  • anonymous
Hmm...why the need to divide by four?.
anonymous
  • anonymous
Oh sorry....... I forgot to put 4 in the numerator.
anonymous
  • anonymous
(4*48C13 * 35C13 * 22C13)/(4*52C13 * 39C13 * 26C13)
anonymous
  • anonymous
u got it?
anonymous
  • anonymous
yup. This is the right answer! nice! though would an analysis of a tree diagram might simplify the whole thing nicely?
Zarkon
  • Zarkon
you can also use the multinomial coefficient \[\frac{{48\choose 9,13,13,13}}{{52\choose 13,13,13,13}}\] \[=\frac{\frac{48!}{9!13!13!13!}}{\frac{52!}{13!13!13!13!}}\] \[=\frac{48!13!}{52!9!}=\frac{11}{4165}\]
anonymous
  • anonymous
Lol this was what I derived from my tree! wow. So, there was a name for that.
Zarkon
  • Zarkon
yep it does... http://en.wikipedia.org/wiki/Multinomial_theorem#Multinomial_coefficients :)
anonymous
  • anonymous
Wow! Lol So all this time I was deriving something that has been derived. Thanks!

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