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Probability Expected Values with Cards

  1. Sep 28, 2009 #1
    1. The problem statement, all variables and given/known data
    Cards from a standard deck of 52 playing cards are laid face up one at a time. If the first card is an ace, the second card is an 2, the third card is a three, ..., the 13th card is a king, the 14th is an ace, etc. we call it a "match." The 13n+1th card does not need to be any particular ace, it just needs to be an ace. Find the expected number of matches that occur.


    2. Relevant equations



    3. The attempt at a solution
    I let the random variable X represent the total number of matches.
    So Xi=1 if the ith card is a match, and 0 if it is not a match.
    Then I summed over the all the cards from 1 to 52.
    The thing that I was having trouble calculating was P(Xi=1). It seems like on any given trial there is a 4/52 chance of getting a match, but this seems oversimplified.
     
  2. jcsd
  3. Sep 28, 2009 #2

    lanedance

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    Homework Helper

    i think you need to think about how you define your random variable..

    If Xi is the probabilty of a match at the ith position, then for i=1, the probability of first drawing an ace will be 4/52

    if X is the total number of matches, then the probabilty of X depends on how you count matches

    do you stop counting matches after your first mismatch...? in that case the probability will be 4/52 multiplied by the probability of not getting a two on the 2nd card (51-4)/51
    if 1st card ace...

    sounds more likely you want to find the total number of matches, then you must count all the occurences of exactly one match anywhere in the draw, ie:
    1-draw an ace, then no other matches...
    1-no ace, 2-draw a 2, then no other matches...
    ... etc.

    unless you can find a smart way to do this, the brute force method could get tricky as the probabilty of drawing cards will be dependent on what has been dealt previously... anyway just stuff to think about
     
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