A man throws 4 dice. What is the probability that same number shows up in at least two faces? I saw a question  in internet , it seems very trivial.I tried to solve it in two different way but i obtained two different result.
I know that my second solution is correct but why is the former wrong ? What am i missing ?
Question: A man throws $4$ dice. What is the probability that same number shows up in at least two faces?
First Solution:
The probability of getting ONLY $2 $ dice showing the same face : $\frac{C(4,2).6.5.4}{6.6.6.6}=\frac{120}{216}$
The probability of getting ONLY $3 $ dice showing the same face : $\frac{C(4,3).6.5}{6.6.6.6}=\frac{20}{216}$
The probability of getting ONLY $4 $ dice showing the same face : $\frac{C(4,4).6}{6.6.6.6}=\frac{1}{216}$
$\therefore \frac{141}{216}=0.652$
Second Solution: All situations - all dice appears different faces
$1- \frac{6.5.4.3}{6.6.6.6}=0.722$
What am i missing ?
 A: Your $\frac{936}{1296}\approx 0.722$ is correct
Your first calculation is missing the case of two faces each appearing on two dice, which has probability $\frac{90}{1296}=\frac{15}{216}\approx 0.069$; this is the difference between your two answers
A: It's worth noting another approach to this solution that imagines each die being rolled individually and looks at the odds of the complimentary event that there are no matching dice.
The first rolled die can be anything, doesn't matter. Imagine its a 1 if it helps.
The second rolled die has a $\frac{1}{6}$ chance of matching the first (which we're not interested in), and a $\frac{5}{6}$ chance of differing from the first.
Given that the first two dice were different, the third rolled die has a $\frac{1}{3}$ chance of matching one of the first two, and a $\frac{2}{3}$ chance of differing from the first two.
Finally, given that we have no matches by the fourth die, there's a $\frac{1}{2}$ chance that the fourth die matches, otherwise, there are no matches. Given that this is the only way to not get a match, we can compute the probability of not having a match as $\frac{5}{6} \cdot \frac{2}{3} \cdot \frac{1}{2}$. This simplifies to $\frac{5}{18}$, and the compliment $\frac{13}{18} \simeq 0.7222$ agrees with the other approaches.
A: Your first solution is 141/216 and your second solution is 156/216. But your first solution doesn't include the case of "two pairs", which accounts for the missing 15/216.
