probability dice consideration of previous results

Question

I am looking for a formula to calculate this problem: If you roll a dice n times and you get y times the number one, how is the probability in try 8 in consideration of the prevois results?

Here an example:

    |   1   2   3   4   5   6 (dice)
------------------------------------
1   |   -   -   -   -   x   -
2   |   -   -   x   -   -   -
3   |   -   x   -   -   -   -
4   |   x   -   -   -   -   -
5   |   -   -   -   -   -   x
6   |   x   -   -   -   -   -
7   |   -   -   -   x   -   -
8   |   ?

P(AuB) = P(A) + P(B) - P(AnB) is not correct, or?

Answer 1

Dice rolls are statistically independent. "Previous outcomes" are irrelevant when calculating the expected outcomes on roll 8. Thus the probability of each number appearing on roll 8 is the same, and equal to 1/6.