I have this simple probability problem that I am not sure I solved correctly. I am not interested in formulas, but rather the thought process of how to solve it.
Suppose we roll six 6-sided dice that are equal. I want to find the probability that at least two dice have the same face.
Let's consider the following roll notation:
$$
[x] [y] [a][b][c][d]
$$
I first started to calculate the probability of the first two dice being the same in an individual 2 die roll, which is $$\frac{1}{6^2}$$
Now considering the next 4 dice as static, the number of permutations of x and y in the 6 die roll would be: $$\frac{6!}{(6-2)!} = 6*5 = 30$$
And with the total number of possible outputs for a->d being 6^4 we would have the probability of rolling the same two dice in a 6 dice roll: $$\frac{\frac{1}{6^2} * 30 * 6^4}{6^6} = \frac{6^2 * 30}{6^6} = \frac{30}{6^4} < \frac{1}{6^2}$$ which doesn't make sense to me, since rolling 6 dice gives you a greater probability of having doubles as opposed to when rolling only 2.
Can anyone pinpoint a correct method of finding this probability?
Best Answer
Hint: The probability that at least two dices have the same face is equal to
$1-P(\texttt{None of the dices have the same face})$