# Homework Help: Can't tell what I am overcounting

1. Oct 15, 2011

### Samuelb88

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:
$P(E_4) = \frac{6 \cdot 5 \cdot 4 \cdot 3}{6^4}$

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

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

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

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

Would some be so kind to tell me how I've counted wrong?

2. Oct 15, 2011

### Samuelb88

So I've identified the problem. I counted (x,x,y,y) twice.