How many times must one roll a pair of dice so that the probability is at least 3/4 that a 7 comes up? We were given this question for homework that the professor couldn't explain how to solve (even in class he had trouble working it out). I'm only aware that we should be using the law of large numbers but I'm not sure how to apply it as the book for the course provides no examples. The answer in class was 10 and the book gave us an 8. Any help would be appreciated.
 A: I don't think the law of large numbers will be much help here, as evidenced by the fact that the answer (or, I guess I should say either of the answers) are not particularly large numbers.
Let $p$ denote the probability that a throw of two dice results in a sum of $7$. (This is a small sub-computation you can do.) Once you have this, consider repeatedly throwing pairs of dice. It will be easier to consider the probability of not seeing a $7$ in this case.


*

*The probability of not seeing a $7$ on your first throw of a pair is $1-p$.

*The probability of not seeing a $7$ by your second throw is $(1-p) \cdot (1 -
 p) = (1-p)^2$.


And so forth. Your task is to determine how many times you must throw the dice so that the probability of not seeing a $7$ dips below $1/4$, because this will imply that the opposite event (i.e. "seeing a $7$") will have a probability of at least $3/4$.
A: The chance of rolling two dice and the result adding to 7 is 1/6. This is given by the 6 ways a 7 can be rolled (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) divided by the 36 possible rolls (6*6).
Therefore, the chance of rolling NO sevens after n rolls is$\ (5/6)^n$ and hence the chance of at least one seven in $\ 1-(5/6)^n$. Solving for $\ 1-(5/6)^n>3/4$:
$\ (5/6)^n<1/4$
$\ n*ln(5/6)<ln(1/4)$
$\ n>ln(1/4)/ln(5/6)$
$\ n>7.603...$
$\ n=8$
