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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?

  • 2 years ago
  • 2 years ago

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  1. terenzreignz Group Title
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    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 years ago
  2. terenzreignz Group Title
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    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 \[\left(\begin{matrix}4 \\ 4\end{matrix}\right)=1\]

    • 2 years ago
  3. Shadowys Group Title
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    The first answer still has something missing out of it...something about the four.

    • 2 years ago
  4. terenzreignz Group Title
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    I'm lost :) I'm not really that good in Statistics... Please enlighten me :D

    • 2 years ago
  5. Shadowys Group Title
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    lol I didn't use statistics. It'll complicated the solution considerably. Though you are invited to try permutations and combinatorics.

    • 2 years ago
  6. terenzreignz Group Title
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    So what was lacking in my answer?

    • 2 years ago
  7. Shadowys Group Title
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    um, quite a lot. specifically the point that the cards of one person can be shuffled around without disturbing his sample space.

    • 2 years ago
  8. terenzreignz Group Title
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    Well, I'm completely lost :) I can do no further, I'm sorry :/

    • 2 years ago
  9. sauravshakya Group Title
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    Is it (4* 49C13 * 36C13 * 23C13)/(52C13 * 39C13 * 26C13)

    • 2 years ago
  10. Shadowys Group Title
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    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...

    • 2 years ago
  11. Shadowys Group Title
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    Sorry, saurav, no...but you did come close...though I'm curious where did you get 49 from? 52-3=49cards?

    • 2 years ago
  12. sauravshakya Group Title
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    Oh Sorry I guess I made a mistake there

    • 2 years ago
  13. sauravshakya Group Title
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    (4* 48C13 * 35C13 * 22C13)/(52C13 * 39C13 * 26C13)

    • 2 years ago
  14. sauravshakya Group Title
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    52-4=48

    • 2 years ago
  15. Shadowys Group Title
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    ah, You've come very very close to it. Just has an extra term, though.

    • 2 years ago
  16. Shadowys Group Title
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    Sorry, change the question to a specific person. Saurva, your answer was right. It was just a mistranslation on my part! lol

    • 2 years ago
  17. sauravshakya Group Title
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    Is it new question?

    • 2 years ago
  18. Shadowys Group Title
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    LOL yup! I translated this out of chinese...so please pardon my mistakes...a specific person gets all four As. What are the chances?

    • 2 years ago
  19. sauravshakya Group Title
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    (48C13 * 35C13 * 22C13)/(4*52C13 * 39C13 * 26C13)

    • 2 years ago
  20. Shadowys Group Title
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    Hmm...why the need to divide by four?.

    • 2 years ago
  21. sauravshakya Group Title
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    Oh sorry....... I forgot to put 4 in the numerator.

    • 2 years ago
  22. sauravshakya Group Title
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    (4*48C13 * 35C13 * 22C13)/(4*52C13 * 39C13 * 26C13)

    • 2 years ago
  23. sauravshakya Group Title
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    u got it?

    • 2 years ago
  24. Shadowys Group Title
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    yup. This is the right answer! nice! though would an analysis of a tree diagram might simplify the whole thing nicely?

    • 2 years ago
  25. Zarkon Group Title
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    you can also use the multinomial coefficient \[\frac{{48\choose 9,13,13,13}}{{52\choose 13,13,13,13}}\] \[=\frac{\frac{48!}{9!13!13!13!}}{\frac{52!}{13!13!13!13!}}\] \[=\frac{48!13!}{52!9!}=\frac{11}{4165}\]

    • 2 years ago
  26. Shadowys Group Title
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    Lol this was what I derived from my tree! wow. So, there was a name for that.

    • 2 years ago
  27. Zarkon Group Title
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    yep it does... http://en.wikipedia.org/wiki/Multinomial_theorem#Multinomial_coefficients :)

    • 2 years ago
  28. Shadowys Group Title
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    Wow! Lol So all this time I was deriving something that has been derived. Thanks!

    • 2 years ago
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