- #1
Samuelb88
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Homework Statement
Compute the probability of achieving 4 distinct face values when you roll 4 dice at most twice.
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?