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 ?
Best Answer
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