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Homework Help: Can't tell what I am overcounting

  1. Oct 15, 2011 #1
    1. The problem statement, all variables and given/known data
    Compute the probability of achieving 4 distinct face values when you roll 4 dice at most twice.

    3. The attempt at a solution
    Let Ei denote the event of rolling i distinct face values. Then:
    [itex]P(E_4) = \frac{6 \cdot 5 \cdot 4 \cdot 3}{6^4}[/itex]

    A roll of 3 distinct face values would look like (x,x,y,z). So there would be [itex]\binom{4}{2}[/itex] choices for the repeat value to appear, and:
    [itex]P(E_3) = \frac{\binom{4}{2} 6 \cdot 5 \cdot 4}{6^4}[/itex]

    A roll of 2 distinct face values would look like (x,x,y,y), or (x,x,x,y). So:
    [itex]P(E_3) = \frac{\binom{4}{2} 6 \cdot 5 + \binom{4}{3} 6 5}{6^4}[/itex]

    There are six ways to roll (x,x,x,x). So:
    [itex]P(E_3) = \frac{6}{6^4}[/itex]

    I need to check that [tex]\sum_{i=1}^4 P(E_i) = 1[/tex]. But:
    [itex]\sum_{i=1}^4 P(E_i) = 77/72[/itex]

    Would some be so kind to tell me how I've counted wrong?
     
  2. jcsd
  3. Oct 15, 2011 #2
    So I've identified the problem. I counted (x,x,y,y) twice.
     
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