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Shadowys
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?
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.
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 \[\left(\begin{matrix}4 \\ 4\end{matrix}\right)=1\]
The first answer still has something missing out of it...something about the four.
I'm lost :) I'm not really that good in Statistics... Please enlighten me :D
lol I didn't use statistics. It'll complicated the solution considerably. Though you are invited to try permutations and combinatorics.
So what was lacking in my answer?
um, quite a lot. specifically the point that the cards of one person can be shuffled around without disturbing his sample space.
Well, I'm completely lost :) I can do no further, I'm sorry :/
Is it (4* 49C13 * 36C13 * 23C13)/(52C13 * 39C13 * 26C13)
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...
Sorry, saurav, no...but you did come close...though I'm curious where did you get 49 from? 52-3=49cards?
Oh Sorry I guess I made a mistake there
(4* 48C13 * 35C13 * 22C13)/(52C13 * 39C13 * 26C13)
ah, You've come very very close to it. Just has an extra term, though.
Sorry, change the question to a specific person. Saurva, your answer was right. It was just a mistranslation on my part! lol
Is it new question?
LOL yup! I translated this out of chinese...so please pardon my mistakes...a specific person gets all four As. What are the chances?
(48C13 * 35C13 * 22C13)/(4*52C13 * 39C13 * 26C13)
Hmm...why the need to divide by four?.
Oh sorry....... I forgot to put 4 in the numerator.
(4*48C13 * 35C13 * 22C13)/(4*52C13 * 39C13 * 26C13)
yup. This is the right answer! nice! though would an analysis of a tree diagram might simplify the whole thing nicely?
you can also use the multinomial coefficient \[\frac{{48\choose 9,13,13,13}}{{52\choose 13,13,13,13}}\] \[=\frac{\frac{48!}{9!13!13!13!}}{\frac{52!}{13!13!13!13!}}\] \[=\frac{48!13!}{52!9!}=\frac{11}{4165}\]
Lol this was what I derived from my tree! wow. So, there was a name for that.
yep it does... http://en.wikipedia.org/wiki/Multinomial_theorem#Multinomial_coefficients :)
Wow! Lol So all this time I was deriving something that has been derived. Thanks!