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Shadowys
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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?
 one year ago
 one year ago
Shadowys Group Title
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?
 one year ago
 one year ago

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terenzreignz Group TitleBest ResponseYou've already chosen the best response.0
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.
 one year ago

terenzreignz Group TitleBest ResponseYou've already chosen the best response.0
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 \[\frac{\left(\begin{matrix}m \\ k\end{matrix}\right)\left(\begin{matrix}Nm \\ nk\end{matrix}\right)}{\left(\begin{matrix}N \\ n\end{matrix}\right)}\] 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 \[\left(\begin{matrix}4 \\ 4\end{matrix}\right)=1\]
 one year ago

Shadowys Group TitleBest ResponseYou've already chosen the best response.0
The first answer still has something missing out of it...something about the four.
 one year ago

terenzreignz Group TitleBest ResponseYou've already chosen the best response.0
I'm lost :) I'm not really that good in Statistics... Please enlighten me :D
 one year ago

Shadowys Group TitleBest ResponseYou've already chosen the best response.0
lol I didn't use statistics. It'll complicated the solution considerably. Though you are invited to try permutations and combinatorics.
 one year ago

terenzreignz Group TitleBest ResponseYou've already chosen the best response.0
So what was lacking in my answer?
 one year ago

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

terenzreignz Group TitleBest ResponseYou've already chosen the best response.0
Well, I'm completely lost :) I can do no further, I'm sorry :/
 one year ago

sauravshakya Group TitleBest ResponseYou've already chosen the best response.3
Is it (4* 49C13 * 36C13 * 23C13)/(52C13 * 39C13 * 26C13)
 one year ago

Shadowys Group TitleBest ResponseYou've already chosen the best response.0
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...
 one year ago

Shadowys Group TitleBest ResponseYou've already chosen the best response.0
Sorry, saurav, no...but you did come close...though I'm curious where did you get 49 from? 523=49cards?
 one year ago

sauravshakya Group TitleBest ResponseYou've already chosen the best response.3
Oh Sorry I guess I made a mistake there
 one year ago

sauravshakya Group TitleBest ResponseYou've already chosen the best response.3
(4* 48C13 * 35C13 * 22C13)/(52C13 * 39C13 * 26C13)
 one year ago

sauravshakya Group TitleBest ResponseYou've already chosen the best response.3
524=48
 one year ago

Shadowys Group TitleBest ResponseYou've already chosen the best response.0
ah, You've come very very close to it. Just has an extra term, though.
 one year ago

Shadowys Group TitleBest ResponseYou've already chosen the best response.0
Sorry, change the question to a specific person. Saurva, your answer was right. It was just a mistranslation on my part! lol
 one year ago

sauravshakya Group TitleBest ResponseYou've already chosen the best response.3
Is it new question?
 one year ago

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

sauravshakya Group TitleBest ResponseYou've already chosen the best response.3
(48C13 * 35C13 * 22C13)/(4*52C13 * 39C13 * 26C13)
 one year ago

Shadowys Group TitleBest ResponseYou've already chosen the best response.0
Hmm...why the need to divide by four?.
 one year ago

sauravshakya Group TitleBest ResponseYou've already chosen the best response.3
Oh sorry....... I forgot to put 4 in the numerator.
 one year ago

sauravshakya Group TitleBest ResponseYou've already chosen the best response.3
(4*48C13 * 35C13 * 22C13)/(4*52C13 * 39C13 * 26C13)
 one year ago

sauravshakya Group TitleBest ResponseYou've already chosen the best response.3
u got it?
 one year ago

Shadowys Group TitleBest ResponseYou've already chosen the best response.0
yup. This is the right answer! nice! though would an analysis of a tree diagram might simplify the whole thing nicely?
 one year ago

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

Shadowys Group TitleBest ResponseYou've already chosen the best response.0
Lol this was what I derived from my tree! wow. So, there was a name for that.
 one year ago

Zarkon Group TitleBest ResponseYou've already chosen the best response.0
yep it does... http://en.wikipedia.org/wiki/Multinomial_theorem#Multinomial_coefficients :)
 one year ago

Shadowys Group TitleBest ResponseYou've already chosen the best response.0
Wow! Lol So all this time I was deriving something that has been derived. Thanks!
 one year ago
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