Dice 10000 probability of not scoring There is a dice game I played recently, called "Dice 10,000", that is very similar to Farkle but a little simpler. In the version I played (house rules) a person scores on any given turn if they get at least three of a kind, or at least one of the die is a 1 or at least 1 of the die is a 5. On a player's first roll, they toss 6 die: I attempted to calculate how likely it is that a person does not score any points on their first roll, but I feel like the final value is too high. Is there anything incorrect following logic?
Let A = The probability of at least one die being a 1
   B = The probability of at least one die being a 5
   C = The probability of getting at least three of a kind
Then, the probability of not scoring is ${P(\neg A \cap \neg B \cap \neg C)}$
${P(\neg A \cap \neg B \cap \neg C) = P((\neg A \cap \neg B)\cap \neg C)}$               
${= P(\neg A \cap \neg B) \times P(\neg C \: |\: \neg A \cap \neg B)}$

The probability of not getting a 1 and not getting a 5, then is:
${P(\neg A \cap \neg B) = \frac{4^{6}}{6^{6}}}$
And the probability of not getting at least 3 of one kind given you do not have any 1's or 5's is:
${P(\neg C \: |\: \neg A \cap \neg B) = 1 - \frac{\sum\limits_{i=3}^6\binom{4}{1}\binom{6}{i}3^{6-i}}{4^{6}}}$
Since there are 6 die, there is a possibility that the player has 3 of a kind for two separate values (for instance three 4's and three 2's are rolled), so I am particularly concerned with my calculation for ${P(\neg C \: |\: \neg A \cap \neg B)}$.

It's easier to directly count the number of ways where you don't get three of a kind using just $2,3,4,6$. There are ${4 \choose 3}  \frac{6!}{(2!)^3} = 360$ ways to do it with three pairs and ${4 \choose 2}  \frac{6!}{(2!)^2} = 1080$ ways to do it with two pairs and two separate rolls. So the final probability becomes $\frac{360+1080}{6^6} \approx 0.0309$.
