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Some probabilities in card game

  1. Sep 19, 2009 #1
    [tex]\sum[/tex]The question is:
    You are involove in a game where two card are dealt consectively. Supposethat the dealer pays you $3of the second card dealt is a club, regardless of the first card and that you pay him $1 if the second card in not a club and the first card is not an ace. (othewise, no money changes hands.) Use [tex]\sum[/tex][tex]f_{i}[/tex][tex]p_{i}[/tex] to compute the mean value of the money you win per game if you play it many time.

    The ans is 5.8cent

    I have spent a lot of time on that, but i still don't understand.
     
  2. jcsd
  3. Sep 19, 2009 #2

    mathman

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    List all different cases and probabilities

    Second card a club - p1=1/4

    Second card not a club and not an ace 36/52, first card not an ace 47/51
    P2=9x47/(13x51)

    Second card not a club and an ace 3/52, first card not an ace 48/51
    p3=3x16/(52x17)

    payoff= 3p1 - (p2+p3)

    I got 5.76 cents.
     
  4. Sep 19, 2009 #3
    Thanks for answering me!
    The following is wrong reasoning,but i still can caculate the answers

    second card a club- p1=1/4

    First card not a club and not an ace 36/52, second card not a club 38/51
    p2=(36X38)/(51X52)

    First card is a club and not an ace 12/52 , second card not a club 39/51
    p3=12x52/(39/51)

    mean = 3p1-(p2+p3) =5.77

    My course TA told me it was wrong reasoning, frustrate me a lot.
     
  5. Sep 20, 2009 #4

    mathman

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    Off hand I would say your TA is wrong.
     
  6. Sep 20, 2009 #5
    I get the same answer, .0576(9) taking the first card first, and the second card dependent. Perhaps the TA didn't understand the ordering you used?
     
  7. Sep 21, 2009 #6

    mathman

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    I can't read your TA's mind. Why don't you ask him (or her)?
     
  8. Sep 21, 2009 #7
    I think you've confused me with the OP, mathman. No problem.
     
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