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anonymous

  • one year ago

Five cards are dealt from a standard 52-card deck. What is the probability that we draw a 1 ace, 1 two, 1 three, 1 four, and 1 five (this is one way to get a “straight”)?

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  1. anonymous
    • one year ago
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    four choices for each up top right? and \[\binom{52}{5}\] for the denominator

  2. anonymous
    • one year ago
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    Yes, I've got the denom!

  3. anonymous
    • one year ago
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    The probability of each on their own is 4/52, but I know that changes as you draw each down to 4/51, 4/50, etc...but how do I express that?

  4. anonymous
    • one year ago
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    I still don't know what goes in the denominator. I can reliably determine the denominator, but I have no clue what's on top!

  5. anonymous
    • one year ago
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    the denomator is the number of ways you can choose 5 cards from a set of 52

  6. anonymous
    • one year ago
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    OH sorry, I meant numerator!!

  7. anonymous
    • one year ago
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    in other words "52 choose 5" written as \[\binom{52}{5}\] or sometimes \[_{52}C_5\]

  8. anonymous
    • one year ago
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    I meant I don't know what goes in the numerator! Silly me! Typo.

  9. anonymous
    • one year ago
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    computed via \[\frac{52\times 51\times 50\times 49\times 48}{5!}\]

  10. anonymous
    • one year ago
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    I was thinking, yes yes I know the denominator. Why is he/she telling me? I realize my mistake. Numerator is the concern.

  11. anonymous
    • one year ago
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    ooh numerator

  12. anonymous
    • one year ago
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    Yeah sorry! Thanks for your patience!

  13. anonymous
    • one year ago
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    4 choices for each, so \(4\times 4\times 4\times 4\times 4\)

  14. anonymous
    • one year ago
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    I don't really get where that's coming from.. hmm. Why can it just be 4s?

  15. anonymous
    • one year ago
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    4 choices of aces right?

  16. anonymous
    • one year ago
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    Yes, but there's more going on it seems. Why is it not including the reducing total or something like 4/52, 4/51, 4/50, /4/49

  17. anonymous
    • one year ago
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    that is taken care of in the denominator

  18. anonymous
    • one year ago
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    It is? Ahh alright...Your answer is right, but I'll have to ask my teacher for clarification about why it's taken care of. It's late and I won't mess with that. Thank you!

  19. anonymous
    • one year ago
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    think of it this way it is all the "counting principle" if the number of ways to do one thing is n and another m, then there are \(n\times m\) ways to go it together

  20. anonymous
    • one year ago
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    4 ways to get an ace, 4 ways to get a 2 etc so \(4^5\) ways all together

  21. anonymous
    • one year ago
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    Okay, I think I understand. The numerator is always the wild card for me

  22. anonymous
    • one year ago
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    then the number of ways to pick the 5 cards is 52 choose 5, which is again really the counting principle in action

  23. anonymous
    • one year ago
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    if it was for example 3 spades and two clubs, the answer would be \[\frac{\binom{13}{3}\binom{13}{2}}{\binom{52}{5}}\]

  24. anonymous
    • one year ago
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    these kind of problems, the question is always "how many ways"

  25. anonymous
    • one year ago
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    I don't get what you've said. I'm sorry.

  26. anonymous
    • one year ago
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    I get how they apply in their own situations, but I'm SO bad at generalizing these sorts of rationales

  27. anonymous
    • one year ago
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    The next question is asking me how ANY straight could be obtained and I don't even know where to start.

  28. anonymous
    • one year ago
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    oh that is ok we are almost there

  29. anonymous
    • one year ago
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    Like what goes on top? No idea.

  30. anonymous
    • one year ago
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    it's no longer just 4s, and it's not comb(13,2) either. So it's a complete effing mystery to my mind.

  31. anonymous
    • one year ago
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    we got one straight right? one that begins with an ace and ends with a 5

  32. anonymous
    • one year ago
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    yep!

  33. anonymous
    • one year ago
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    ok so now we know the probability of that particular straight (that is why that question came first)

  34. anonymous
    • one year ago
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    now all we have to do is figure out how many different possible straights there are

  35. anonymous
    • one year ago
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    no idea!

  36. anonymous
    • one year ago
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    so we count them!

  37. anonymous
    • one year ago
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    not a poker player i guess huh?

  38. anonymous
    • one year ago
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    Not in a million years.

  39. anonymous
    • one year ago
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    ace to 5 2 to 6 3 to 7 4 to 8 5 to 9 6 to 10 7 to jack 8 to queen 9 to king 10 to ace

  40. anonymous
    • one year ago
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    i count ten

  41. anonymous
    • one year ago
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    Okay.

  42. anonymous
    • one year ago
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    btw you can't have for example jack to 2

  43. anonymous
    • one year ago
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    so we are done

  44. anonymous
    • one year ago
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    multiply the above answer by 10

  45. anonymous
    • one year ago
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    oh. So .000394 *10

  46. anonymous
    • one year ago
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    i.e. \[\frac{10\times 4^5}{\binom{52}{5}}\]

  47. anonymous
    • one year ago
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    http://www.wolframalpha.com/input/?i=%2810*4^5%29%2F%2852+choose+5%29

  48. anonymous
    • one year ago
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    oh if you had the previous answer then yeah move the decimal one to the left

  49. anonymous
    • one year ago
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    right

  50. anonymous
    • one year ago
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    whatever

  51. anonymous
    • one year ago
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    That's not quite right.

  52. anonymous
    • one year ago
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    The answer is .00355

  53. anonymous
    • one year ago
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    i get \[0.00394\]

  54. anonymous
    • one year ago
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    So did I, from what you told me. The textbook says it's .00355

  55. anonymous
    • one year ago
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    they are wrong

  56. anonymous
    • one year ago
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    umm..

  57. anonymous
    • one year ago
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    ok?

  58. anonymous
    • one year ago
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    unless...

  59. anonymous
    • one year ago
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    you are supposed to subtract the number of flushes which i doubt

  60. anonymous
    • one year ago
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    I have no idea. This is all bullcrap to me. I don't like math.

  61. anonymous
    • one year ago
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    Let STATA do it, that's what I say.

  62. anonymous
    • one year ago
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    But I'll put down the answer we worked out and ask later.

  63. anonymous
    • one year ago
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    Thanks for your help.

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