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Difficult probability problems

  1. Nov 1, 2008 #1
    Here are two probability problems which are more difficult than they look.

    1) A coach is training 15 girls. He wants to form 5 lines of 3 forwards each (left-wing, center,and right-wing). Assume that the order of assigning these positions matters. What is the probability that both Ann and May are in the same line?


    2) There is a standard deck of 52 playing cards. EACH person is dealt 13 cards. What is the probability that one of the 4 people gets ALL 4 aces?
     
  2. jcsd
  3. Nov 14, 2008 #2
    #2, I'm going with 1/256...
     
  4. Nov 14, 2008 #3

    Office_Shredder

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    For number 1, the order of assigning the positions doesn't matter, since it doesn't matter which positions Ann and May play as long as they're on the same line. Then Ann will be on a line, and May is assigned to any other slot, so has a 2/(4*3+2) = 1/7 chance of being on the same line as Ann
     
  5. Nov 14, 2008 #4

    gel

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    #2 I'll go with 44/4165
     
  6. Oct 17, 2010 #5
    hmm...

    1) [1/15 x 1/14] +
    [14/15 x 13/14 x 1/12 x 1/11] +
    [14/15 x 13/14 x 11/12 x 10/11 x 1/9 x 1/8] +
    [14/15 x 13/14 x 11/12 x 10/11 x 8/9 x 7/8 x 1/6 x 1/5] +
    [14/15 x 13/14 x 11/12 x 10/11 x 8/9 x 7/8 x 5/6 x 4/5 x 1/3 x 1/2]
    = 10.25%

    OMG I am so slow!!
    (lesson: never fall in love!)

    will do 2nd q a little later...
     
  7. Oct 17, 2010 #6
    2) [4/52 x 3/51 x 2/50 x 1/49] x 4 = 1.48 x 10^-3 %
     
  8. Oct 17, 2010 #7
    oh no wait!
     
  9. Oct 17, 2010 #8
    the cards get distributed to each person in turn: so each guy gets his turn 13 times and the aces can get dealt at any one of the instances! damn!

    let's work this step by step:
    ok so for the guy with whom you start the dealing:
    a) if they happen consecutively (if only life were so simple!): 4/52 x 3/48 x 2/44 x 1/40
    b) if the 2nd opportunity comes at the 3rd time: 4/52 x 3/44 x 2/40 x 1/36
    ..............
    note: the numerator is always 4x3x2x1

    the denominator varies: you have 52/4 = 13 choices e.g. 52, 48, 44, 40, etc which can be arranged in any way i.e. 13C4 = 715 combinations (!)
    now the question is how do we find out the product for each of these combinations?

    hmm..some algebra...
    so it's always 52-4x where x varies from 1-13 (i.e. y=52-4x is a triangle in the +ve quadrant)

    ok, so i think we need a computer/matlab program to solve this for us

    the logic is now pretty straightforword:
    1) numerator for answer= 4x3x2x1 x 4
    2) denominator is the sum of the inverse product of the combinations at each dealing

    wow! i would write the code if it weren't for the time constraint right now and also, now i think you can take over?
     
  10. Oct 17, 2010 #9
    but if i may say, the answer to 2) is going to be tiny...which is why i so strongly believe in fate lol
     
  11. Oct 18, 2010 #10
    2)

    [tex]\frac{4 \binom{48}{9}}{\binom{52}{13}} \approx 0.010564[/tex]
     
    Last edited: Oct 18, 2010
  12. Oct 23, 2010 #11
    awkward is correct
     
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